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\title{Profile decompositions for critical\\ Lebesgue and Besov space embeddings}
\author{{\sc Gabriel S. Koch}\\ \small Oxford University\\ \small {\em koch@maths.ox.ac.uk}}
\maketitle
\begin{abstract}
Profile decompositions for ``critical" Sobolev-type embeddings are established, allowing one to regain some compactness despite the non-compact nature of the embeddings.  Such decompositions have wide applications to the regularity theory of nonlinear partial differential equations, and have typically been established for spaces with Hilbert structure.  Following the method of S. Jaffard, we treat settings of spaces with only Banach structure by use of wavelet bases.  This has particular applications to the regularity theory of the Navier-Stokes equations, where many natural settings are non-Hilbertian.
\end{abstract}
\ \\

In this article, we characterize the causes for the lack of compactness in certain embeddings of Banach spaces.  In particular, one can re-write a bounded sequence in such a way that it is obvious why there fails to be a convergent subsequence (even in the weaker space).  Once the defects of compactness are thus identified, it is often possible to isolate them and regain some aspects of compactness.  \\

This useful tool for re-writing bounded sequences is known as a ``decomposition into profiles" (typically after passing to a subsequence), the ``profiles" being the typical obstacle to compactness -- specifically, norm-invariant transformations of fixed non-zero elements of the space.  Since these elements are fixed (for the whole sequence) one can think of them as ``limits" which can replace the need for an actual convergent subsequence.  This is particularly useful in the study of nonlinear partial differential equations, where the natural setting is often that of such Banach spaces, and the convergence of certain sequences corresponds to establishing the existence of certain solutions to the equation being considered.  In fact, it is primarily for this purpose that such decompositions have historically been established.\\

As we have just intimated, such a program is not new.  Patrick G\' erard \cite{gerard} was probably the first to do this and treated the embedding $\dot H^s(\rd) \hookrightarrow L^q(\rd)$ for $s>0$ and an appropriate $q>1$.  It is typical that such decompositions are established for spaces of functions defined on $\rd$, since the invariance of $\rd$ under translations and dilations leads to the typical defect of compactness of the embedding (see below). Later, St\' ephane Jaffard \cite{jaffard} extended the work of Gerard (although with slightly less specific results) to $\dot H^{s,p}(\rd) \hookrightarrow L^q(\rd)$ for $s>0$ and appropriate $p,q >1$.  His proof was somewhat technically different from \cite{gerard}, and he was able to move to the non-Hilbertian setting (i.e., spaces not based on $L^2$) by use of wavelet bases.  There have of course been many other works on profile decompositions, see for example \cite{BG1,bmm,bc,km1,keraani,keraani2,mv}, to name only a few. \\

In this article we follow the method of Jaffard to prove similar decompositions for the embedding $L^p \hookrightarrow \bfin$ for $p\geq 2$, where $\bfin$ is any one of a family of Besov spaces for $r,q>p$ with a particular negative index of regularity $s_{p,r}$.  Of course, in general these spaces are non-Hilbertian and are therefore well suited to the method described in \cite{jaffard}.  As an indication of the generality of the method (and with specific applications in mind which we'll describe momentarily), we show how similar results may be obtained as well for embeddings of the type $\besa \hookrightarrow \besb$. We also clarify, correct and improve some of the arguments in \cite{jaffard}, for example showing that a compactly supported wavelet basis is not needed and more carefully detailing the relationships between the ``orthogonal cubes" in the construction.\\

The method of establishing profile decompositions using wavelet bases is actually quite general and can be applied to many embeddings between spaces of Besov or Triebel-Lizorkin type, which we plan to present in a unified and hopefully simpler way in the upcoming work \cite{ck}.  Our motivation for studying the particular embeddings in this work now are their applications to the regularity theory of the Navier-Stokes equations, which we'll now proceed to describe.\\

