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\begin{document}

\begin{abstract}
In this paper we study conditions assuring that the Bishop-Phelps-Bollob\'{a}s property (BPBp, for short) is inherited by absolute summands of the range space or of the domain space. Concretely, given a pair $(X,Y)$ of Banach spaces having the BPBp,
\begin{itemize}
\item[(a)] if $Y_1$ is an absolute summand of $Y$, then $(X,Y_1)$ has the BPBp;
\item[(b)] if $X_1$ is an absolute summand of $X$ of type $1$ or $\infty$, then $(X_1,Y)$ has the BPBp.
\end{itemize}
Besides, analogous results for the BPBp for compact operators and for the density of norm attaining operators are also given. We also show that the Bishop-Phelps-Bollob\'{a}s property for numerical radius is inherited by absolute summands of type $1$ or $\infty$. Moreover, we provide analogous results for numerical radius attaining operators and for the BPBp for numerical radius for compact operators.
\end{abstract}

\maketitle

\section{Introduction \& Preliminaries}

Let $X$ be a Banach space. We denote by $B_X$ and $S_X$ the unit ball and the unit sphere of $X$, respectively. We consider the topological dual space of $X$ and we denote it by $X^*$. We say that $x^* \in X^*$ {\it attains its norm} if there is $x_0 \in S_X$ such that $|x^*(x_0)| = \|x^*\| = \sup_{x \in S_X} |x^*(x)|$. The famous Bishop-Phelps theorem \cite{BP} says that given $\eps > 0$ and $x^* \in X^*$, there exists $x_0^* \in X^*$ such that $|x_0^*(x_0)| = \| x_0^* \|$ for some $x_0 \in S_X$ and $\|x_0^* - x^*\| < \eps$. It is natural to ask if it is true also for bounded linear operators. Given two Banach spaces $X$ and $Y$, we denote by $\mathcal{L}(X, Y)$ the set of all continuous linear operators. When $Y = X$, we denote it simply by $\mathcal{L}(X)$. We say that $T \in \mathcal{L}(X, Y)$ {\it attains its norm} when there exists $x_0 \in S_X$ such that $\|Tx_0\| = \|T\| = \sup_{x \in S_X} \|T(x)\|$. We denote by $\NA(X, Y)$ the set of all norm attaining operators from $X$ to $Y$. Then, the Bishop-Phelps theorem states that $\NA(X,\K)$ is dense in $X^*$ for every Banach space $X$ (where $\K$ denotes the base field ($=\R$ or $\C$)). Trying to extend the Bishop-Phelps theorem for bounded linear operators, J.~Lindenstrauss \cite{L} showed that there are operators which can not be approximated by norm attaining ones. Therefore, in general, there is no version of the Bishop-Phelps theorem for operators. On the other hand, if $X$ is reflexive, then $\NA(X,Y)$ is dense in $\mathcal{L}(X,Y)$ for every Banach space $Y$ (actually, this holds for Banach spaces $X$ with the Radon-Nikod\'{y}m property by a result of J.~Bourgain \cite{Bour}); if $Y$ is a closed subspace of $\ell_\infty$ containing the canonical copy of $c_0$, then $\NA(X,Y)$ is dense in $\mathcal{L}(X,Y)$ for every Banach space $X$. We refer the reader to the survey \cite{Acosta-RACSAM} for a detailed account on norm attaining operators.

In 1970, Bollob\'as \cite{Bol} proved a quantitative version of the Bishop-Phelps theorem which turned out to be very useful in numerical range theory. Nowadays, this result is known as the Bishop-Phelps-Bollob\'as theorem. It can be enunciated as follows. Let $X$ be a Banach space, let $0<\eps <2$ and suppose that $x \in B_X$ and $x^* \in B_{X^*}$ satisfy $\re x^*(x) > 1 - \frac{\eps^2}{2}$. Then, there are $y \in S_X$ and $y^* \in S_{X^*}$ such that $y^*(y) = 1$, $\|y - x\| < \eps$, and $\|y^* - x^*\| < \eps$  (see \cite{CKMM} for this slightly improved version of the original one \cite{Bol}).

This result motivated M.~Acosta, R.~Aron, D.~Garc\'ia and M.~Maestre \cite{AAGM} to introduce in 2008 the following property.

\begin{definition}[\mbox{\cite[Definition 1.1]{AAGM}}]
A pair $(X, Y)$ of Banach spaces has the {\it Bishop-Phelps-Bollob\'as property} ({\it BPBp}, for short) if given $\eps > 0$, there exists $\eta(\eps) > 0$ such that whenever $T \in \mathcal{L}(X, Y)$ with $\|T\| = 1$ and $x_0 \in S_X$ are such that
	\begin{equation*}
	\|Tx_0\| > 1 - \eta(\eps),
	\end{equation*}
	there are $S \in \mathcal{L}(X, Y)$ with $\|S\| = 1$ and $x_1 \in S_X$ such that
	\begin{equation*}
	\|Sx_1\| = 1, \ \ \ \|x_1 - x_0\| < \eps, \ \ \ \text{and} \ \ \ \|S - T\| < \eps.
	\end{equation*}
In this case, we say that the pair $(X, Y)$ has the BPBp with the function $\eps \longmapsto \eta(\eps)$.

If we restrict the operators $T$ and $S$ to be compact in the above definition, then the corresponding property is called the {\it Bishop-Phelps-Bollob\'{a}s property for compact operators} ({\it BPBp for compact operators}, for short) (see \cite{DGMM}).
\end{definition}

The aim of the authors of \cite{AAGM} was to study the conditions that $X$ and $Y$ must satisfy to get a Bishop-Phelps-Bollob\'as type theorem for bounded linear operators. They characterized when the pair $(\ell_1,Y)$ has the BPBp via a geometric property of the Banach space $Y$ which is satisfied by many Banach spaces as $C(K)$, $L_1(\mu)$, but not for all Banach spaces. They also proved that $(X, Y)$ has the BPBp when $X$ and $Y$ are finite-dimensional, or when $X$ is arbitrary and $Y$ is a closed subspace of $\ell_\infty$ containing the canonical copy of $c_0$. There is a vast literature about this topic and we invite the reader to take a look at the papers cited here and the references therein, as the already cited \cite{AAGM} and \cite{AMS, ACKLM,Cho-Choi,DGMM,DKL,DKKLM,KimLeeMartin-JMAA2015}.

In this paper we are interested in the behavior of the Bishop-Phelps-Bollob\'as property and other related properties with respect to absolute sums. Our main motivation is the similar study done in \cite{ACKLM} for $c_0$-, $\ell_1$- and $\ell_\infty$-sums, which was a very useful technique to produce some important results and examples.

Before we continue, let us give the proper terminology and notation.

An {\it absolute norm} is a norm $| \cdot |_a$ in $\R^2$ such that $|(1, 0)|_a = |(0, 1)|_a = 1$ and $|(s, t)|_a = |(|s|, |t|)|_a$ for every $s, t \in \R$. Given two Banach spaces $W$ and $Z$ and an absolute norm $| \cdot |_a$, the {\it absolute sum} of $W$ and $Z$ with respect to $| \cdot |_a$, denoted by $W \oplus_a Z$, is the Banach space $W \times Z$ endowed with the norm
\begin{equation*}
\|(w, z)\|_a = |(\|w\|, \|z\|)|_a \ \ \ (w \in W, z \in Z).
\end{equation*}
It is immediate to see that $\|(w, 0)\|_a = \|w\|$ for all $w \in W$, so $W$ is isometric to the subspace $\{(w,0)\colon w\in W\}$ of $W \oplus_a Z$. It is also easy to show that
\begin{equation} \label{ine1}
\max\{\|w\|,\|z\|\} \leq \| (w,z) \|_a \leq \|w\|+\|z\|
\end{equation}
for every $(w,z)\in W\oplus_a Z$ and every absolute norm $\| \cdot \|_a$. We will say that the Banach space $W$ is an {\it absolute summand} of the Banach space $X$, if there are another Banach space $Z$ and an absolute norm $| \cdot |_a$ in $\R^2$ such that $X = W \oplus_a Z$.  This terminology extends the well-known concepts of $L$-summand and $M$-summand (see \cite{HWW}): $W$ is a {\it $L$-summand} of $X$ if there is another Banach space $Z$ such that $X = W \oplus_1 Z$; analogously, if $X = W \oplus_{\infty} Z$ for some Banach space $Z$, then we say that $W$ is a {\it $M$-summand} of $X$. For background on absolute norms and absolute sums, we refer the reader to \cite{Bonsall1,MPR1,MPR2, MPR3, Paya, Paya1}. For a more recent reference, we suggest \cite{Hardtke}, where the author studies the stability of some geometrical properties of Banach spaces by absolute sums. Examples of absolute sums are the $\ell_p$-sums for $1 \leq p \leq \infty$ associated to the $\ell_p$-norms in $\R^2$.

In his doctoral dissertation \cite{Paya}, R.~Pay\'a proposed an intuitive classification of absolute norms defined through its behavior at the unit vector $(1, 0)$ of $\R^2$ (see also \cite[p.~38]{MPR1}). Some of our results depend on this classification, so we include it here. Let us first recall some necessary definitions. For $x \in X$, let $D(X, x)$ be the set of all $x^* \in S_{X^*}$ such that $x^*(x) = \|x\|$, which is convex and nonempty by the Hahn-Banach theorem. We say that $x \in S_X$ is a {\it vertex} of $B_X$ if $D(X, x)$ separates the points of $X$ and we say that $x$ is a {\it smooth point} of $B_X$ if $D(X, x)$ is a singleton subset of $X^*$. A vertex of the unit ball is an extreme point (see, for example, the remark after \cite[Theorem 4.6]{Bonsall0}).

\begin{definition} Let $| \cdot |_a$ be an absolute norm in $\R^2$. We say that $| \cdot |_a$ is of
	\begin{itemize}
		\item[(i)] {\it type $1$} if the vector $(1, 0)$ is an vertex of $B_{(\R^2,\|\cdot\|_a)}$;
		\item[(ii)] {\it type $2$} if the vector $(1, 0)$ is a smooth and extreme point of $B_{(\R^2,\|\cdot\|_a)}$;
		\item[(iii)] {\it type $\infty$} if the vector $(1, 0)$ is not extreme point of $B_{(\R^2,\|\cdot\|_a)}$.
	\end{itemize}
\end{definition}
The $\ell_p$-norm is of type $1$ for $p=1$, of type $\infty$ for $p=\infty$, and of type $2$ for $1<p<\infty$. In subsection \ref{subsection:absolute-sum}, at the end of this introduction, we will give an account on the results of absolute sums that we will need in this paper.

In section \ref{range} we show that if $Y_1$ is an absolute summand of $Y$ and a pair $(X,Y)$ has the BPBp (resp.\ BPBp for compact operators), then so does $(X,Y_1)$. The analogous result for the density of norm attaining operators and norm attaining compact operators also hold. For domain spaces, we show in section \ref{domain} that analogous results hold for type $1$ and type $\infty$ absolute norms: if $X_1$ is an absolute summand of type $1$ or $\infty$ of a Banach space $X$ and a pair $(X,Y)$ has the BPBp (resp.\ BPBp for compact operators), then so does $(X_1,Y)$. The corresponding results for the density of norm attaining operators and norm attaining compact operators also hold for type $1$ absolute norms.

The last section of the paper (\S \ref{numericalradius}) is devoted to the study of the Bishop-Phelps-Bollob\'{a}s property for numerical radius. Let us recall the relevant notation and terminology about this. Let $X$ be a Banach space and consider the set
\begin{equation*}
\Pi(X) := \{ (x, x^*) \in S_X \times S_{X^*}\colon x^*(x) = 1 \}.
\end{equation*}
The {\it numerical radius} of an operator $T \in \mathcal{L}(X)$ is defined by
\begin{equation*}
v(T) := \sup \{ |x^*(Tx)|\colon (x, x^*) \in \Pi(X) \}.
\end{equation*}
It is clear that $v(T) \leq \|T\|$ for all $T \in \mathcal{L}(X)$ and that $v(\cdot)$ is a seminorm in $\mathcal{L}(X)$. We say that $T \in \mathcal{L}(X)$ {\it attains its numerical radius} (or it is a {\it numerical radius attaining operator}) if there is $(x_0, x_0^*) \in \Pi(X)$ such that $|x_0^*(Tx_0)| = v(T)$. We denote by $\NRA(X)$ the set of all numerical radius attaining operators on $X$.
We refer the reader to the classical books \cite{Bonsall0,Bonsall1} for background on numerical radius of operators and to \cite{Acosta-RACSAM,CapMM,Paya1} and the references therein for background on the study of the density of the set of numerical radius attaining operators.

Let us give the definition of two properties related to the Bishop-Phelps-Bollob\'{a}s property. We take the definitions from \cite{KLM1} although they had appeared earlier for concrete Banach spaces (see \cite{GK}). We refer to \cite{AcoFakSole,AGR,GK, KLM1,KLM} and references therein for background.

\begin{definition} \label{defBPBp-nu}
Let $X$ be a Banach space. We say that
\begin{itemize}
\item[(a)] $X$ has the \emph{BPBp for numerical radius} (\emph{BPBp-nu}, for short) if given $\eps>0$, there is $\eta(\eps) > 0$ such that whenever $T \in \mathcal{L}(X)$ with $v(T) = 1$ and $(x, x^*) \in \Pi(X)$ satisfy
\begin{equation*}
|x^*(Tx)| > 1 - \eta(\eps),
\end{equation*}
there are $S \in \mathcal{L}(X)$ with $v(S) = 1$ and $(x_0, x_0^*) \in \Pi(X)$ such that
\begin{equation*}
|x_0^*(Sx_0)| = 1, \ \ \ \|x_0^*-x^*\| < \eps, \ \ \ \|x_0 - x\| < \eps, \ \ \ \text{and} \ \ \ \|S - T\| < \eps.
\end{equation*}
\item[(b)] $X$ has the {\it weak BPBp for numerical radius} (\emph{weak BPBp-nu}, for short) if given $\eps>0$, there is $\eta(\eps) > 0$ such that whenever $T \in \mathcal{L}(X)$ with $v(T) = 1$ and $(x, x^*) \in \Pi(X)$ satisfy
\begin{equation*}
|x^*(Tx)| > 1 - \eta(\eps),
\end{equation*}
there are $S \in \mathcal{L}(X)$ and $(x_0, x_0^*) \in \Pi(X)$ such that
\begin{equation*}
|x_0^*(Sx_0)| = v(S), \ \ \ \|x_0^*-x^*\| < \eps, \ \ \ \|x_0 - x\| < \eps, \ \ \ \text{and} \ \ \ \|S - T\| < \eps.
\end{equation*}
\end{itemize}
\end{definition}

Observe that the only difference between the BPBp-nu and the weak BPBp-nu is the normalization of the numerical radius of the operator $S$ given in the first definition. Both properties imply the density of the set of numerical radius attaining operators (see Lemma \ref{lemmaNRA}). When $v(\cdot)$ is a norm, equivalent to the operator norm (and actually in more situations, see \cite{KLM1, KLM2}), both properties are equivalent. As far as we know, it is not known whether both properties are always equivalent. Let us also say that both properties have their corresponding versions for compact operators, defined in the obvious way.

In section \ref{numericalradius} we will show that if $X$ is a Banach space with the BPBp-nu and $W$ is an absolute summand of type $1$ or $\infty$ of $X$, then $W$ has the BPBp-nu. The analogous result for the weak BPBp-nu, for the BPBp-nu for compact operators, and for the weak BPBp-nu for compact operators also hold. Furthermore, we show that if $X$ is a Banach space such that $\NRA(X)$ is dense in $\mathcal{L}(X)$ and $W$ is an absolute summand of type $1$ or $\infty$ of $X$, then $\NRA(W)$ is dense in $\mathcal{L}(W)$. The same result holds for compact operators.

Let us finally say that some of our results were previously known for the particular case of $L$-summands and/or $M$-summands, but other ones are new even in this context. We will highlight in the main part of the paper of which kind is each result. Let us also mention that for $L$-summands of the domain and for $M$-summands of the range, there is a formula for the norm of the operators involving the norms of the restrictions or projections (see the proof of \cite[Lemma 2]{PS}, for instance) which makes things easier. This is no longer true for arbitrary absolute summands.