In \cite{kk}, regularity results for Navier-Stokes were established based on the profile decomposition of Isabelle Gallagher \cite{gallagher} (based in turn on \cite{gerard}) in the setting of $\hdhalf(\rt)$.  Perhaps a more natural setting for these equations is the space $L^d(\rd)$, however in that work (and others, see e.g. \cite{sverakrusin}) the more special setting of $\hdhalf$ was investigated.  This is partly because much theory (such as \cite{gerard}) has been developed in the $L^2$ setting, and partly because the Hilbert structure of $L^2$-based spaces makes the analysis somewhat simpler.    In the non-Hilbertian setting, somewhat different methods are in general required. The results and methods presented below will make it possible to move to the more natural (non-Hilbertian) settings in the Navier-Stokes framework (see, e.g., the upcoming work \cite{gkp} where we shall give extensions of \cite{kk} and \cite{sverakrusin} in the more natural setting), and hopefully in the study of other PDEs as well for which $H^s$ is not the only interesting or natural setting.
\\

Navier-Stokes is of course not the only application of interest.  As we mentioned above, using profile decompositions has become quite a common approach in the study of many nonlinear partial differential equations in ``critical" function-space settings, see e.g. \cite{ckm,km1,km2,km3,kk,ks,mxz3} to name only a few, where the existence of a profile decomposition implies various regularity results for solutions of the PDE.  Until the recent work \cite{kk}, the methods had been applied exclusively to dispersive and hyperbolic PDE where the Hilbertian $H^s$ settings were the natural ones.  It was therefore natural to perform the analysis of the parabolic Navier-Stokes equations in \cite{kk} in a similar setting, namely $\hdhalf(\rt)$, and moreover until now the more general $L^p$ theory was not available.  The application of the results in this paper to Navier-Stokes in the natural setting of $L^d(\rd)$ will be addressed soon in \cite{gkp}, in the same spirit as \cite{kk}.  This will give a different proof of the well-known and difficult result in \cite{ess} (see also \cite{dongdu}), that the $L^3(\rt)$ norm of a solution to Navier-Stokes must become infinite near a singularity (the result of \cite{kk} treated the special case of blow-up of the stronger $\hdhalf$ norm).
\\

It is interesting to note as well that profile decompositions (which we will describe below) can also be used (see \cite{gerard}) to recover the existence of minimizers in the classical Sobolev embeddings, a fact established by P. L. Lions \cite{lions3} in the '80s using his method of ``concentration-compactness".
\\

Let us finally outline now what a profile decomposition is.  The principle defect in the lack of compactness of many embeddings  $X \hookrightarrow Y$ of homogeneous spaces of functions defined on $\rd$ comes from the following example:  For some non-zero element $f\in X$ and $\alpha >0$, for any $n\in \N$ and $x\in \rd$ define
$$f_n(x):=\frac{1}{(\l_n)^\alpha}f \left(\frac{x-x_n}{\l_n}\right)\ ,$$
where $\l_n >0$ and $x_n \in \rd$.  The spaces and the scaling constant $\alpha$ will be such that $\|f_n\|_X \equiv \|f\|_X$ and $\|f_n\|_Y \equiv \|f\|_Y$ (when the scalings of the spaces coincide in this way we call the embedding ``critical" -- in the context of PDE this usually means moreover that the scaling is a natural one for the equation).  Moreover, $X$ and $Y$ are such that if either $\lim_{n\to\infty}\l_n \in \{0,\infty\}$ or $\l_n \equiv 1$ and $\lim_{n\to\infty}|x_n| = \infty$, then $f_n \rightharpoonup 0$ in $X$ and $Y$.  Then one has $\sup_{n\in\N}\|f_n\|_X = \|f\|_X < \infty$, yet clearly one cannot have any subsequence which is strongly convergent (necessarily to zero) in $Y$ since $\|f_n\|_Y \equiv \|f\|_Y >0$.  A profile decomposition for bounded sequences in $X$ is a way to show that the lack of compactness of such embeddings comes in general from linear combinations of the above type of ``profile" example.  Moreover, the profiles do not interact with each other in ways which might ``cancel" the defect of compactness caused by each one.
\\
\pagebreak