\subsection{Some background on absolute sums}\label{subsection:absolute-sum}
Let us recall some known facts on absolute sums which will be relevant in our discussion. Let $W$, $Z$ be Banach spaces and let $\|\cdot\|_a$ be an absolute norm. There exists an isometric isomorphism between $[W \oplus_a Z]^*$ and $W^* \oplus_{a^*} Z^*$, where $| \cdot |_{a^*}$ is the dual norm associated to $| \cdot |_a$, which is also absolute. The action of a functional $(w^*, z^*) \in W^* \oplus_{a^*} Z^*$ at a point $(w, z) \in W \oplus_a Z$ is given by
\begin{equation*}
\langle (w, z), (w^*, z^*) \rangle = w^*(w) + z^*(z).
\end{equation*}

We will profusely use the following useful results which were proved in \cite{Paya}.

\begin{lemma}[\mbox{\cite[Propositions 5.3, 5.5, and 5.6]{Paya}}]\label{charac} Let $| \cdot |_a$ be an absolute norm in $\R^2$. Then,
	\begin{itemize}
		\item[(a)] $| \cdot |_a$ is of type $1$ if and only if there exists $K > 0$ such that $|x| + K |y| \leq |(x, y)|_a$ for every $x,y\in \R$.
		\item[(b)] $|\cdot |_a$ is of type $\infty$ if and only if there exists $b_0 > 0$ such that $|(1, b_0)|_a = 1$ (so, $|(1,b)|_a=1$, $\forall b\leq b_0$).
		\item[(c)] $| \cdot |_a$ is of type $1$ if and only $| \cdot |_{a^*}$ is of type $\infty$.
		\item[(d)] $| \cdot |_a$ is of type $\infty$ if and only $| \cdot |_{a^*}$ is of type $1$.
	\end{itemize}
\end{lemma}

Finally, we state the following easy result (for its proof see, for example, \cite[Lemma 2.2]{G}).

\begin{lemma} \label{state0} Let $W$ and $Z$ be Banach spaces and $\oplus_a$ be any absolute sum in $\R^2$. If $(w, z) \in S_{W \oplus_a Z}$ and $(w^*, z^*) \in S_{W^* \oplus_{a^*} Z^*}$ are such that
\begin{equation*}
\langle (w, z), (w^*, z^*) \rangle = 1,
\end{equation*}
then
\begin{equation*}
w^*(w) = \|w^*\|\|w\| \ \ \ \text{and} \ \ \ z^*(z) = \|z^*\| \|z\|.
\end{equation*}
\end{lemma}

\section{Results on Range Spaces} \label{range}
We start this section by showing that the BPBp passes from $(X, Y)$ to $(X, Y_1)$, when $Y_1$ is an absolute summand of $Y$. This result extends \cite[Propositions 2.3 and 2.7]{ACKLM}, where the results were done for $L$- and $M$-summands, and \cite[Theorem 2.3]{G}, where it was done for the particular case of $X=\ell_1$.

\begin{theorem} \label{range1} Let $X$, $Y$ be Banach spaces and let $Y_1$ be an absolute summand of $Y$. If the pair $(X, Y)$ has the BPBp, then so does $(X, Y_1)$.
\end{theorem}

\begin{proof} Given $\eps \in (0 , 1)$, consider $\eta (\eps) > 0$ to be the BPBp function for the pair $(X, Y)$ and let $Y_2$ be such that $Y = Y_1 \oplus_a Y_2$. Let $T_1 \in \mathcal{L}(X, Y_1)$ with $\|T_1\| = 1$ and $x_0 \in S_X$ be such that
\begin{equation*}
\|T_1x_0\| > 1 - \eta (\eps).
\end{equation*}
Define $\widetilde{T} \in \mathcal{L}(X, Y)$ by $\widetilde{T}(x) = (T_1x, 0)$ for all $x \in X$. Then $\|\widetilde{T}\| = \|T\| = 1$ and
\begin{equation*}
\|\widetilde{T} x_0\|_a = \|(T_1 x_0, 0)\|_a = \|T_1 x_0\| > 1 - \eta (\eps).
\end{equation*}
Since $\|\widetilde{T}\| = 1$, $x_0 \in S_X$ and the pair $(X, Y)$ has the BPBp with $\eta$, there are $\widetilde{S} \in \mathcal{L}(X, Y)$ with $\|\widetilde{S}\| = 1$ and $x_1 \in S_X$ such that
\begin{equation*}
\| \widetilde{S}x_1\|_a = 1, \ \ \ \|x_1 - x_0\| < \eps \ \ \ \mbox{and} \ \ \ \|\widetilde{S} - \widetilde{T} \| < \eps.
\end{equation*}

Write $\widetilde{S} = (\widetilde{S}_1, \widetilde{S}_2)$, where $\widetilde{S}_j \in \mathcal{L}(X, Y_j)$ for $j = 1, 2$. By using (\ref{ine1}), for all $x \in B_X$, we have
\begin{equation*}
\|(\widetilde{S}_1x - T_1x, \widetilde{S}_2x)\|_{\infty} \leq \|(\widetilde{S}_1x - T_1x, \widetilde{S}_2x)\|_a \leq \|\widetilde{S} - \widetilde{T}\| < \eps.
\end{equation*}
Then, $\|\widetilde{S}_1 - T_1\| < \eps$ and $\|\widetilde{S}_2\| < \eps$. Now we consider $y^* = (y_1^*, y_2^*) \in Y_1^* \oplus_{a^*} Y_2^*$ with $\|y^*\|_{a^*} = 1$ to be such that
\begin{equation*}
1 = \|\widetilde{S}x_1\| = y^*(\widetilde{S}x_1) = y_1^* (\widetilde{S}_1x_1) + y_2^* (\widetilde{S}_2x_1).
\end{equation*}
Lemma \ref{state0} gives that $y_1^* (\widetilde{S}_1x_1) = \|y_1^*\| \|\widetilde{S}_1 x_1\|$ and $y_2^* (\widetilde{S}_2x_1) = \|y_2^*\| \|\widetilde{S}_2 x_1\|$. Since
\begin{equation*}
\|y_1^*\| \| \widetilde{S}_1x_1\| =  y_1^*(\widetilde{S}_1 x_1) = 1 -  y_2^*(\widetilde{S}_2 x_1) \geq 1 - \|\widetilde{S}_2\| > 0,
\end{equation*}
we have that $y_1^* \not= 0$ and $\|\widetilde{S}_1 x_1\| \not= 0$. Define $S \in \mathcal{L}(X, Y_1)$ by
\begin{equation}\label{eq:S1}
S_1(x) := \|y_1^*\| \widetilde{S}_1x + y_2^* (\widetilde{S}_2x) \frac{\widetilde{S}_1x_1}{\|\widetilde{S}_1x_1\|} \ \ (x  \in X).
\end{equation}
Then, for every $x \in B_X$, we have that
\begin{align*}
\|S_1x\| &\leq \|y_1^*\|\|\widetilde{S}_1x\| + \|y_2^*\|\|\widetilde{S}_2x\| \\
&= \langle ( \| \widetilde{S}_1x\|, \|\widetilde{S}_2x\|), ( \|y_1^*\|, \|y_2^*\|) \rangle \\
&\leq | ( \| \widetilde{S}_1x\|, \| \widetilde{S}_2x\|)|_a | (\|y_1^*\|, \|y_2^*\|)|_{a^*} \\
&= \| (\widetilde{S}_1x, \widetilde{S}_2x)\|_a \|(y_1^*, y_2^*)\|_{a^*} \\
&= \|\widetilde{S}x\|_a \|(y_1^*, y_2^*)\|_{a^*} \leq \|\widetilde{S}\| \|y^*\|_{a^*} = 1.
\end{align*}
So, $\|S_1\| \leq 1$. On the other hand,
\begin{equation*}
\|S_1x_1\| \geq \frac{y_1^*}{\|y_1^*\|} \left(\|y_1^*\| \widetilde{S}_1x_1 + y_2^*(\widetilde{S}_2x_1) \frac{\widetilde{S}_1x_1}{\|\widetilde{S}_1x_1\|} \right)
= y_1^*(\widetilde{S}_1x_1) +  y_2^*(\widetilde{S}_2x_1) = 1.
\end{equation*}
This shows that $\|S_1\| = \|S_1x_1\| = 1$. It remains to prove that $\|S_1 - T_1\| < \eps$. Indeed, since
\begin{equation*}
1 - \|y_1^*\| \leq 1 -  y_1^*(\widetilde{S}_1x_1) =  y_2^*(\widetilde{S}_2x_1) \leq \|\widetilde{S}_2\| < \eps,
\end{equation*}
we have for all $x \in B_X$ that
\begin{equation*}
\|S_1x - T_1x\| \leq \| \|y_1^*\| \widetilde{S}_1x - \widetilde{S}_1x\| + \|\widetilde{S}_1 - T_1\| + \|S_2\| < 3 \eps
\end{equation*}
Therefore, $\|S_1 - T_1\| < 3 \eps$. Since we already have $\|x_1 - x_0\| < \eps$, we conclude that the pair $(X, Y_1)$ has the BPBp as desired.
\end{proof}

There is a property related to the BPBp for which we may also give an analogous result. A  pair $(X, Y)$ of Banach spaces has the \emph{pointwise BPB property} (see \cite[Definition 1.2]{DKL} or \cite[Definition 1.1]{DKKLM}) if given $\eps > 0$, there is $\eta(\eps) > 0$ such that whenever $T \in \mathcal{L}(X, Y)$ with $\|T\| = 1$ and $x_0 \in S_X$ satisfy $\|Tx_0\| > 1 - \eta(\eps)$, there is $S \in \mathcal{L}(X, Y)$ such that
$$
\|S\|=\|Sx_0\| = 1 \quad \text{ and } \quad \|S - T\| < \eps.
$$
It is clear that this property is stronger than the BPBp. It was proved in \cite{DKL} that if $(X, Y)$ has the pointwise BPB property for some $Y$, then the space $X$ must be uniformly smooth. The proof of Theorem \ref{range1} can be obviously adapted to the case of the pointwise BPB property. Therefore, we can state the following result.

\begin{prop} \label{pointwiserange} Let $X$, $Y$ be Banach spaces and let $Y_1$ be an absolute summand of $Y$. If the pair $(X, Y)$ has the pointwise BPB property, then so does $(X, Y_1)$.
\end{prop}

We would like to notice also that, in Theorem \ref{range1}, if one starts with a compact operator $T_1: X \longrightarrow Y_1$ and assume that the pair $(X, Y)$ has the BPBp for compact operators, then the operator $\widetilde{T}$ defined in the proof is compact, so we can continue the proof getting a compact operator $\widetilde{S}$ and, therefore, the operator $S_1: X \longrightarrow Y_1$ defined in \eqref{eq:S1} is also compact. Thus, we have the following analogous result for this class of operators. This generalizes \cite[Lemma 2.6.ii]{DGMM}, where the result was enunciated for $L$- and $M$-summands.

\begin{prop} \label{range2} Let $X$, $Y$ be Banach spaces and let $Y_1$ be an absolute summand of $Y$. If the pair $(X, Y)$ has the BPBp for compact operators, then so does $(X, Y_1)$.
\end{prop}

Let $D$ be a bounded closed convex subset of a Banach space $X$ and let $Y$ be another Banach space. Define $$\|T\|_D := \sup \{\|Tx\|: x \in D\}$$ for every $T \in \mathcal{L}(X, Y)$. In \cite{Cho-Choi} the following version of the BPBp was defined: the pair $(X, Y)$ has the {\it BPBp on $D$} if for every $\eps > 0$, there is $\eta_D (\eps) > 0$ such that whenever $T \in \mathcal{L}(X, Y)$ with $\|T\|_D = 1$ and $x \in D$ satisfy that $\|Tx\| > 1 - \eta_D(\eps)$, there are $S \in \mathcal{L}(X, Y)$ with $\|S\|_D = 1$ and $z \in D$ such that $$\|Sz\| = 1, \qquad \|z - x\| < \eps \quad \text{ and } \quad \|S - T\| < \eps$$ (the distance $\|S - T\|$ being calculated in the usual operator norm). Note that in the proof of Theorem \ref{range1}, we can work with any bounded closed convex subset $D$ of $X$ such that $D \subset B_X$ instead of $B_X$. So, we also obtain the following result which is an extension of \cite[Propositions 4.3 and 4.5]{Cho-Choi}, where the result was done for $L$- and $M$-summands.

\begin{prop} \label{range3} Let $X$, $Y$ be Banach spaces, let $D$ be a bounded closed convex subset of $X$ such that $D \subset B_X$ and let $Y_1$ be an absolute summand of $Y$. If the pair $(X, Y)$ has the BPBp on $D$, then so does $(X, Y_1)$.
\end{prop}

We next consider analogous results for norm attaining operators.

\begin{prop} \label{NA4} Let $X$, $Y$ be Banach spaces and let $Y_1$ be an absolute summand of $Y$. If $\NA(X, Y)$ is dense in $\mathcal{L}(X, Y)$, then $\NA(X, Y_1)$ is dense in $\mathcal{L}(X, Y_1)$.
\end{prop}

\begin{proof} Let $\eps \in (0, 1)$ and $T_1 \in \mathcal{L}(X, Y_1)$ with $\|T_1\| = 1$ be given. Let $Y_2$ be such that $Y = Y_1 \oplus_a Y_2$. Define $\widetilde{T} \in \mathcal{L}(X, Y_1 \oplus_a Y_2)$ by $\widetilde{T}(x) := (T_1x, 0)$ for all $x \in X$. Then, $\|\widetilde{T}\| = \|T\| = 1$. Since $\NA(X, Y_1 \oplus_a Y_2)$ is dense in $\mathcal{L}(X, Y_1 \oplus_a Y_2)$, there are $x_0 \in S_X$ and $\widetilde{S} \in \mathcal{L}(X, Y_1 \oplus_a Y_2)$ with $\|\widetilde{S}\| = 1$ such that $\|\widetilde{S}x_0\| = 1$ and $\|\widetilde{S} - \widetilde{T}\| < \eps$. Write $\widetilde{S} = (\widetilde{S}_1, \widetilde{S}_2)$, where $\widetilde{S}_j: X \longrightarrow Y_j$ for $j = 1, 2$. By (\ref{ine1}), we have that $\|\widetilde{S}_1 - \widetilde{T}_1\| < \eps$ and $\|\widetilde{S}_2\| < \eps$. Now, take $y^* = (y_1^*, y_2^*) \in Y_1^* \oplus_{a^*} Y_2^*$ with $\|y^*\|_{a^*} = 1$ to be such that $1 = \|\widetilde{S}x_0\| = y^*(\widetilde{S}x_0)$. Then, by Lemma \ref{state0}, $y_1^*(\widetilde{S}_1x_0) = \|\widetilde{S}_1x_0\|$ and $y_2^*(\widetilde{S}_2x_0) = \|\widetilde{S}_2x_0\|$. Since
	\begin{equation*}
	\|y_1^*\| \|\widetilde{S}_1x_0\| = y_1^*(\widetilde{S}_1x_0) = 1 - y_2^*(\widetilde{S}_2x_0) \geq 1 - \|\widetilde{S}_2\| > 1 - \eps > 0,
	\end{equation*}
	we may defined $S_1 \in \mathcal{L}(X, Y_1)$ by
	\begin{equation*}
	S_1(x) := \|y_1^*\| \widetilde{S}_1x + y_2^* (\widetilde{S}_2x) \frac{\widetilde{S}_1x_1}{\|\widetilde{S}_1x_1\|} \ \ (x  \in X).
	\end{equation*}
	This operator attains its norm at $x_0$ and it its close to $T_1$ (see the end of the proof of Theorem \ref{range1}).
\end{proof}

The above result was known for $L$-summands \cite[Proposition 2.9]{ACKLM} and for $M$-summands \cite[Lemma 2]{PS}.

Notice that the proof of Proposition \ref{NA4} also works for compact operators, providing the following result.

\begin{prop} \label{NA5} Let $X$, $Y$ be Banach spaces and let $Y_1$ be an absolute summand of $Y$. If $\NA(X, Y) \cap K(X, Y)$ is dense in $\mathcal{K}(X, Y)$, then $\NA(X, Y_1) \cap \mathcal{K}(X, Y_1)$ is dense in $\mathcal{K}(X, Y_1)$.
\end{prop}

We finish the section with an small discussion about the validity of somehow  reciprocal results. It is shown in \cite[Proposition 2.4]{ACKLM} that if $X$, $Y_1$, $Y_2$ are Banach spaces and the pairs $(X,Y_1)$ and $(X,Y_2)$ have the BPBp, then so does the pair $(X,Y_1\oplus_\infty Y_2)$; the same result holds for the BPBp for compact operators \cite[Lemma 3.16]{DGMM} and for the density of norm attaining operators \cite[Lemma 2]{PS}. We do not know whether any of these reciprocal results is also true for arbitrary absolute sums, even for the case of $\ell_1$-sum.

\section{Results on Domain Spaces} \label{domain}

We first prove that if $X_1$ is an absolute summand of type $1$ or $\infty$, and the pair $(X, Y)$ has the BPBp, so does $(X_1, Y)$. This extends \cite[Proposition 2.6]{ACKLM}, where the result was shown for $L$- and $M$-summands.

\begin{theorem} \label{domain1} Let $X$, $Y$ be Banach spaces and let $X_1$ be an absolute summand of $X$ of type $1$ or $\infty$. If the pair $(X, Y)$ has the BPBp, then so does $(X_1, Y)$.
\end{theorem}

\begin{proof} Let $\eps \in (0, 1)$ be given and suppose that the pair $(X, Y)$ has the BPBp with some function $\eps \longmapsto \eta(\eps)$. Let $X_2$ be the Banach space such that $X = X_1 \oplus_a X_2$. Pick any $T \in \mathcal{L} (X_1, Y)$ with $\|T\| = 1$ and $x_1 \in S_{X_1}$ such that
\begin{equation*}
\|T x_1\| > 1 - \eta(\eps).
\end{equation*}
Define $\widetilde{T} \in \mathcal{L} (X, Y)$ by $\widetilde{T} (z_1, z_2) := Tz_1$ for $(z_1, z_2) \in X$. Then $\|\widetilde{T}\| = 1$. Since $(x_1, 0) \in S_X$,
\begin{equation*}
\|\widetilde{T} (x_1, 0)\| = \|Tx_1\| > 1 - \eta(\eps)
\end{equation*}
and the pair $(X, Y)$ has the BPBp with $\eta$, there are $\widetilde{S} \in \mathcal{L} (X, Y)$ with $\| \widetilde{S}\| = 1$ and $(x_1', x_2') \in S_X$ such that
\begin{equation*}
\| \widetilde{S} (x_1', x_2') \| = 1, \ \ \|(x_1', x_2') - (x_0, 0)\|_a < \eps \ \ \mbox{and} \ \ \| \widetilde{S} - \widetilde{T} \| < \eps.
\end{equation*}
Using (\ref{ine1}), we get that $\|x_1' - x_0\| < \eps$ and $\|x_2'\| < \eps$. Define $S \in \mathcal{L} (X_1, Y)$ by
\begin{equation*}
S(z_1) := \widetilde{S} (z_1, 0)  \ \ \ (z_1 \in X_1).
\end{equation*}
Then $\|S\| \leq \|\widetilde{S}\| = 1$ and $\|S - T\| \leq \|\widetilde{S} - \widetilde{T}\| < \eps$. To finish the proof, we will prove that $S$ attains its norm at $x_1'$ since we already have $\|x_1' - x_0\| < \eps$ and $\|S - T\| < \eps$. To do so, we divide the proof in two cases.

\noindent
{\it Case 1}: Suppose that $\oplus_a$ is an absolute norm of type $1$. By Lemma \ref{charac}.a, there exists $K > 0$ such that
\begin{equation*}
\|x_1'\| + K \| x_2'\| \leq | (\|x_1'\|, \|x_2'\|)|_a = \|(x_1', x_2')\|_a = 1.
\end{equation*}
We prove that $x_2' = 0$. Note that for all $z_2 \in B_{X_2}$, we have
\begin{equation*}
\| \widetilde{S} (0, z_2) \| = \| \widetilde{S} (0, z_2) - \widetilde{T} (0, z_2) \| \leq \| \widetilde{S} - \widetilde{T} \| < \eps.
\end{equation*}
Therefore, if we assume that $x_2' \not= 0$, we get for all $\eps \in (0, K)$ that
\begin{align*}
1 = \| \widetilde{S} (x_1', x_2') \| &= \|x_1'\| \left\| \widetilde{S} \left( \frac{x_1'}{\|x_1'\|}, 0 \right) \right\| + \|x_2'\| \left\| \widetilde{S} \left( 0, \frac{x_2'}{\|x_2'\|} \right) \right\| \\
&\leq \|x_1'\| + \eps \|x_2'\| \\
&< \|x_1'| + K \|x_2'\| \leq 1
\end{align*}

which is a contradiction. Then, $\|S\| = \| S(x_1') \| = \|\widetilde{S} (x_1', 0) \| = 1$.

\noindent
{\it Case 2}: Now assume that $\oplus_a$ is an absolute norm of type $\infty$. By Lemma \ref{charac}.b, there is $b_0 > 0$ such that $|(1, b_0)|_a = 1$. Set $\rho = \frac{b_0}{\eps} > 0$ and consider the vector $(x_1', \rho x_2') \in X$. Note that since $\|x_2'\| < \eps$, $\| \rho x_2'\| = \rho \|x_2'\| < \rho \eps = b_0$. Therefore, since $\|(x_1', \rho x_2')\|_a = | (\|x_1'\|, \| \rho x_2' \|)|_a$, $\|x_1'\| \leq 1$ and $\| \rho x_2'\| < b_0$, we have by the definition of $b_0$ that
$\|(x_1', \rho x_2')\|_a \leq |(1, b_0)|_a = 1$ and then $(x_1', \rho x_2') \in B_X$. So, writting
\begin{equation*}
(x_1', x_2') = \left( 1 - \frac{\eps}{b_0} \right) (x_1', 0) + \frac{\eps}{b_0} (x_1', \rho x_2'),
\end{equation*}
we get
\begin{equation*}
1 = \|(x_1', x_2')\|_a \leq \left( 1 - \frac{\eps}{b_0}\right) \|(x_1', 0)\|_a + \frac{\eps}{b_0} \|(x_1', \rho x_2')\|_a \leq 1
\end{equation*}
and
\begin{equation*}
1 = \|\widetilde{S} (x_1', x_2') \| \leq \left( 1 - \frac{\eps}{b_0} \right) \|\widetilde{S} (x_1', 0) \| + \frac{\eps}{b_0} \| \widetilde{S} (x_1, \rho x_2')\| \leq 1.
\end{equation*}
Then, $\|x_1'\| = \|(x_1', 0)\|_a = 1$ and $\|Sx_1'\| = \|\widetilde{S}(x_1', 0)\| = 1 = \|S\|$.
\end{proof}

We would like to notice that there is a more general result than Theorem \ref{domain1} for the pointwise BPB property (see \cite[Proposition 2.1]{DKKLM}), which says that if $X_1$ is one-complemented in $X$ and $(X, Y)$ has the pointwise BPB property, then so does $(X_1, Y)$. We do not know if it is possible to get such a general result for the BPBp.

On the other hand, we can easily adapt the proof of Theorem \ref{domain1}  in order to get an analogous result for the BPBp for compact operators. This extends \cite[Lemma 2.6.i]{DGMM}, where the result was enunciated for $L$- and $M$-summands.

\begin{prop} \label{domain2} Let $X$, $Y$ be Banach spaces and let $X_1$ be an absolute summand of $X$ of type $1$ or $\infty$. If the pair $(X, Y)$ has the BPBp for compact operators, then so does $(X_1, Y)$.
\end{prop}

Let us present now the version of Theorem \ref{domain1} for norm attaining operators, but in this case we may only deal with type $1$ absolute norms. The result was previously known for $L$-summands (see \cite[Lemma 2]{PS}).

\begin{prop} \label{NA1} Let $X$, $Y$ be Banach spaces and let $X_1$ be an absolute summand of $X$ of type $1$. If $\NA(X, Y)$ is dense in $\mathcal{L}(X, Y)$, then $\NA(X_1, Y)$ is dense in $\mathcal{L}(X_1, Y)$.
\end{prop}

\begin{proof} Let $\eps \in (0, 1)$ and $T \in \mathcal{L}(X_1, Y)$ with $\|T\| = 1$ be given. Consider $X_2$ to be a Banach space such that $X = X_1 \oplus_a X_2$. Define $\widetilde{T} \in \mathcal{L}(X, Y)$ by $\widetilde{T}(z_1, z_2) := (Tz_1, 0)$ for all $(z_1, z_2) \in X_1 \oplus_a X_2$. Then $\|\widetilde{T}\| = \|T\| = 1$. Since $\NA(X, Y)$ is dense in $\mathcal{L}(X, Y)$, there are $(\widetilde{x}_1, \widetilde{x}_2) \in S_{X_1 \oplus_a X_2}$ and $\widetilde{S} \in \mathcal{L}(X, Y)$ with $\|\widetilde{S}\| = 1$ such that $\|\widetilde{S}(\widetilde{x}_1, \widetilde{x}_2)\| = 1$ and $\|\widetilde{S} - \widetilde{T}\| < \eps$. Define $S \in \mathcal{L}(X_1, Y)$ by $S(z_1) := \widetilde{S}(z_1, 0)$ for all $z_1 \in X_1$. Then, $\|S\| \leq 1$ and for all $z_1 \in S_{X_1}$, we have
	\begin{equation*}
	\|Sz_1 - Tz_1\| = \|\widetilde{S}(z_1, 0) - \widetilde{T}(z_1, 0)\| \leq \|\widetilde{S} - \widetilde{T}\| < \eps.
	\end{equation*}
	So, $\|S - T\| < \eps$. It remains to prove that $S$ attains its norm. Indeed, first notice that for all $z_2 \in B_{X_2}$, we have
	\begin{equation*}
	\|\widetilde{S}(0, z_2)\| = \| \widetilde{S}(0, z_2) - \widetilde{T}(0, z_2)\| \leq \|\widetilde{S} - \widetilde{T}\| < \eps.
	\end{equation*}
	This implies that $\widetilde{x}_1 \not= 0$, otherwise, we would have $1 = \|\widetilde{S}(\widetilde{x}_1, \widetilde{x}_2)\| = \|\widetilde{S}(0, \widetilde{x}_2)\| < \eps$, which is a contradiction.	Since $\oplus_a$ is of type $1$, by Lemma \ref{charac}.a, there is $K > 0$ such that $\|\widetilde{x}_1\| + K \|\widetilde{x}_2\| \leq |(\|\widetilde{x}_1\|, \|\widetilde{x}_2\|)|_a = \| (\widetilde{x}_1, \widetilde{x}_2)\|_a = 1$. If $\widetilde{x}_2 \not= 0$, we have that, for every $\eps \in (0, K)$,
	\begin{equation*}
	1 = \| \widetilde{S}(\widetilde{x}_1, \widetilde{x}_2)\| \leq \|\widetilde{x}_1\| + \|\widetilde{x}_2\| \left\| \widetilde{S} \left( 0, \frac{\widetilde{x}_2}{\|\widetilde{x}_2\|} \right)\right\| < \|\widetilde{x}_1\| + \eps \| \widetilde{x}_2 \| < \|\widetilde{x}_1\| + K \|\widetilde{x}_2\| \leq 1,
	\end{equation*}
	which is a new contradiction. So, $\widetilde{x}_2 = 0$ and, then, $\|S\widetilde{x}_1\| = \| \widetilde{S}(\widetilde{x}_1, 0)\| = 1 = \|\widetilde{x}_1\|$. Therefore, $\NA(X_1, Y)$ is dense in $\mathcal{L}(X_1, Y)$.
\end{proof}

With the same proof, when one restricts it to compact operators, we get the following result.

\begin{prop} \label{KNA-domain} Let $X$, $Y$ be Banach spaces and let $X_1$ be an absolute summand of type $1$ of $X$. If the set $\NA(X, Y) \cap K(X, Y)$ is dense in $\mathcal{K}(X, Y)$, then $\NA(X_1, Y) \cap \mathcal{K}(X_1, Y)$ is dense in $\mathcal{K}(X_1, Y)$.
\end{prop}

We do not know if the analogous result of Proposition \ref{NA1} holds true also for absolute norms of type $\infty$. Actually, we do not know what happens even for $M$-summands.

As in the previous case, we finish the section with an small discussion about the validity of reciprocal results. Let $X_1$, $X_2$, $Y$ be Banach spaces. It is shown in \cite[Lemma 2]{PS} that if $\NA(X_1,Y)$ and $\NA(X_2,Y)$ are dense in their respective spaces of operators, then $\NA(X_1\oplus_1 X_2,Y)$ is dense in $\mathcal{L}(X_1\oplus_1 X_2,Y)$. The validity of the analogous result for the BPBp is not true: the pair $(\R,Y)$ has the BPBp for every Banach space $Y$ (trivial), while there are $Y$'s such that $(\R\oplus_1 \R,Y)$ does not have the BPBp (see \cite[Corollary 3.3]{ACKLM} for instance). As $\R\oplus_1\R \equiv \R\oplus_\infty \R$, the same example shows that the reciprocal result is not true for the $\ell_\infty$-sum. We do not know what is the situation for $\ell_p$-sums with $1<p<\infty$.

\section{Results for Numerical Radius} \label{numericalradius}

We would like now to tackle the analogous questions of the previous sections for numerical radius. Our first result in this line is the following one which extends \cite[Lemma 19]{KLM1}, where the result was proved for $L$- and $M$-summands.

\begin{theorem} \label{numericalradius1} Let $X$ be a Banach space and let $W$ be an absolute summand of type 1 or $\infty$ of $X$. If $X$ has the BPBp-nu, so does $W$.
\end{theorem}

We will profusely use in this section the following result which is a particular case of  \cite[Lemma 3.3]{CMM}.

\begin{lemma} \label{radius} Let $W$, $Z$ be Banach spaces and let $|\cdot|_a$ be an absolute norm. Given $T \in \mathcal{L}(W)$, we define $\widetilde{T} \in \mathcal{L}(W \oplus_a Z)$ by $\widetilde{T}(w, z) := (Tw, 0)$ for every $(w, z) \in W \oplus_a Z$. Then $v(\widetilde{T}) = v(T)$ and $\|\widetilde{T}\|=\|T\|$.
\end{lemma}

We are now able to provide the proof of Theorem \ref{numericalradius1}.