The main example we study here is the following: Suppose $2\leq p < q,r \leq \infty$ and set $s_{p,r}:= d(\frac{1}{r} - \frac{1}{p}) < 0$.  Then $\lprd \hookrightarrow \brq$ and we have the following profile decomposition for bounded sequences in $\lprd$:
\begin{thm}\label{thm:a}
Let $\{u_n\}_{n=1}^\infty$ be a bounded sequence in $\lprd$.  There exists a sequence of profiles $\{\phi_l\}_{l=1}^\infty \subset \lprd$ and for each $l\in \N$ a sequence $\{(j_n^l,k_n^l)\}_{n=1}^\infty \subset \Z \times \Z^d$, both depending on $\{u_n\}$, such that, after possibly passing to a subsequence in $n$,
\begin{equation}\label{profiles}
u_n(x) = \sum_{l=1}^L (2^{j^l_n})^{{d/p}} \phi_l(2^{j^l_n}x - k^l_n) + r_n^L(x)
\end{equation}
for any $L \in \N$ where the following properties hold:
\begin{enumerate}
\item For $l\neq l'$, the sequences $\{(j_n^l,k_n^l)\}$ and $\{(j_n^{l'},k_n^{l'})\}$ are orthogonal in the following sense:
\begin{equation}\label{orth1}
\left|\log \left(2^{(j_n^l - j_n^{l'})}\right)\right| + \left|2^{(j_n^l - j_n^{l'})}k_n^{l'} - k_n^l\right|  \xrightarrow[n\to\infty]{} +\infty
\end{equation}
\item The remainder $r_n^L$ satisfies the following smallness condition:
\begin{equation}\label{orth2}
\lim_{L\to\infty} \left(\varlimsup_{n\to \infty} \|r_n^L\|_\brq \right) = 0
\end{equation}
\item There is a norm $\| \cdot \tilde \|_\lprd$ which is equivalent to $\|\cdot \|_\lprd$ such that for each $n \in \N$,
\begin{equation}\label{orth3}
\sum_{l=1}^\infty \|(2^{j^l_n})^{{d/p}} \phi_l(2^{j^l_n}x - k^l_n) \tn_\lprd^p \leq \varliminf_{n'\to\infty} \|u_{n'}\tn_\lprd^p
\end{equation}
and for any $L \in \N$,
\begin{equation}\label{orth4}
 \|r_n^L\tn_\lprd \leq \|u_n\tn_\lprd + \circ(1) \quad \textrm{as} \quad n\to\infty\ .
\end{equation}
\end{enumerate}
\end{thm}

The orthogonality property (\ref{orth1}) implies exactly that the interaction between the profiles $\{\phi_l\}$ (after the norm-invariant transformations) becomes negligible as $n\to\infty$ (see e.g. \cite{gallagher,kk} as well as discussions below).
Properties (\ref{orth3}) and (\ref{orth4}) can be thought of as ``stability properties" of the profile decomposition in terms of the sequences from which they are derived.  We also comment here that the norm introduced in (\ref{orth3}) - (\ref{orth4}) may not have the same transformational invariance as $\| \cdot \|_\lprd$, but (\ref{orth3}) does imply that there exists some universal constant $C>0$ such that
\begin{equation}\label{altorth}
\sum_{l=1}^\infty \|\phi_l \|_\lprd^p \leq C \varliminf_{n\to\infty} \|u_{n}\|_\lprd^p
\end{equation}
due to the invariance of the usual $\lprd$ norm under the transformations on $\phi_l$ in (\ref{orth3}).  One could alternatively define a new $\tilde{\phi}_l:=(2^{j^l_1})^{{d/p}} \phi_l(2^{j^l_1}\cdot - k^l_1)$ (and adjust $\{(j_n^l,k_n^l)\}$ in the obvious way) to obtain (\ref{altorth}) with the norm $\| \cdot \tn_\lprd$ and $C=1$ (plus all the previous properties), which gives a more standard and useful property (see e.g. \cite{gkp}).  With minimal adjustments, a similar theorem about bounded sequences in Besov norms (which also has applications to Navier-Stokes, see \cite{gkp}) is also seen to be true, see Theorem \ref{thm:aa}.
\\