\begin{proof}[Proof of Theorem \ref{numericalradius1}] Let $\eps \in (0, 1)$ be given and suppose that $X$ has the BPBp-nu with some function $\eta(\eps) > 0$. Consider $Z$ to be a Banach space with $X = W \oplus_a Z$. We will prove that $W$ satisfies the BPBp-nu with $\eta$. Let $T \in \mathcal{L}(W)$ with $v(T) = 1$ and $(w_0, w_0^*) \in \Pi(W)$ be such that
	\begin{equation*}
	|w_0^*(Tw_0)| > 1 - \eta(\eps).
	\end{equation*}
Consider $\widetilde{T} \in \mathcal{L}(W \oplus_a Z)$ to be defined by $\widetilde{T}(w, z) = (Tw, 0)$ for every $(w, z) \in W \oplus_a Z$. Lemma \ref{radius} gives $v(\widetilde{T}) = v(T) = 1$. Also,
\begin{equation*}
|\langle \widetilde{T}(w_0, 0), (w_0^*, 0) \rangle| = |w_0^*(Tw_0)| > 1 - \eta(\eps).
\end{equation*}
Since $((w_0, 0), (w_0^*, 0)) \in \Pi(W \oplus_a Z)$, $v(\widetilde{T}) = 1$ and $W \oplus_a Z$ has the BPBp-nu with $\eta$, there are $\widetilde{S} \in \mathcal{L}(W \oplus_a Z)$ with $v(\widetilde{S}) = 1$ and $((w_1, z_1), (w_1^*, z_1^*)) \in \Pi(W \oplus_a Z)$ such that
\begin{itemize}
\item[(a)] $|\langle \widetilde{S}(w_1, z_1), (w_1^*, z_1^*) \rangle | = 1$,
\item[(b)] $\|(w_1^*, z_1^*) - (w_0^*, 0)\|_{a^*} < \eps$,
\item[(c)] $\|(w_1, z_1) - (w_0, 0)\|_a < \eps$, and
\item[(d)] $\|\widetilde{S} - \widetilde{T}\| < \eps$.
\end{itemize}
\noindent
Write $\widetilde{S} = (\widetilde{S}_1, \widetilde{S}_2)$, where $\widetilde{S}_1: W \oplus_a Z \longrightarrow W$ and $\widetilde{S}_2: W \oplus_a Z \longrightarrow Z$. Define $S \in \mathcal{L}(W)$ by
\begin{equation*}
S(w) := \widetilde{S}_1(w, 0) \ \ \ \  \ (w \in W).
\end{equation*}
Then, for every $(w, w^*) \in \Pi(W)$, we have
\begin{equation} \label{eq1}
|w^*(Sw)| = |w^*(\widetilde{S}_1(w, 0))| = | \langle \widetilde{S}(w, 0), (w^*, 0) \rangle| \leq v(\widetilde{S}) = 1.
\end{equation}
So, $v(S) \leq 1$. Now, since $\|(w_1^*, z_1^*) - (w_0^*, 0)\|_{a^*} < \eps$ and $\|(w_1, z_1) - (w_0, 0)\|_a < \eps$, by using (\ref{ine1}), we get that
\begin{equation*}
\|w_1^* - w_0^*\| < \eps \ \ \ \mbox{and} \ \ \ \|w_1 - w_0\| < \eps.
\end{equation*}
Moreover, for every $w \in S_W$, we have
\begin{align*}
\|Sw - Tw\| &= \|\widetilde{S}_1(w, 0) - Tw\| \\
&\leq \max \{ \|\widetilde{S}_1(w, 0) - Tw\|, \|\widetilde{S}_2(w, 0)\| \} \\
&= \|\widetilde{S}(w, 0) - \widetilde{T}(w, 0)\|_{\infty}
\leq \|\widetilde{S}(w, 0) - \widetilde{T}(w, 0)\|_a < \eps.
\end{align*}
So, $\|S - T\| < \eps$. It remains to prove that $w_1^*(w_1) = 1$ and $|w_1^*(Sw_1)| = 1$. To do so, we divide the proof in two cases.

\noindent
{\it Case 1:} Suppose that $\oplus_a$ is of type $\infty$. We will prove that $z_1^* = 0$. To do so, suppose that it is not true. By Lemma \ref{charac}.d, $\oplus_{a^*}$ is of type $1$. So, there is $K > 0$ such that
\begin{equation*}
\|w_1^*\| + K \|z_1^*\| \leq |(\|w_1^*\|, \|z_1^*\|)|_{a^*} = \|(w_1^*, z_1^*)\|_{a^*} = 1.
\end{equation*}
Notice that, since $\|(w_1, z_1) - (w_0, 0)\|_a < \eps$, we have that $\|z_1\| < \eps$. So, if $\eps \in (0, K)$, then
\begin{align*}
1 = w_1^*(w_1) + z_1^*(z_1) &\leq \|w_1^*\|\|w_1\| + \|z_1^*\|\|z_1\| \\
&< \|w_1^*\| + \eps \|z_1^*\| < \|w_1^*\| + K \|z_1^*\| \leq 1,
\end{align*}
which is a contradiction. So, $z_1^* = 0$. This implies that $(w_1, w_1^*) \in \Pi(W)$.

Now, by Lemma \ref{charac}.b, there is $b_0 > 0$ such that $|(1, b_0)|_a = 1$. Put $\rho = \frac{b_0}{\eps} > 0$. Then, $\|\rho z_1\| = \rho \|z_1\| < \rho \eps = b_0$, so
\begin{equation*}
\|(w_1, \rho z_1)\|_a = |(\|w_1\|, \|\rho z_1\|)|_a \leq |(1, b_0)|_a = 1.
\end{equation*}
Writing
\begin{equation*}
(w_1, z_1) = \left(1 - \frac{\eps}{b_0}\right) (w_1, 0) + \frac{\eps}{b_0} (w_1, \rho z_1)
\end{equation*}
we get that
\begin{equation*}
1 = \|(w_1, z_1)\|_a \leq \left(1 - \frac{\eps}{b_0} \right) \|(w_1, 0)\|_a + \frac{\eps}{b_0}\|(w_1, \rho z_1)\|_a \leq 1,
\end{equation*}
which implies that $\|(w_1, \rho z_1)\|_a = 1$. So, $((w_1, \rho z_1), (w_1^*, 0)) \in \Pi(W \oplus_a Z)$ and
\begin{equation*}
|\langle \widetilde{S}(w_1, \rho z_1), (w_1^*, 0) \rangle| \leq v(\widetilde{S}) = 1.
\end{equation*}
Therefore,
\begin{align*}
1 &= |\langle \widetilde{S}(w_1, z_1), (w_1^*, 0) \rangle| \\
&= \left| \left\langle \widetilde{S} \left(	\left(1 - \frac{\eps}{b_0}\right) (w_1, 0) + \frac{\eps}{b_0} (w_1, \rho z_1) \right), (w_1^*, 0) \right\rangle \right| \\
&\leq \left(1 - \frac{\eps}{b_0}\right) | \langle \widetilde{S}(w_1, 0), (w_1^*, 0) \rangle| + \frac{\eps}{b_0} | \langle \widetilde{S}(w_1, \rho z_1), (w_1^*, 0) \rangle| \leq 1.
\end{align*}
This shows that $|w_1^*(Sw_1)| = | \langle \widetilde{S}(w_1, 0), (w_1^*, 0) \rangle| = 1$. By (\ref{eq1}), we get that $v(S) = |w_1^*(Sw_1)| = 1$.

\noindent
{\it Case 2:} Now suppose that $\oplus_a$ is of type $1$. Since $\|(x_0^*, y_0^*) - (x, 0)\|_{a^*} < \eps$, we have that $\|y_0^*\| < \eps$. We will prove that $z_1 = 0$. Suppose not. By Lemma \ref{charac}.a, there is $K > 0$ such that
\begin{equation*}
\|w_1\| + K \|z_1\| \leq | (\|w_1\|, \|z_1\|)|_a = \|(w_1, z_1)\|_a = 1.
\end{equation*}
Therefore, if $\eps \in (0, K)$, then
\begin{align*}
1 = w_1^*(w_1) + z_1^*(z_1) &\leq \|w_1^*\|\|w_1\| + \|z_1^*\|\|z_1\| \\
&< \|w_1\| + \eps \|z_1\| < \|w_1\| + K \|z_1\| \leq 1,
\end{align*}
which is a contradiction. So, $z_1 = 0$ and $(w_1, w_1^*) \in \Pi(W)$.

By Lemma \ref{charac}.c, $\oplus_{a^*}$ is of type $\infty$. So, there is $b_0 > 0$ such that $|(1, b_0)|_{a^*} = 1$. Put $\rho = \frac{b_0}{\eps} > 0$. Then, $\|\rho z_1^*\| = \rho \|z_1^*\| < \rho \eps = b_0$. So,
\begin{equation*}
\|(w_1^*, \rho z_1^*)\|_{a^*} = |(\|w_1^*\|, \|\rho z_1^*\|)|_{a^*} \leq |(1, b_0)|_{a^*} = 1.
\end{equation*}
Write
\begin{equation*}
(w_1^*, z_1^*) = \left( 1 - \frac{\eps}{b_0} \right) (w_1^*, 0) + \frac{\eps}{b_0} (w_1^*, \rho z_1^*).
\end{equation*}
So, $\|(w_1^*, \rho z_1^*)\|_{a^*} = 1$ and then $((w_1, 0), (w_1^*, \rho z_1^*)) \in \Pi(W \oplus_a Z)$. This implies that
\begin{equation*}
| \langle \widetilde{S} (w_1, 0), (w_1^*, \rho z_1^*) \rangle| \leq v(\widetilde{S}) = 1
\end{equation*}
and
\begin{equation*}
1 = | \langle \widetilde{S}(w_1, 0), (w_1^*, z_1^*) \rangle|
\leq \left(1 - \frac{\eps}{b_0} \right) |(w_1^*, 0)(\widetilde{S}(w_1, 0))| + \frac{\eps}{b_0} |(w_1^*, \rho z_1^*)(\widetilde{S}(w_1, 0))| \leq 1.
\end{equation*}
So, $|w_1^*(Sw_1)| = | \langle \widetilde{S}(w_1, 0), (w_1^*, 0) \rangle| = 1$ and this finishes the proof.
\end{proof}

We would like to point out that the same proof of Theorem \ref{numericalradius1} works also for compact operators and we have the analogous result for the BPBp-nu for compact operators.

\begin{prop} \label{numericalradius2} Let $X$ be a Banach space and let $W$ be an absolute summand of type 1 or $\infty$ of $X$. If $X$ has the BPBp-nu for compact operators, so does $W$.
\end{prop}

Arguing as in Theorem \ref{numericalradius1}, we get the following result for the weak BPBp-nu. Recall that the only difference between the BPBp-nu and the weak BPBp-nu is the normalization of the numerical radius of the operator by its numerical radius (see Definition \ref{defBPBp-nu}). We include an sketch of the proof for completeness.

\begin{prop} \label{weaknumericalradius1} Let $X$ be a Banach space and let $W$ be an absolute summand of type 1 or $\infty$ of $X$. If $X$ has the weak BPBp-nu, so does $W$.
\end{prop}

\begin{proof} Let $\eps \in (0, 1)$ be given and consider the weak BPBp-nu function $\eta(\eps) > 0$ for the Banach space $X = W \oplus_a Z$. Let $T \in \mathcal{L}(W)$ with $v(T) = 1$ and $(w_0, w_0^*) \in \Pi(W)$ be such that $|w_0^*(Tw_0)| > 1 - \eta(\eps)$. Using the same notation of Theorem \ref{numericalradius1} and applying our hypothesis, there are $\widetilde{S} \in \mathcal{L}(W \oplus_a Z)$  and $((w_1, z_1), (w_1^*, z_1^*)) \in \Pi(W \oplus_a Z)$ such that $|\langle \widetilde{S}(w_1, z_1), (w_1^*, z_1^*) \rangle | = v(\widetilde{S})$, $\|(w_1^*, z_1^*) - (w_0^*, 0)\|_{a^*} < \eps$, $\|(w_1, z_1) - (w_0, 0)\|_a < \eps$, and $\|\widetilde{S} - \widetilde{T}\| < \eps$. Then, $\|w_1^* - w_0^*\| < \eps$ and $\|w_1 - w_0\| < \eps$. Writing again $\widetilde{S} = (\widetilde{S}_1, \widetilde{S}_2)$, we define $S \in \mathcal{L}(W)$ by
$S(w) := \widetilde{S}_1(w, 0)$ for every $w \in W$. Then, $v(S) \leq v(\widetilde{S})$ and $\|S - T\| < \eps$.

Suppose that $\oplus_a$ is of type $\infty$. As in Theorem \ref{numericalradius1} (Case 1), we have that $z_1^* = 0$ and then $(w_1, w_1^*) \in \Pi(W)$. Since $\oplus_a$ is of type $\infty$, there is $b_0 > 0$ such that $|(1, b_0)|_a = 1$. Put $\rho = \frac{b_0}{\eps} > 0$. Since $\|z_1\| < \eps$, we have that $\|(w_1, \rho z_1)\|_a \leq |(1, b_0)|_a = 1$. Then, $((w_1, \rho z_1), (w_1^*, 0)) \in \Pi(W \oplus_a Z)$. This implies that $|\langle \widetilde{S}(w_1, \rho z_1), (w_1^*, 0) \rangle| \leq v(\widetilde{S})$. Therefore,
\begin{align*}
	v(\widetilde{S}) &= |\langle \widetilde{S}(w_1, z_1), (w_1^*, 0) \rangle| \\
	&= \left| \left\langle \widetilde{S} \left(	\left(1 - \frac{\eps}{b_0}\right) (w_1, 0) + \frac{\eps}{b_0} (w_1, \rho z_1) \right), (w_1^*, 0) \right\rangle \right| \\
	&\leq \left(1 - \frac{\eps}{b_0}\right) | \langle \widetilde{S}(w_1, 0), (w_1^*, 0) \rangle| + \frac{\eps}{b_0} | \langle \widetilde{S}(w_1, \rho z_1), (w_1^*, 0) \rangle| \\
	&= \left(1 - \frac{\eps}{b_0}\right) |w_1^*(Sw_1)| + \frac{\eps}{b_0} | \langle \widetilde{S}(w_1, \rho z_1), (w_1^*, 0) \rangle| \leq v(\widetilde{S}).
\end{align*}
This shows that $|w_1^*(Sw_1)| = v(\widetilde{S}) \geq v(S)$. So, $|w_1^*(Sw_1)| = v(S)$ and $W$ has the weak BPBp-nu.

Now suppose that $\oplus_a$ is of type 1. As in Theorem \ref{numericalradius1} (Case 2), $z_1 = 0$ and then $(w_1, w_1^*) \in \Pi(W)$. Since $\oplus_{a^*}$ is of type $\infty$ (see Lemma \ref{charac}.(c)), there is $b_0 > 0$ such that $|(1, b_0)|_{a^*} = 1$. Put $\rho = \frac{b_0}{\eps} > 0$. Since $\|z_1^*\| < \eps$, we have that $\|(w_1^*, \rho z_1^*)\|_{a^*} \leq |(1, b_0)|_{a^*} = 1$. Then $((w_1, 0), (w_1^*, \rho z_1^*)) \in \Pi(W \oplus_a Z)$. This implies that $|\langle \widetilde{S}(w_1, 0), (w_1^*, \rho z_1^*) \rangle | \leq v(\widetilde{S})$. Therefore,
\begin{align*}
v(\widetilde{S}) &= | \langle \widetilde{S} (w_1, 0), (w_1^*, z_1^*) \rangle| \\
&= \left| \left \langle \widetilde{S}(w_1, 0), \left( 1 - \frac{\eps}{b_0} \right) (w_1^*, 0) + \frac{\eps}{b_0} (w_1^*, \rho z_1^*) \right \rangle \right| \\
&\leq \left(1 - \frac{\eps}{b_0} \right) | \langle  \widetilde{S}(w_1, 0), (w_1^*, 0) \rangle| + \frac{\eps}{b_0} | \langle \widetilde{S}(w_1, 0), (w_1^*, \rho z_1^* ) \rangle | \\
&= \left(1 - \frac{\eps}{b_0} \right) | w_1^*(Sw_1)| + \frac{\eps}{b_0} | \langle \widetilde{S}(w_1, 0), (w_1^*, \rho z_1^* ) \rangle | \leq v(\widetilde{S}).
\end{align*}
This shows that $|w_1^*(Sw_1)| = v(\widetilde{S}) \geq v(S)$. So, $|w_1^*(Sw_1)| = v(S)$ and $W$ has the weak BPBp-nu.
\end{proof}

Again, the above proof can be adapted to compact operators to get the following result.

\begin{prop} Let $X$ be a Banach space and let $W$ be an absolute summand of type 1 or $\infty$ of $X$. If $X$ has the weak BPBp-nu for compact operators, so does $W$.
\end{prop}

Next, we have interest to investigate the density of the set of numerical radius attaining operators. To do so, we will prove the following easy lemma which says that, in order to prove the denseness of the set $\NRA(X)$ for a Banach space $X$, it is enough to consider operators with numerical radius one. Observe that $T \in \mathcal{L}(X)$ attains its numerical radius if and only if the operator $\lambda T$ does for every $\lambda \in \R$.

\begin{lemma} \label{lemmaNRA} Let $X$ be a Banach space. The following statements are equivalents.
\begin{itemize}
\item[(a)] The set $\NRA(X)$ is dense in $\mathcal{L}(X)$.
\item[(b)] For every $T \in \mathcal{L}(X)$ with $v(T) = 1$, there is a sequence $\{S_n\} \subset \NRA(X)$ with $v(S_n) = 1$ for every $n \in \N$ and such that $S_n \longrightarrow T$ in norm.
\item[(c)] For every $T \in \mathcal{L}(X)$ with $v(T) = 1$ and every $\eps>0$, there is $S \in \mathcal{L}(X)$ with $v(S) = 1$ such that $\|S - T\| < \eps$.
\end{itemize}
\end{lemma}

\begin{proof} (a) $\Rightarrow$ (b). Let $T \in \mathcal{L}(X)$ with $v(T) = 1$. By hypothesis, there is a sequence $\{S_n'\} \subset \NRA(X)$ such that $S_n' \longrightarrow T$ in norm. This implies that there is $n_0 \in \N$ such that $|v(S_n') - v(T)| \leq \|S_n' - T\| < 1$ for all $n \geq n_0$. Since $v(T) = 1$, we have that $v(S_n') > 0$ for all $n \geq n_0$. Consider then the sequence $(S_n)$ defined by $$S_n := \frac{1}{v(S_{n+n_0}')}S_{n+n_0}' \in \mathcal{L}(X) \qquad (n\in \N).$$ Then $\{S_n\} \subset \NRA(X)$, $v(S_n) = 1$ for all $n \in \N$ and, since $v(S_n') \longrightarrow v(T) = 1$, $S_n \longrightarrow T$ in norm. This proves (b).

\noindent
(b) $\Leftrightarrow$ (c) is immediate.

\noindent
(b) $\Rightarrow$ (a). Let $T \in \mathcal{L}(X)$ be given. If $v(T) = 0$, then $T$ attains its numerical radius and we are done. Otherwise, $v(T) \not= 0$ and we may consider the operator $T' := \frac{T}{v(T)}$ which satisfies that $v(T') = 1$. By hypothesis, there is a sequence $\{S_n\} \subset \NRA(X)$ with $v(S_n) = 1$ for all $n \in \N$ and such that $S_n \longrightarrow T'$ in norm. This implies that $v(T) S_n \longrightarrow v(T)T'=T$ and we are done since $v(T) S_n \in \NRA(X)$ for every $n \in \N$.
\end{proof}

Now we are ready to provide a result for the denseness of the operators which attain their numerical radius.

\begin{prop} \label{NRA1} Let $X$ be a Banach space and let $W$ be an absolute summand of $X$ of type 1 or $\infty$. If $\NRA(X)$ is dense in $\mathcal{L}(X)$, then $\NRA(W)$ is dense in $\mathcal{L}(W)$.
\end{prop}

\begin{proof} Let $Z$ be a Banach space such that $X = W \oplus_a Z$. Let $\eps \in (0, 1)$ and $T \in \mathcal{L}(W)$ be given. We may consider $v(T) = 1$ by using Lemma \ref{lemmaNRA}. Define $\widetilde{T} \in \mathcal{L}(W \oplus_a Z)$ by $\widetilde{T}(w, z) := (Tw, 0)$ for every $(w, z) \in W \oplus_a Z$. By Lemma \ref{radius}, $v(\widetilde{T}) = v(T) = 1$. Since $\NRA(W \oplus_a Z)$ is dense in $\mathcal{L}(W \oplus_a Z)$, there are $\widetilde{S} \in \mathcal{L}(W \oplus_a Z)$ with $v(\widetilde{S}) = 1$ and $((w_0, z_0), (w_0^*, z_0^*)) \in \Pi(W \oplus_a Z)$ such that
\begin{equation*}
|\langle \widetilde{S}(w_0, z_0), (w_0^*, z_0^*) \rangle| = v(\widetilde{S}) = 1 \ \ \ \mbox{and} \ \ \ \|\widetilde{S} - \widetilde{T}\| < \eps.
\end{equation*}
Set $\widetilde{S} = (\widetilde{S}_1, \widetilde{S}_2)$, where $\widetilde{S}_1: W \oplus_a Z \longrightarrow W$ and $\widetilde{S}_2: W \oplus_a Z \longrightarrow Z$. By Lemma \ref{state0}, we have
\begin{equation} \label{state}
w_0^*(w_0) = \|w_0^*\| \|w_0\| \ \ \ \mbox{and} \ \ \ z_0^*(z_0) = \|z_0^*\|\|z_0\|.
\end{equation}

\noindent
Now for all $(w, z) \in B_{W \oplus_a Z}$, we have that
\begin{align*}
\max \{ \| \widetilde{S}_1 (w, z) - Tw \|, \| \widetilde{S}_2 (w, z)\| \} &= \| ( \widetilde{S}_1(w, z) - Tw, \widetilde{S}_2(w, z))\|_{\infty} \\
&= \|((\widetilde{S}_1(w, z), \widetilde{S}_2(w, z)) - (Tw, 0))\|_{\infty} \\
&=\| \widetilde{S}(w, z) - \widetilde{T}(w, z)\|_{\infty} \\
&\leq\| \widetilde{S}(w, z) - \widetilde{T}(w, z)\|_a
\leq \|\widetilde{S} - \widetilde{T} \| < \eps.
\end{align*}
This implies that, for all $w \in B_W$,
\begin{equation} \label{ineq1}
	\|\widetilde{S}_1(w, 0) - Tw\| < \eps \ \ \ \mbox{and} \ \ \ \|\widetilde{S}_2\| < \eps.
\end{equation}
On the other hand, for all $z \in B_Z$, we get that
\begin{equation} \label{ineq}
	\|\widetilde{S}(0, z)\|_a = \|\widetilde{S} (0, z) - \widetilde{T}(0, z)\|_a \leq \|\widetilde{S} - \widetilde{T}\| < \eps.
\end{equation}
In particular, $\|\widetilde{S}_1(0, z)\| < \eps$ and $\|\widetilde{S}_2(0, z)\| < \eps$ for all $z \in B_Z$.

\noindent
{\it Claim:} $w_0 \not= 0$.
Otherwise, by using (\ref{ineq}), we would have
\begin{equation*}
1 = | \langle \widetilde{S}(0, z_0), (w_0^*, z_0^*) \rangle| \leq \|\widetilde{S}(0, z_0)\|_a < \eps,
\end{equation*}
which is a contradiction.

\noindent
{\it Claim:} $w_0^* \not= 0$.
Otherwise, by using (\ref{ineq1}), we would have
\begin{equation*}
	1 = | \langle \widetilde{S}(w_0, z_0), (0, z_0^*) \rangle| = |z_0^*(\widetilde{S}_2(w_0, z_0))| \leq \| \widetilde{S}_2\| < \eps,
\end{equation*}
which is a new contradiction.

\vspace{0.3cm}

Therefore, since $w_0, w_0^* \not=0$, by using (\ref{state}), we have that
\begin{equation} \label{state1}
\left( \left(\frac{w_0}{\|w_0\|}, 0 \right),  \left(\frac{w_0^*}{\|w_0^*\|}, 0 \right) \right) \in \Pi(W \oplus_a Z).
\end{equation}
Define the operator $S \in \mathcal{L}(W)$ by
\begin{equation*}
S(w) := \widetilde{S}_1(w, 0) \ \ \ (w \in W).
\end{equation*}
Note that for all $(w, w^*) \in \Pi(W)$, we have that $((w, 0), (w^*, 0)) \in \Pi(W \oplus_a Z)$ and then,
\begin{equation*}
|w^*(Sw)| = |w^*(\widetilde{S}_1(w, 0))| = | \langle \widetilde{S}(w, 0), (w^*, 0) \rangle| \leq v(\widetilde{S}) = 1.
\end{equation*}
This shows that $v(S) \leq 1$. Also, by using (\ref{ineq1}), note that for all $w \in S_W$,
\begin{equation*}
\|Sw - Tw\| = \|\widetilde{S}_1(w, 0) - T(w)\| < \eps.
\end{equation*}
So, $\|S - T\| < \eps$. It remains to prove that $S$ attains its numerical radius, and we do this separating the proof in two cases.

\noindent
{\it Case 1:} Assume first that $\oplus_a$ is of type $1$. We will prove that $z_0 = 0$. Suppose not. Since $\oplus_a$ is of type $1$, by Lemma \ref{charac}.a there is $K > 0$ such that $\|w_0\| + K\|z_0\| \leq \|(w_0, z_0)\|_a = 1$. On the other hand, being $\oplus_{a^*}$ of type $\infty$ (see Lemma \ref{charac}.c), there is $b_0 > 0$ such that $|(1, b_0)|_{a^*} = 1$ and then
\begin{equation*}
\left\| \left( \frac{w_0^*}{\|w_0^*\|}, b_0 z_0^* \right) \right\|_{a^*} \leq |(1, b_0)|_{a^*} = 1.
\end{equation*}
Using (\ref{state}), we have that
\begin{equation*}
\left \langle \left( \frac{w_0}{\|w_0\|}, 0 \right), \left( \frac{w_0^*}{\|w_0^*\|}, b_0 z_0^* \right) \right \rangle = 1.
\end{equation*}
Then,
\begin{equation*}
\left\| \left( \frac{w_0^*}{\|w_0^*\|}, b_0 z_0^* \right) \right\|_{a^*} = 1 \ \ \ \mbox{and} \ \ \ \left(\left( \frac{w_0}{\|w_0\|}, 0 \right), \left( \frac{w_0^*}{\|w_0^*\|}, b_0 z_0^* \right)\right) \in \Pi(W \oplus_a Z).
\end{equation*}
This implies that
\begin{equation} \label{NRAineq}
	\left| \left \langle \widetilde{S} \left( \frac{w_0}{\|w_0\|}, 0 \right), \left( \frac{w_0^*}{\|w_0^*\|}, b_0 z_0^* \right) \right \rangle \right| \leq v(\widetilde{S}) = 1.
\end{equation}
Now, set
\begin{equation*}
(w_0^*, z_0^*) = \left(1 - \frac{1}{b_0\|w_0^*\|} \right)(w_0^*, 0) + \frac{1}{b_0 \|w_0^*\|} (w_0^*, b_0 \|w_0^*\| z_0^*).
\end{equation*}
So, we have that
\begin{align*}
|\langle \widetilde{S}(w_0, 0), (w_0^*, z_0^*) \rangle| &= \left| \left \langle \widetilde{S}(w_0, 0), \left( 1 - \frac{1}{b_0 \|w_0^*\|} \right) (w_0^*, 0) + \frac{1}{b_0 \|w_0^*\|} (w_0^*, b_0 \|w_0^*\| z_0^*) \right \rangle \right| \\
&\leq \left(1 - \frac{1}{b_0 \|w_0^*\|} \right) |\langle \widetilde{S}(w_0, 0), (w_0^*, 0) \rangle| + \frac{1}{b_0 \|w_0^*\|} |\langle \widetilde{S}(w_0, 0), (w_0^*, b_0 \|w_0^*\| z_0^*) \rangle|
\end{align*}
By using (\ref{state1}), we get that
\begin{align*}
\left(1 - \frac{1}{b_0 \|w_0^*\|} \right) | \langle \widetilde{S}(w_0, 0), (w_0^*, 0) \rangle| &= \left(1 - \frac{1}{b_0 \|w_0^*\|} \right) \|w_0^*\| \|w_0\| \left| \left \langle \widetilde{S} \left( \frac{w_0}{\|w_0\|}, 0 \right), \left( \frac{w_0^*}{\|w_0^*\|}, 0 \right) \right \rangle \right| \\ &\leq \left(1 - \frac{1}{b_0 \|w_0^*\|}\right) \|w_0\| v(\widetilde{S}) = \left(1 - \frac{1}{b_0 \|w_0^*\|}\right) \|w_0\|.
\end{align*}
Now, using (\ref{NRAineq}),
\begin{align*}
\frac{1}{b_0 \|w_0^*\|} | \langle \widetilde{S}(w_0, 0), (w_0^*, b_0 \|w_0^*\| z_0^*) \rangle | &= \frac{1}{b_0 \|w_0^*\|} \|w_0^*\|\|w_0\| \left| \left \langle \widetilde{S} \left( \frac{w_0}{\|w_0\|}, 0 \right), \left( \frac{w_0^*}{\|w_0^*\|}, \frac{b_0 \|w_0^*\| z_0^*}{\|w_0^*\|} \right) \right \rangle \right| \\
&= \frac{1}{b_0\|w_0^*\|} \|w_0^*\| \|w_0\| \left| \left \langle \widetilde{S} \left( \frac{w_0}{\|w_0\|}, 0 \right), \left( \frac{w_0^*}{\|w_0^*\|}, b_0 z_0^*\right) \right \rangle \right| \\
&\leq \frac{1}{b_0 \|w_0^*\|} v(\widetilde{S} )\|w_0\| = 	\frac{1}{b_0 \|w_0^*\|} \|w_0\|
\end{align*}
Then,
\begin{equation*}
| \langle \widetilde{S}(w_0, 0), (w_0^*, z_0^*) \rangle| \leq \left(1 - \frac{1}{b_0 \|w_0^*\|}\right) \|w_0\| + \frac{1}{b_0\|w_0^*\|}\|w_0\| = \|w_0\|.
\end{equation*}
On the other hand, by using (\ref{ineq}), we have that
\begin{align*}
| \langle \widetilde{S}(0, z_0), (w_0^*, z_0^*) \rangle| &= \|z_0\|   \left| \left \langle \widetilde{S} \left(0, \frac{z_0}{\|z_0\|} \right), (w_0^*, z_0^*) \right \rangle \right|	\\
&\leq 	\|z_0\| \left\| \widetilde{S} \left(0, \frac{z_0}{\|z_0\|} \right) \right\| < \eps \|z_0\|.
\end{align*}
Using these inequalities, for all $\eps \in (0, K)$, we get that
\begin{align*}
1 &= | \langle \widetilde{S}(w_0, z_0), (w_0^*, z_0^*) \rangle| \\
&\leq |\langle \widetilde{S}(w_0, 0), (w_0^*, z_0^*) \rangle| + |\langle \widetilde{S}(0, z_0), (w_0^*, z_0^*) \rangle | \\
&< \|w_0\| + \eps \|z_0\| < \|w_0\| + K \|z_0\| \leq 1,
\end{align*}
which is a contradiction. So, $z_0 = 0$.

\vspace{0.3cm}

\noindent
Being $z_0 = 0$, we have that $(w_0, w_0^*) \in \Pi(W)$ and $| \langle \widetilde{S}(w_0, 0), (w_0^*, z_0^*) \rangle | = 1$. Let us show that $v(S)=1$ and that it is attained. Indeed, since $(w_0, w_0^*) \in \Pi(W)$, we have that
\begin{equation*}
	| \langle \widetilde{S}(w_0, 0), (w_0^*, b_0 z_0^*) \rangle | \leq v(\widetilde{S}) \leq 1.
\end{equation*}
Then,
\begin{align*}
1 = | \langle \widetilde{S}(w_0, 0), (w_0^*, z_0^*) \rangle|
&= \left| \left( 1 - \frac{1}{b_0} \right) \langle \widetilde{S}(w_0, 0), (w_0^*, 0) \rangle + \frac{1}{b_0} \langle \widetilde{S}(w_0, 0), (w_0^*, b_0 z_0^*) \rangle \right| \\
&\leq 	 \left( 1 - \frac{1}{b_0} \right) | \langle \widetilde{S}(w_0, 0), (w_0^*, 0) \rangle| + \frac{1}{b_0} |\langle \widetilde{S}(w_0, 0), (w_0^*, b_0 z_0^*) \rangle | \\
&= \left( 1 - \frac{1}{b_0} \right) |  w_0^*(\widetilde{S}_1(w_0, 0)) | + \frac{1}{b_0} |\langle \widetilde{S}(w_0, 0), (w_0^*, b_0 z_0^*) \rangle | \\
&= \left( 1 - \frac{1}{b_0} \right) | w_0^*(Sw_0)| + \frac{1}{b_0} |\langle \widetilde{S}(w_0, 0), (w_0^*, b_0 z_0^*) \rangle | \\
&\leq \left(1 - \frac{1}{b_0} \right) v(S) + \frac{1}{b_0} v(\widetilde{S}) \leq 1.
\end{align*}
This implies that $|w_0^*(Sw_0)| = v(S) = 1$.

\noindent
{\it Case 2:} Now assume that $\oplus_a$ is of type $\infty$. We will prove that $z_0^* = 0$. Suppose not. By Lemma \ref{charac}.b, there is $b_0 > 0$ such that $|(1, b_0)|_a = 1$. Then,
\begin{equation*}
\left\| \left(\frac{w_0}{\|w_0\|}, b_0 z_0 \right) \right\|_a \leq |(1, b_0)|_a = 1.
\end{equation*}
By (\ref{state1}), we have that
\begin{equation*}
\left \langle \left( \frac{w_0}{\|w_0\|}, b_0 z_0 \right), \left( \frac{w_0^*}{\|w_0^*\|} , 0 \right) \right \rangle = 1
\end{equation*}
So,
\begin{equation*}
\left\| \left(\frac{w_0}{\|w_0\|}, b_0 z_0 \right) \right\|_a = 1 \qquad \text{and} \qquad \left( \left( \frac{w_0}{\|w_0\|}, b_0 z_0 \right), \left( \frac{w_0^*}{\|w_0^*\|} , 0 \right) \right) \in \Pi(W \oplus_a Z).
\end{equation*}
This implies that
\begin{equation*}
\left| \left \langle \widetilde{S} \left( \frac{w_0}{\|w_0\|}, b_0 z_0 \right), \left( \frac{w_0^*}{\|w_0^*\|}, 0 \right) \right \rangle \right| \leq v(\widetilde{S}) = 1.
\end{equation*}

\noindent
Since $\oplus_{a^*}$ is of type $1$ (see Lemma \ref{charac}.d), there is $K > 0$ such that $\|w_0^*\| + K \|z_0^*\| \leq \|(w_0^*, z_0^*) \|_{a^*} = 1$. Set
\begin{equation*}
(w_0, z_0) = \left(1 - \frac{1}{b_0 \|w_0\|} \right) (w_0, 0) + \frac{1}{b_0 \|w_0\|} (w_0, b_0 \|w_0\| z_0).
\end{equation*}
Then,
\begin{align*}
1 &= | \langle \widetilde{S}(w_0, z_0), (w_0^*, z_0^*) \rangle | \\
&= | \langle \widetilde{S}(w_0, z_0), (w_0^*, 0) \rangle + \langle \widetilde{S}(w_0, z_0), (0, z_0^*) \rangle | \\
&\leq \left|  \left \langle \widetilde{S} \left( \left( 1 - \frac{1}{b_0 \|w_0\|} \right) (w_0, 0) + \frac{1}{b_0 \|w_0\|}  (w_0, b_0 \|w_0\| z_0) \right), (w_0^*, 0) \right \rangle \right| + |\langle \widetilde{S}(w_0, z_0), (0, z_0^*) \rangle | \\
&\leq \left( 1 - \frac{1}{b_0 \|w_0\|} \right) | \langle \widetilde{S} (w_0, 0), (w_0^*, 0) \rangle| + \frac{1}{b_0 \|w_0\|} | \langle \widetilde{S}(w_0, b_0 \|w_0\| z_0), (w_0^*, 0) \rangle| + |\langle \widetilde{S}(w_0, z_0), (0, z_0^*) \rangle |
\end{align*}
But we have that
\begin{align*}
\left( 1 - \frac{1}{b_0 \|w_0\|} \right) | \langle \widetilde{S} (w_0, 0), (w_0^*, 0) \rangle| &= \left( 1 - \frac{1}{b_0 \|w_0\|} \right) \|w_0^*\|\|w_0\| \left| \left\langle \widetilde{S} \left( \frac{w_0}{\|w_0\|}, 0\right), \left(\frac{w_0^*}{\|w_0^*\|}, 0 \right) \right\rangle \right| \\
&\leq\left( 1 - \frac{1}{b_0 \|w_0\|} \right) \|w_0^*\|
\end{align*}
and
\begin{align*}
\frac{1}{b_0 \|w_0\|} | \langle \widetilde{S}(w_0, b_0 \|w_0\| z_0), (w_0^*, 0) \rangle | &= \frac{1}{b_0 \|w_0\|} \|w_0^*\|\|w_0\| \left| \left \langle \widetilde{S} \left( \frac{w_0}{\|w_0\|}, \frac{b_0 \|w_0\| z_0}{\|w_0\|} \right), \left( \frac{w_0^*}{\|w_0^*\|}, 0 \right) \right \rangle \right| \\
&= \frac{1}{b_0 \|w_0\|} \|w_0^*\|\|w_0\| \left| \left \langle \widetilde{S} \left( \frac{w_0}{\|w_0\|}, b_0 z_0 \right), \left( \frac{w_0^*}{\|w_0^*\|}, 0 \right) \right \rangle \right| \\
&\leq	\frac{1}{b_0 \|w_0\|}\|w_0^*\| v(\widetilde{S}) = \frac{1}{b_0 \|w_0\|}\|w_0^*\|.
\end{align*}
Moreover, by using (\ref{ineq1}), we have that
\begin{equation*}
| \langle \widetilde{S}(w_0, z_0), (0, z_0^*) \rangle| = \|z_0^*\| 	\left| \left\langle \widetilde{S}(w_0, z_0), \left(0, \frac{z_0^*}{\|z_0^*\|} \right) \right\rangle \right| = \|z_0^*\| \left| \left(\frac{z_0^*}{\|z_0^*\|}\right) \widetilde{S}_2(w_0, z_0) \right|  \leq \|z_0^*\| \|\widetilde{S}_2\| < \eps \|z_0^*\|.
\end{equation*}
Therefore, for all $\eps \in (0, K)$, we have
\begin{align*}
1 &\leq \left( 1 - \frac{1}{b_0 \|w_0\|} \right) | \langle \widetilde{S} (w_0, 0), (w_0^*, 0) \rangle| + \frac{1}{b_0 \|w_0\|} | \langle \widetilde{S}(w_0, b_0 \|w_0\| z_0), (w_0^*, 0) \rangle| + |\langle \widetilde{S}(w_0, z_0), (0, z_0^*) \rangle | \\
&\leq \left(1 - \frac{1}{b_0 \|w_0\|} \right) \|w_0^*\| + \frac{1}{b_0 \|w_0\|} \|w_0^*\| + \eps \|z_0^*\|	\\
&= \|w_0^*\| + \eps \|z_0^*\| < \|w_0^*\| + K \|z_0^*\| \leq 1.
\end{align*}
This contradiction gives $z_0^* = 0$. So, $(w_0, w_0^*) \in \Pi(W)$ and
\begin{equation*}
1 = | \langle \widetilde{S}(w_0, z_0), (w_0^*, 0) \rangle| = |w_0^*(\widetilde{S}_1(w_0, z_0))|.
\end{equation*}
Now, since $((w_0, b_0 z_0), (w_0^*, 0)) \in \Pi(W)$, we have that
\begin{align*}
1 &= | \langle \widetilde{S}(w_0, z_0), (w_0^*, 0) \rangle | \\
&\leq \left(1 - \frac{1}{b_0} \right) |\langle \widetilde{S}(w_0, 0), (w_0^*, 0) \rangle| + \frac{1}{b_0} | \langle \widetilde{S}(w_0, b_0 z_0), (w_0^*, 0) \rangle| \\
&\leq \left(1 - \frac{1}{b_0} \right) |w_0^*(Sw_0)| + \frac{1}{b_0} v(\widetilde{S}) \\
&\leq 	\left(1 - \frac{1}{b_0} \right) v(S) + \frac{1}{b_0} v(\widetilde{S}) \leq 1.
\end{align*}
So, $|w_0^*(Sw_0)| = v(S) = 1$ and this finishes the proof.
\end{proof}

As far as we know, Proposition \ref{NRA1} is new even for $L$- and $M$-summands.

\begin{cor} \label{NR1} Let $X$ be Banach space.
\begin{itemize}
\item[(a)] If $W$ is an $L$-summand of $X$ and $\overline{\NRA(X)} = \mathcal{L}(X)$, then $\overline{\NRA(W)} = \mathcal{L}(W)$
\item[(b)] If $W$ is an $M$-summand of $X$ and $\overline{\NRA(X)} = \mathcal{L}(X)$, then $\overline{\NRA(W)} = \mathcal{L}(W)$
\end{itemize}
\end{cor}

Adapting Proposition \ref{NRA1} to the compact case, we have the following result.

\begin{prop} \label{NRAK}  Let $X$ be a Banach space and let $W$ be an absolute summand of $X$ of type 1 or $\infty$. If $\NRA(X) \cap \mathcal{K}(X)$ is dense in $\mathcal{K}(X)$, then $\NRA(W) \cap \mathcal{K}(W)$ is dense in $\mathcal{K}(W)$.
\end{prop}

Let us finally discuss on the validity of reciprocal results. In \cite[Remark 2.5]{KLM1} that the sets $\NRA(C[0, 1] \oplus_1 L_1[0, 1])$ and $\NRA(C[0, 1] \oplus_\infty L_1[0, 1])$ are not dense, respectively, in $\mathcal{L}(C[0, 1] \oplus_1 L_1[0, 1])$ and $\mathcal{L}(C[0, 1] \oplus_\infty L_1[0, 1])$ was observed. As both $C[0,1]$ and $L_1[0,1]$ have the BPBp for numerical radius \cite{AGR,KLM1}, this shows that there is no possible valid reciprocal results for Theorem \ref{numericalradius1} and Proposition \ref{NRA1}. On the other hand,  the sets $\NRA(C[0, 1] \oplus_1 L_1[0, 1]) \cap \mathcal{K}(C[0, 1] \oplus_{1} L_1[0, 1])$ and $\NRA(C[0, 1] \oplus_{\infty} L_1[0, 1]) \cap \mathcal{K}(C[0, 1] \oplus_{\infty} L_1[0, 1])$ are dense in $\mathcal{K}(C[0, 1] \oplus_1 L_1[0, 1])$ and $\mathcal{K}(C[0, 1] \oplus_{\infty} L_1[0, 1])$, respectively (see \cite[Example 3.4]{CapMM}). Nevertheless, we do not know if there is some reciprocal result for the BPBp-nu for compact operators.

config: arxiv


TITLE: Define $f(x)=mx+b$ for all $x$. Prove that the function $f:\mathbb{R}\rightarrow\mathbb{R}$ is uniformly continuous.
QUESTION [0 upvotes]: Define $f(x)=mx+b$ for all $x$. Prove that the function $f:\mathbb{R}\rightarrow\mathbb{R}$ is uniformly continuous.

Let $\{u_n\}$ and $\{v_n\}$ be sequences of $\mathbb{R}$ and converges to $x_0$, then we have $$\lim\limits_{n\rightarrow\infty} (u_n-v_n)=\lim\limits_{n\rightarrow\infty} u_n-\lim\limits_{n\rightarrow\infty} v_n = x_0-x_0=0$$ next we have: $$\lim\limits_{n\rightarrow\infty} (f(u_n)-f(v_n))=\lim\limits_{n\rightarrow\infty} f(u_n)-\lim\limits_{n\rightarrow\infty} f(v_n)=(mx_0+b)-(mx_0+b)=0$$
Hence, $f(x)$ is continuous.

Can anyone check my solution? I am not sure that I have show enough information, I just follow the definition. Thanks

REPLY [2 votes]: You've already shown continuity. I'll stick to your definition. Let $(u_n)$ and $(v_n)$ be real sequences with $\lim_{n\to\infty} \vert u_n-v_n \vert=0$. We have:
$$\vert f(u_n)-f(v_n)\vert=\vert mu_n+b-mv_n-b\vert = \vert m(u_n-v_n) \vert \to 0$$
as $n\to\infty$.


TITLE: Complete orthonormal system in $L_2(a,b)$
QUESTION [2 upvotes]: I want to prove that if $(u_j)_{j=1}^\infty$ is a complete orthonormal system in $L_2(a,b)$, then $(\overline{u_j})_{j=1}^\infty$ is also a complete orthonormal system in $L_2(a,b)$. Any ideas on how to approach this proof?
Proving $(\overline{u_j},\overline{u_i})=\delta_{ij}$ seems to be simple:
$$(\overline{u_j},\overline{u_i})=\int_a^b\overline{u_j}(t)\overline{\overline{u_i}}(t)dt=\int_a^bu_i(t)\overline{u_j}(t)dt=(u_i,u_j)=\delta_{ij}.$$
Any ideas on how to prove completeness?

REPLY [2 votes]: Suppose that $(f,\bar{u}_j)=0$ for all $j$. Then by definition
$$ \int_a^bf(x)u_j(x)\;dx=0 $$
for all $j$, and taking the complex conjugate of the integral we get
$$ 0=\overline{\int_a^bf(x)u_j(x)\;dx}=\int_a^b\overline{f(x)}\overline{u_j(x)}\;dx=(\bar{f},u_j) $$
for all $j$.
Since $\{u_j\}$ is by hypothesis an orthonormal basis, this implies that $\bar{f}=0$ almost everywhere, hence $f=0$ almost everywhere. Therefore $\{\bar{u}_j\}$ is an orthonormal basis as well.


TITLE: Discrete Maths - Range of function
QUESTION [0 upvotes]: I'm currently studying CS, and as i didn't do maths A level i'm finding the module particularly difficult. We've now changed topics and lecturer, going onto discrete maths; and i'm refusing to fall behind :P. So, i'm going to post regularly/daily questions, just to make sure i have an understanding.
Hopefully some of you guys have the time to give detailed answers to give me some sort of foundation
Question - Range of a Functon
Let $X = \{a,b,c,d\}$ and $Y = \{1,2,3,4,5\}$ and define $f:X\to Y$ by $f(a) =1$, $f(b) =2$, $f(c) = 5$, $f(d) = 2$.
Find the domain, codomain and range. If someone could explain this question in detail so i can do some revision on it, i'd be grateful. Thanks

REPLY [1 votes]: By definition the domain of $f$ is the set of all inputs for which $f$ is defined; in this case that’s $\{a,b,c,d\}=X$. This is actually implicit in the notation $f:X\to Y$, which almost always implies that that the domain of $f$ is $X$. (I say almost because in some areas of mathematics one deals with so-called partial functions from $X$ to $Y$, whose domains may not be all of $X$. I would not worry about this: it should not come up in what you’re doing.)
The codomain can also be read straight from the notation $f:X\to Y$: it’s the target set $Y$, which here is $\{1,2,3,4,5\}$. The range is always a subset of the codomain: it’s the set of values that the function actually assumes (or if you prefer — and in CS you might! — outputs). For your function $f$ those values are $1,2$, and $5$, so the range of $f$ is the set $\{1,2,5\}$.
That’s all there is to it.

REPLY [0 votes]: Domain is a set on which valid inputs to $f$ are defined. Since $f$ is defined on $a,b,c,d$, the domain is the entire $X$.
Range is a set to which $f$ maps the input -- in other words, all possible outputs of $f$. Here, $\{1,2,5\}$. Sometimes range is called image.
Codomain is the bigger set where every output must be defined, here it is $Y$
For another example consider $f:\mathbb{R} \to \mathbb{R}$ given by $f(x) = 1/x^2$.

The domain are all possible inputs, so $0$ is not a part of it since $1/0$ is undefined. So the domain are all non-zero reals.
The codomain is where $f$ maps, here is defined as $\mathbb{R}$.
The image or range of $f$ is the actual outputs of $f$, here all positive reals.


TITLE: Equivalent Expressions
QUESTION [0 upvotes]: Basically I'm trying to get this expression:
$${(-1)^nn(n+1)\over2} + (-1)^{(n+1)}(n+1)^2$$
In this form:
$$(-1)^{(n+1)}(n+1)(n+2)\over2$$
This is for a proof using mathematical induction, and I'm 99% sure that they DO equal each other. For some reason I'm having trouble doing this. What I've tried is setting 2 as the common denominator, and then factoring out a $(n+1)$ term and a $(-1)^{(n+1)}$ term.
After doing these steps, I ended up with:
$$(-1)^{(n+1)}(n+1)(2n+1)\over2$$
I'm not sure if I'm making a simple error, missing something, or if these two expressions are not equal to each other (but I seriously think that they are, because of the nature of the assignment that this question is from). Sometimes the simplest solutions are so hard to find.

REPLY [3 votes]: Try this:
$$
\frac{(-1)^{n}n(n+1)}{2}+(-1)^{n+1}(n+1)^2 = (-1)^n\bigg(\frac{n(n+1)}{2}-\frac{2(n+1)^2}{2}\bigg)\\
=\frac{(-1)^n(n+1)}{2}(n-2(n+1))\\
=\frac{(-1)^{n+1}(n+1)(n+2)}{2}
$$


TITLE: Do mirrors have a "resolution" quality to them?
QUESTION [8 upvotes]: I would like to know if mirrors have a quality of "resolution" to them like a regular photograph might, or like a JPEG does.
For example, if you looked to closely, or magnified a photograph, you would soon come to a point where the image blurs, or that which was too far away when photographed is not discernible.
If a high quality microscope was applied to a mirror, would it be possible to observe far-away objects reflected by the mirror, taking only the quality and scope of the microscope into account?
Are there different quality mirrors, all other things (e.g. shape) being the same?
Could you also talk about any loss of luminescence if possible?
Also, do all plane mirror have the same converging/diverging power i.e, the image size will be same on all plane mirrors ?

REPLY [0 votes]: In response to your last question, all plane mirrors have the same converging/diverging power: zero.  The size of the virtual image in a plane mirror is identical to the the size of the object.  If you hold up a mirror and look at the reflection, in it, of the moon, for example, you are looking at a virtual image that is some 3500 km across, and is about 385000 km away, behind the mirror.  If you use an auto focus camera to take a picture of this image, the camera will focus on infinity and create a in-focus photo of the moon.  If you force the camera to focus on the surface of the mirror, you will get an in-focus picture of dust, scratches, pits, finger smudges, etc on the surface.  There will be no image of the moon in the shot...


TITLE: Is this integral representation of $\zeta(2n+1)$ known?
QUESTION [43 upvotes]: Background: I'm an undergraduate at an institution with no researchers in analytic number theory, and no ties to the analytic number theory community.  I believe I have found what is, as far as I can tell after some googling, a new family of integral representations of $\zeta(2n+1)$.  I was told I should post this here by an algebraic number theorist at my university, to see if it was a known result or not.
Feel free to look at Equations (4), (5), and (7), to see if you recognize them before reading the entire text.  Most of this text is for the special case of $\zeta(3)$, but all results generalize quite readily to the case of $\zeta(2n+1)$, which I discuss at the end.
Beginning of setup
As a preliminary step, we will split the sum $\zeta(3)$ into even and odd parts. Observe that $$
\sum_{n=1}^\infty \frac{1}{(2n)^3} = \frac{1}{8} \sum_{n=1}^\infty \frac{1}{n^3} = \frac{1}{8}\zeta(3)
$$
Therefore $$
\sum_{n=1}^\infty \frac{1}{(2n+1)^3} = \frac{7}{8}\zeta(3) \tag{1}
$$
We now write $$
\frac{1}{(2n+1)^3} = \left(\int_0^1 x^{2n} dx\right)^3 = \int_0^1 \int_0^1 \int_0^1 x^{2n} y^{2n} z^{2n} \tag{2} dx dy dz
$$
and plug this triple integral into Equation (1), obtaining $$
\zeta(3) = \frac{8}{7} \sum_{n=0}^\infty \left[ \int_0^1 \int_0^1 \int_0^1 x^{2n} y^{2n} z^{2n} dxdydz \right]
$$
Using the absolute convergence of the geometric series on $(0,1)$, we can interchange the sum and integrals, obtaining $$
\zeta(3) = \frac{8}{7} \int_0^1 \int_0^1 \int_0^1 \frac{1}{1-x^2y^2z^2} dxdydz \tag{3}
$$
End of setup
With this part out of the way, I will now describe the integral representations that form the meat of my question.
Consider the $u$-substitution for the integral given by Equation (3) $$
x = \frac{\sinh{u}}{\cosh{v}}, \qquad y = \frac{\sinh{v}}{\cosh{w}}, \qquad z = \frac{\sinh{w}}{\cosh{u}}
$$
Some computation shows that $$
dxdydz = [1-(\tanh{u}\tanh{v}\tanh{w})^2]dudvdw
$$
But dividing both sides by $1-(\tanh{u}\tanh{v}\tanh{w})^2$ and rewriting the left-hand side in terms of $(x,y,z)$ gives $$
\frac{1}{1-x^2y^2z^2}dxdydz = dudvdw
$$
After changing limits, the transformed integral then reads $$
\zeta(3) = \frac{8}{7} \int_0^\infty \int_0^{\sinh^{-1}(\cosh(w))} \int_0^{\sinh^{-1}(\cosh(v))} dudvdw \\
= \frac{8}{7} \int_0^\infty \int_0^{\sinh^{-1}(\cosh(w))} \sinh^{-1}(\cosh(v)) dv dw \tag{4}
$$
But the most interesting part of the above is the following generalization. Let $f : (0,\infty) \to (0,\infty)$ be a function satisfying the following three properties.

$f$ is surjective, with $\lim\limits_{x\to 0} f(x) = 0$ and $\lim\limits_{x \to \infty} f(x) = \infty$
$f$ is invertible
$f$ is differentiable

Now consider the $u$-substitution $$
x = \frac{\sinh{f(u)}}{\cosh{f(v)}}, \qquad y = \frac{\sinh{f(v)}}{\cosh{f(w)}}, \qquad z = \frac{\sinh{f(w)}}{\cosh{f(u)}}
$$
We then find that $$
dxdydz = f'(u)f'(v)f'(w)[1-(\tanh{f(u)}\tanh{f(v)}\tanh{f(w)})^2]dudvdw
$$
and hence $$
\zeta(3) = \frac{8}{7} \int_0^\infty \int_0^{g(w)} \int_0^{g(v)} f'(u)f'(v)f'(w) dudvdw \tag{5}
$$
where $g(x) = f^{-1}(\sinh^{-1}(\cosh(f(x))))$.
All of the above can be generalized to $\zeta(2n+1)$.  Equation (3) becomes $$
\zeta(2n+1) = \frac{2^{2n+1}}{2^{2n+1}-1} \int_0^1 \int_0^1 \cdots \int_0^1 \frac{1}{1-x_1^2 x_2^2 \cdots x_{2n+1}^2} dx_1 dx_2 \cdots dx_{2n+1} \tag{6}
$$
Our $u$-substitution becomes (where $u_{(2n+1)+1} = u_1$) $$
x_i = \frac{\sinh(f(u_i))}{\cosh(f(u_{i+1}))}
$$
which transforms Equation (6) to $$
\zeta(2n+1) = \frac{2^{2n+1}}{2^{2n+1}-1} \int_0^\infty \int_0^{g(u_{2n})} \int_0^{g(u_{2n-1})} \cdots \int_0^{g(u_1)} f'(u_1)f'(u_2)\cdots f'(u_{2n+1}) du_1 du_2 \cdots du_{2n+1} \tag{7}
$$
My questions are then:

Are Equations (4), (5), and (7) known results?
Equation (4) is a volume integral. Exploring the region of integration in Mathematica numerically, it looks like an octant of a hyperbolic cube with vertices at infinity. Is this in fact the case?

REPLY [23 votes]: As indicated in my comment, some of these integrals are essentially known, and involve the hyperbolic "Beukers-Kolk-Calabi" change of variables.
In particular, in this paper, Z. Silagadze shows in (27) that
\begin{equation*}
 \zeta(n) = \frac{2^n}{2^n-1}\int_0^1\cdots\int_0^1 \frac{dx_1\cdots dx_n}{1-x_1^2\cdots x_n^2},
\end{equation*}
which clearly yields (7) above.
The hyperbolic change of variables is used later in that paper to obtain a related formula over the $2n$-simplex $\Delta_{2n}$, namely
\begin{equation*}
  \zeta(2n+1) = -\frac{1}{2n}\frac{2^{2n+1}}{2^{2n+1}-1}\int_{\Delta_{2n}} \log(\tan x_1 \tan x_2\ldots \tan x_{2n})dx_1\cdots dx_{2n}.
\end{equation*}
This integral is amenable to further interesting modifications.
EDIT:

The arXiv paper cited above has been published in the journal "Resonance"
Another paper by the same author (Z. Silagadze) comments on these integrals.
The "Beukers-Kolk-Calabi" change of variables was also considered in this paper.
This paper (cited below by Benjamin Dickman) is also relevant here.
The relation of these representations to Amoebas is quite interesting, and worthy of further exploration.

REPLY [8 votes]: I recommend this recent preprint "Sums of generalized harmonic series for kids from five to fifteen" by Z.K. Silagadze. The author explains identities for $\zeta(n)$ that come from your equation (6) by using hyperbolic function substitutions. It is pointed out that the Beukers-Kolk-Calabi substitution and its hyperbolic analogue can be explained by relating them to the amoeba's of a certain Laurent polynomial (the regions you have been computing in mathematica). Unfortunately I can't retell that part of the story beyond $\zeta(2)$, but this is somewhat discussed in the last section.


TITLE: Riemann Zeta function's analytic continuation relationship with simple closed formulas
QUESTION [6 upvotes]: Consider the famous output of the analytic continuation of the Riemann zeta function at negative values :
$$
\zeta(-1) = 1+2+3+4... \hspace{3mm} =  \dfrac{-1}{12}
$$
Now consider the closed formula which Gauss discovered for the sum of the first $n$ integers($i,n \in \mathbb{N}$):
$$
\sum_{i=1}^{n}i = 1+2+3+4...+n =  \dfrac{n(n+1)}{2} = \dfrac{n^2}{2} + \dfrac{n}{2}
$$
We Find the roots of the closed form ( using my special analytic Theorem which is based on the principal : " I decide the function is continuous now ") :
$$
n_1 = -1
$$
$$
n_2 = 0
$$
Now we find the right to left definite integral between the two roots (by leaning on the same analytic method I described above):
$$
\int_{0}^{-1} \dfrac{n^2}{2}+\dfrac{n}{2} dn  =  \dfrac{n^3}{6}+\dfrac{n^2}{4}\Biggr|_{0}^{-1} = 0 - \left(\dfrac{-1}{6}+\dfrac{1}{4}\right) = \dfrac{-1}{12}
$$
I can't prove it for all negative values of $\zeta$, I have verified the same relationship is true for $\zeta(-3)$ and the closed form of $n^3$ which integrates into $\dfrac{1}{120}$ if we use my "special analytic method".
Now on a more serious note, can anybody please explain this to me? When I look at the functional equation for the analytic continuation of the zeta function, it seems like an impenetrable fortress, yet it turns out it is possible to produce these results with simple (illogical) assumptions regarding continuity of discrete functions?

REPLY [5 votes]: The general formula for the partial sums of the $n^{th}$ powers of integers can be written as
$$\displaystyle S(n,x)=\sum_{i=1}^{x} i^n
=\frac{(-1)^n B_{n+1}}{n+1 } \\+ \frac{ \sum_{j=0}^{n+1} \,B_{n+1 - j} \,  \binom {n+1}{j} (x+1)^j    }{n+1} \tag{*}$$
where $B_z$ indicates the $z^{th}$ Bernoulli number. Substituting $n$ with integer values in (*), we get the classical formulas
$$ S(1,x)=\sum_{i=1}^{x} i^1=\frac 12 x(x+1)$$
$$ S(2,x)=\sum_{i=1}^{x} i^2=\frac 16 x(2x^2+3x+1)$$
$$ S(3,x)=\sum_{i=1}^{x} i^3=\frac 14 x^2(x+1)^2\\...$$
and so on. The first term of the RHS in (*) corresponds to the explicit value of the zeta function for negative integers, given by
$${\displaystyle \zeta (-n)={\frac{(-1)^n \,B_{n+1}}{n+1}}}$$
So we have
$$\displaystyle S(n,x)
=\zeta(-n) \\ + \frac{ \sum_{j=0}^{n+1} \,B_{n+1 - j} \,  \binom {n+1}{j} (x+1)^j    }{n+1} \tag{**}$$
Now we can take the definite integral. To avoid confusion, I assume that the definite integral described in the OP corresponds to that calculated in the interval from $-1$ to $0$ (or equivalently, to the opposite value of that calculated in the range from $0$ to $-1$). Accordingly, this is what has been done in the OP to obtain $$-\int_0^{-1} \left(\frac{x^2}{2}- \frac x2 \right) dx \\ =-(-\frac 16 + \frac 14 )=-\frac{1}{12}$$
Taking the definite integral, the equation (**) reduces to
$$\displaystyle \int_{-1}^{0} S(n,x)
=\zeta(-n) \\ + \sum_{j=0}^{n+1} \frac{ B_{n+1-j} \, \binom {n+1}{j}   }{(n+1)(j+1)}$$
So the hypothesis proposed in the OP, that is to say
$$\displaystyle \int_{-1}^{0} S(n,x)
=\zeta(-n) $$
is true for any $n$ satisfying
$$ \sum_{j=0}^{n+1} \frac{ B_{n+1-j}  \binom {n+1}{j}  }{(n+1)(j+1)}=0$$
or equivalently
$$ \sum_{j=0}^{n+1} \frac{ \,B_{n+1-j}  \binom {n+1}{n+1-j}  }{j+1}=0$$
Now we can make the substititions $N=n+1$ and $K=n+1-j$ to get
$$ \sum_{K=0}^{N} \frac{ \,B_{K}  \binom {N}{K}  }{(N-K+1)}=0$$
Multiplying both terms of the fraction to $N+1$, we get
$$ \sum_{K=0}^{N} \frac{ \,B_{K} (N+1)N! }{(N+1)K!(N-K)!(N-K+1)}\\
=\sum_{K=0}^{N} \frac{ \,B_{K} (N+1)! }{(N+1)K!(N-K+1)!}\\
= \frac{1}{N+1} \sum_{K=0}^{N}  \,B_{K}  \binom {N+1}{K} =0$$
which is true for any positive integer $N$ since a well known property of Bernoulli numbers is
$$\sum_{K=0}^{M-1}  \,B_{K}  \binom {M}{K} =0$$


TITLE: Crystal Field Theory
QUESTION [1 upvotes]: I am literally lost with this question:

Suppose that within the set of (2L+1)(2S+1) lowest-lying ionic states the crystal field can be represented in the form a(L_x)^2 + b(L_y)^2 + c(L_z)^2, with a, b, and c all different. Show in the special case L = 1 that if the crystal field is the dominant perturbation (compared with the spin-orbit coupling), then it will yield a (2S+1)-fold degenerate set of ground states in which every matrix element of every component of L vanishes.

REPLY [1 votes]: The physical context would have been helpful. I am guessing that this is some sort of once ionized atom, whose electron is in an L=1 state, sitting inside an atomic crystal, which breaks the rotational symmetry by the crystal axes.
First, the S is a decoupled electron spin, which is completely irrelevant. You can ignore the 2S+1 factors and all the misdirecting jawboning about spin in the question. This is a reformulation:
You have a quantum mechanical system with angular momentum 1. This means that there are 3 different rotated versions of the state, with $L_z=-1,0,+1$, which have exactly the same energy, they are degenerate, because the world is rotationally invariant.
Now you put this system inside a crystal that breaks the rotational symmetry, and the leading perturbation from the crystal between these three states and no other, is
$$ a L_x^2 + b L_y^2 + cL_z^2 $$
The operators $L_x$,$L_y$ and $L_z$ are specific 3 by 3 matrices acting on these three states. You probably were given their form from the raising and lowering operators for z-angular momentum: $L_x - iL_y$, $L_x + iL_y$.

Find the new ground state,
show that all the "matrix elements" of each component of L vanish in this ground state.

The first part is easy diagonalization of the 3 by 3 matrix given to you above. This is clearly a homework exercize, so you should do it yourself. It is straightforward.
The second depends on the intended interpretation of the word "matrix elements": after comments by jojo, I learn that the "matrix elements" are just supposed to be the expected value of the $L_x$, $L_y$ an $L_z$ between the ground state copies with different spin. This is a terrible abuse of the term "matrix elements", because these matrix elements are just the expectation value of $L_x$,$L_y$ and $L_z$ in the ground state, the spin is completely decoupled. So all you need to show is that the expected value of $L_x$,$L_y$ and $L_z$ are zero.
I misinterpreted the wording initially to mean something impossible, namely:

Show that the matrix elements between the ground states and the other $L_z$ states are all zero, for each of the three operators $L_x,L_y,L_z$.

There are no states of L=1 system for which the matrix elements of each component of L vanishes. Such a state would be a rotationally invariant vector, which does not exist--- the only rotationally invariant objects are scalars.
Because the actual wording is potentially misleading, this problem is not very good. It is better to say explicitly: "Show that all the matrix elements between the components of L between the different ground states is zero."


text:
\section{Formulating Graph Polynomials}
\label{sec:formulations}

We have seen two methods for formulating a graph polynomial with the
linear recursion (deletion/contraction) and generating function
definitions of the Tutte polynomial in the previous chapter.  Here we
will see several more.  We begin with one of the earliest graph
polynomials, the edge-difference polynomial, a multivariable
polynomial defined as a product and originally studied by
Sylvester~\cite{Syl78} and Peterson~\cite{Pet91} in the late 1800s.
More recently, it has been used to address list coloring questions
(see Alon and Tarsi~\cite{AT92} and Ellingham and Goddyn~\cite{EG96}),
where a list coloring of a graph is a proper coloring of the vertices
of a graph with the color of each vertex selected from a predetermined
list of colors assigned to that vertex.

\begin{definition}\label{edgediff}
\index{edge-difference polynomial}
\index{polynomial!edge-difference}
The edge-difference polynomial.
Let $\left( {v_1 , \ldots ,v_n } \right)$  be an ordering of the
vertices of a graph $G$.  Then $D\left( {G;x_1 , \ldots ,x_n } \right)
= \prod_{i < j} {( {x_i  - x_j } )} $, where the
product is over all edges $\left( {v_i ,v_j } \right)$  of G.
\end{definition}

Note that a proper coloring of $G$ corresponds to finding positive
integer values $N_i $  (not necessarily distinct) for each of the
$x_i$'s so that $D\left( {G;N_1 , \ldots ,N_n } \right)$ $\ne$ 0.

There are also several polynomials based on various determinants (or
even permanents; see Pathasarthy~\cite{Par89} for a survey) involving
the adjacency matrix of a graph.  Recall that $A(G)$, the adjacency
matrix of a graph, has entries $a_{ij}  = 1$ if $\left( {i,j} \right)$
is an edge of the graph and 0 if it is not. The characteristic
polynomial is the classic example of a such a graph polynomial, and
will be discussed further in Section~\ref{sec:polynomials}.

\begin{definition}\label{adjpolys} The characteristic polynomial. Let
  $A(G)$ be the adjacency matrix of a graph $G$. Then
  \index{characteristic polynomial}\index{polynomial!characteristic}
         $f( {G;x} ) = |xI-A(G)|$.
\end{definition}

 Other examples of such polynomials are the \emph{idiosyncratic polynomial}
 \index{idiosyncratic polynomial}\index{polynomial!idiosyncratic}
 introduced by   Tutte, see~\cite{Tut79}, that is defined by
 $\nu ( {G;x,y} ) = |A(G) + y(J-I-A) - xI |$, where $J$ is  the matrix
 having all  entries equal to 1. Also
 $\mu (G;x,y ) = | xI-D(G)+A(G) |$ a polynomial introduced in
 \cite{Kel65},  where $D(G)$ is  the \emph{degree matrix} of  $G$ that
 is the diagonal matrix with  $\operatorname{deg}(i)$ in the position
  $(i,i)$. Note that $J-I-A$ is the adjacency matrix of the complement
 of $G$ and when   $G$ is a simple graph, $D(G)-A(G)$ is just
 the Laplacian matrix $L(G)$ of the  graph.

A number of important graph polynomials may be defined by state model
formulas. Loosely speaking, a state of a graph \index{graph state} is
some configuration resulting from making local assignments for
substructures (e.g. the edges or vertices) of the graph.  These
assignments may be, for example, associating an element of a given set
to each vertex, or even the result of reconfiguring the edges incident
with a vertex.  A graph polynomial is formed by associating an
expression, often a weighted monomial, to each state of the graph, and
then summing over all possible graph states. The language comes from
physics, and is also found in knot theory.  We will see several state
model graph polynomials among those surveyed below, as well as an
application of this method in the Potts model of statistical mechanics
in Section~\ref{sec:applications}.

An early example of a graph polynomial given by a state model
formulation \index{state model} is $P(G;x)$, the \emph{Penrose
polynomial}.\index{Penrose polynomial}\index{polynomial!Penrose}  This
polynomial graph invariant for planar graphs was defined implicitly by
Penrose~\cite{Pen69} in the context of tensor diagrams in physics, but
an excellent graph theoretical exposition can be found in
Aigner~\cite{Aig97}.  To compute $P(G;x)$, let $G$ be a plane graph,
and let $G_m $ be its medial graph, face two-colored with the
unbounded face colored white.  At each vertex, we consider three
possible local reconfigurations, as in Figure~\ref{Fig:vertexfig}.
A state $S$ of $G_m$ then results from choosing one of these three
reconfigurations at each vertex of $G_m$ and consists of a set of
disjoint closed curves (like a knot diagram).  Furthermore, to each
local reconfiguration at a vertex $v$, we assign a weight $\omega
(S,v)$ that is $+1$, 0, or $-1$ for a white, black, or crossing
configuration, respectively.
    \begin{figure}[hbtp]
      \begin{center}
       \includegraphics{vertexfig.pdf}
        \caption{The three possible local reconfigurations at a vertex
        $v$, identified, from left to right, as white, black, and
        crossing.  Which strand passes over which in the crossing
        configuration does not affect the
        computation}\label{Fig:vertexfig}
      \end{center}
      \end{figure}

\begin{definition}\label{Penrose} The Penrose polynomial.  Let $G$ be
  a planar graph with medial graph $G_m$, and let $St(G_m)$ be the set
  of states of $G_m$ and let $St'(G_m)$ be the set of states with no
  black configurations.  Then,
\[
P(G;x) = \sum\limits_{St(G_m )}^{} {\left( {\left( {\prod\limits_{v
            \in G_m } {\omega (S,v)} } \right)x^{k(S)} } \right)}  =
         \sum\limits_{St'(G_m )}^{}
            {\left( {\left( { - 1} \right)^{cr\left( S\right)}
            x^{k(S)} } \right)},
\]
where $k(S)$ is the number of components in the graph state $S$, and
$cr(S)$ is
the number of crossing vertex configurations chosen in the state
$S$.
\end{definition}

For example, if $G$ is the $\theta$-graph consisting of two vertices
joined by three edges in parallel,
 then $P(G;x)=x^3-3x^2+2x$, as in~Figure~\ref{Fig:Penrosefig}. The
 Penrose polynomial may also be computed via a linear recursion
 relation (see Jaeger~\cite{Jae90} for example).
    \begin{figure}[hbtp]
      \begin{center}
       \includegraphics[width=0.9\textwidth]{Penrosefig.pdf}
        \caption{Computing the Penrose polynomial of a graph $G$ from
          the states of its medial graph}\label{Fig:Penrosefig}
      \end{center}
      \end{figure}

The Penrose polynomial has some surprising properties, particularly
with respect to graph coloring.  The Four Color Theorem is equivalent
to showing that every planar, cubic, connected graph can be properly
edge-colored with three colors.  The Penrose polynomial, when applied
to planar, cubic, connected graphs, encodes exactly this information
(see Penrose~\cite{Pen69}):
\[
P\left( {G;3} \right) = \left( {\frac{{ - 1}}{4}}
\right)^{\frac{{\left| V \right|}}{2}} P\left( {G; - 2} \right) =
\text{the number of edge-3-colorings of } G.
\]


TITLE: If rref$([A|b])=[B|d]$, find $B$ and $d$.
QUESTION [1 upvotes]: Let $A$ be a $3\times3$ matrix and let the complete solution of $Ax=b$ be $x=\begin{bmatrix}
4\\
0\\
0
\end{bmatrix}+$ $\alpha\begin{bmatrix}
2\\
1\\
0
\end{bmatrix}+$ $\beta\begin{bmatrix}
5\\
0\\
1
\end{bmatrix}$, where $\alpha$ and $\beta$ are any two scalars.
If rref$([A|b])=[B|d]$. Find $B$ and $d$.
The first part of this question asks for a basis of $N(A)$ and from the complete solution, it is obvious what a basis of $N(A)$ is. How does one go about using the particular and special solutions to find what the question is asking for, i.e., $B$ and $d$?

REPLY [1 votes]: If $v_1+\alpha v_2 + \beta v_3$ is the solution to $Ax=b$ with the vectors linearly independent, then $v_1$ and $v_2$ form a basis for the null space.  The key insight is that $A$ and $\operatorname{rref}(A)$ have the same null space.  So $\operatorname{rref}(A)$ must have two free columns, and one pivot column.  This forces there to be only one non-zero row, and it must be orthogonal to both $v_2$ and $v_3$.  This will tell you what $B$ must be.  Then, because the row reduction does not change the solution space, we must have that $d=Bv_1$.
The task, now, is to unravel this paragraph and fill in the details.


TITLE: Proving linearity of the differential operator $a_n(x)D^n+a_{n-1}(x)D^{n-1}+\ldots+a_1(x)D+a_0(x)$.
QUESTION [1 upvotes]: I am taking a first course on ordinary differential equations(with background in linear algebra). I would like to know, if the proof to the below claim is correct (and that I am not using advanced facts to prove basic facts).

Prove that every expression of the form:
$$a_n(x)D^n+a_{n-1}(x)D^{n-1}+\ldots+a_1(x)D+a_0(x)$$
defines a linear transformation from $C^n[a,b]$ to $C[a,b]$ whenever $a_0(x),\ldots,a_n(x)$ are continuous on $[a,b]$.

Proof.
(1) Consider $\frac{d^n}{dx^n}(y) = h$. Any function $y$ that satisfies this equation must have atleast $n$ derivatives.
In other words, $D^n$ sends a function $f$ in $C^n[a,b]$ to $h$ in $C[a,b]$.
(2) From high-school calculus, if $u,v$ functions $n$ times differentiable,
$\frac{d^n}{dx^n}(u + v)=\frac{d^n}{dx^n}(u)+\frac{d^n}{dx^n}(v)$
$\frac{d^n}{dx^n}(\alpha u)=\alpha \frac{d^n}{dx^n}(u)$
Therefore, $I,D,D^2,\ldots,D^n$ are linear maps. In fact, $a_0(x)I, a_1(x)D, a_2(x)D^2,\ldots,a_n(x)D^n$ are also linear transformations.
(3) Since, $C^n[a,b]$ and $C[a,b]$ are finite-dimensional vector spaces, the set of all such linear transformations from $\{T:C^n[a,b]\rightarrow C[a,b]\}$ is a vector space. Vector addition is defined in the usual way functions are added point-wise. $(S+T)(v)=S(v)+T(v)$. Scalar multiplication is defined by multiplying the scalar with each of the components. $(\alpha T)(v)=\alpha\cdot T(v)$.
Additivity.
Let $T = a_n(x)D^n + a_{n-1}D^{n-1}+\ldots+a_1(x)D + a_0(x)$.
\begin{align}
T(f+g)=
&[a_n(x)D^n + a_{n-1}D^{n-1}+\ldots+a_1(x)D + a_0(x)](f+g) \\
= &a_n(x)D^n (f+g) + a_{n-1}D^{n-1}(f+g)+\ldots+a_1(x)D(f+g) + a_0(x)(f+g) \hspace{3mm}...(S+T)(v)=S(v)+T(v)\\
= &a_n(x)D^n (f) + a_{n-1}D^{n-1}(f)+\ldots+a_1(x)D(f) + a_0(x)(f) \\
+ &a_n(x)D^n (g) + a_{n-1}D^{n-1}(g)+\ldots+a_1(x)D(g) + a_0(x)(g)  \hspace{5mm} ...D^n(u+v)=D^n(u)+D^n(v)\\
=& T(f) + T(g)
\end{align}
Homogeneity.
\begin{align}
T(\alpha f) &= [a_n(x)D^n + a_{n-1}D^{n-1}+\ldots+a_1(x)D + a_0(x)](\alpha f) \\
& = a_n(x)D^n(\alpha f) + a_{n-1}D^{n-1}(\alpha f)+\ldots+a_1(x)D(\alpha f) + a_0(x)(\alpha f)\\
& = \alpha a_n(x)D^n (f)+ \alpha a_{n-1}D^{n-1} (f) +\ldots + \alpha a_1(x)D(f) + \alpha a_0(x)(f)\\
& = \alpha T(f)
\end{align}
This completes the proof.
Q.E.D

REPLY [1 votes]: Yes this proof is correct. One thing to note is that the vector spaces in question are very much not finite dimensional. Luckily, that doesn't impact your proof in any meaningful way.
Part (3) can also be shortened a lot by just observing that sums and multiples of linear maps are always linear. That removes a lot of complicated looking symbols.


TITLE: Finding bounds for recurrence relation
QUESTION [0 upvotes]: Prove that for the recuurence relation $f(n) = 2f(n-1) + f(n-2)$
with the starting values: f(1) = 1, f(2) = 6, $\forall n \ge 2$
$$(7/3)^n \le f(n) \le (5/2)^n$$
I tried using the fact that $f(n) \gt 2f(n-1)$

REPLY [0 votes]: For $n = 1$ you get $7/3 \le 1 \le 5/2$. This is false. So the claim is wrong.
If you want bounds, try looking for an upper bound of the form $f(n) \le a_2 \cdot b^n$, substitute into the recurrence and see what happens:
$\begin{align*}
   f(n)
     &=    2 f(n - 1) + f(n - 2) \\
     &\le  2 a_2 b^{n - 1} + a_2 b^{n - 2} \\
     &=    a_2 b^{n - 2} (2 b + 1)
\end{align*}$
We want $a_2 b^n \le a_2 b^{n - 2} (2 b + 1)$, i.e. $b^2 = 2 b + 1$, positive zero is $1 + \sqrt{2} < 5/2$. Adjust $a_2$ to satisfy the bound for the initial values, i.e. $1 \le a_2 (1 + \sqrt{2})$ and $6 \le a_2 (1 + \sqrt{2})^2 = a_2 (3 + 2 \sqrt{2})$. The first gives $a_2 \ge \sqrt{2} - 1$, the second one $a_2 \ge 18 - 12 \sqrt{2}$. The later is larger, a bit more than 1. Thus we can say $f(n) \le (5/2)^n$
Do the same, but looking for a lower bound. As above, for $f(1) = 1$ you get $a_1 b < 1$, $a_1 < 2/5$, bound $b < 12/5$, giving $2/5 \cdot (12/5)^n < f(n)$; it would be simpler to just say $(2/5)^n \le f(n)$.
The bounds can be checked by induction.


TITLE: Analytic function with arbitrary, countable zero set
QUESTION [1 upvotes]: Let $S\subset\mathbb{C}$ be a countable set with no accumulation points. Does there exist an analytic function $f(z)$ such that $f(z_i)=0$ for all $z_i\in S$ and $f(z)\neq 0$ for $z\neq S$?
Work so far: If we consider analytic to be equivalent to infinitely differentiable (at least for complex functions) then the analogous case has an affirmative answer for real functions. Basically, chain smooth transition functions together such that the construction is zero on the desired points non-zero everywhere else. For an example, see here for a function that is zero at $0$ and $3$ using the above construction. Since there are no accumulation points among the zeros, you are always assured that you can stretch and shrink these functions to your desired fit. The difficulty I am having is that this construction is not analytic, and does not extend to complex functions.

REPLY [1 votes]: Below is Theorem 4.1 of Section 4 (Weierstrass infinite products) of Complex Analysis by Stein and Shakarchi:

Given any sequence $\{a_n\}$ of complex numbers with $|a_n|\to\infty$, there exists an entire function $f$ that vanishes at all $z=a_n$ and nowhere else. Any other such entire function is of the form $f(z)e^{g(z)}$ where $g$ is entire.

Since your $S$ is countable, you can write them as a sequence. Moreover, since $S$ has no accumulation points, in any finite disk there are only finitely many points of $S$. You can rearrange your prescribed zeros so that they form a sequence that tends to $\infty$. The theorem is thus applicable.