As mentioned above, the method of proof we will use follows that of \cite{jaffard}, using the ``wavelet" norms $\|\cdot \tn_\lprd$, $\| \cdot \tn_\brq$, etc., which are based on the expansion of functions in all spaces in {\em one} unconditional wavelet basis (and are equivalent to the standard norms).  Roughly, one extracts and categorizes the ``largest" wavelet components of $u_n$ in an exhaustive manner, which decreases the norms of $u_n$ in a ``largest" space $Z$ (consisting simply of functions with bounded wavelet coefficients) while the norms in the ``smallest" space $X$ (e.g., $\lprd$) remain bounded.  In fact, the remainder must tend to zero in $Z$ due to properties of ``nonlinear (wavelet) projections".  Then an interpolation inequality of the form $\|f\|_Y \lesssim \|f\|_X^\alpha \|f\|_Z^{1-\alpha}$ for some ``middle" space $Y$ (e.g. $Y=\brq$ and $X \hookrightarrow Y \hookrightarrow Z$) completes the main part of the proof.\\

{\bf Acknowledgements:}
\quad
The author would like to thank Professors Albert Cohen, Isabelle Gallagher and St\' ephane Jaffard for many helpful discussions.  \\

This work was supported by the EPSRC Science and Innovation award to the Oxford Centre for Nonlinear PDE (EP/E035027/1).
\section{Preliminaries}
\noindent
We'll start with the following facts regarding wavelet bases (see e.g. \cite{daubechies}, \cite{meyer2}; see also \cite{bf} and \cite{lr} for divergence-free wavelet bases):
\\\\
Fix any $m\in \N$.  Denote
$$\l = (i,j,k) \in \{1, \dots , 2^d -1\} \times \Z \times \Z^d =: \LL\ .$$
There exists a real-valued set of functions\footnote{In fact, one can take $\varphi \in \mathcal{C}^m_0$, but we do not presently require compact support of the basis functions.} $\{ \varphi^{(i)} \}_{1\leq i \leq 2^d -1} \subset \mathcal{C}^m (\R^d) \cup (\cap_{p>1}L^p(\R^d))$ such that $\varphi_\l = \varphi^{(i)}_{j,k}$ defined by $\varphi_\l(x) = 2^\frac{jd}{2} \varphi^{(i)}(2^j x -k)$ is an orthonormal basis of $L^2$:
$$f\in L^2 (\R^d) \iff f= \sum_{\l \in \LL} c_\l \varphi_\l, \qquad \textrm{where} \quad c_\l = c^{(i)}_{j,k} := \int_{\R^d} \varphi_\l \cdot f\ .$$
Moreover, we have the following equivalences of norms for other function spaces:
\begin{equation}\label{a}
\|f\|_{L^p(\R^d)} \simeq \|f\tilde\|_{L^p(\R^d)} := \left\| \left\{ \sum_{\l \in \LL} |c_{j,k}^{(i)}|^2 2^{dj} \chi_\d(x) \right\}^\frac{1}{2}\right\|_{L^p(\R^d)}
\end{equation}
for $1< p < \infty$ where $\chi_\d$ is the indicator function of the
cube\footnote{Note that these cubes correspond roughly to the ``supports" of the $\psi_\l$s.}
$\d = \d(\l)  =\d^{j,k}= {\{x \in \R^d \ | \ 2^j x -k \in [0,1)^d \}}$ (note that $|\d^{j,k}| = 2^{-dj}$), and
\begin{equation}\label{b}
\|f\|_{\dot B^s_{a,b}} \simeq \|f\tilde\|_{\dot B^s_{a,b}} := \left\| 2^{j(s + d(\frac{1}{2} - \frac{1}{a}))} \|c^{(i)}_{j,k}\|_{\ell^a_{i,k}} \right\|_{\ell^b_j}
\end{equation}
for $1 \leq a,b \leq \infty$ so long as $|s| < m$.  Here and in what follows we use the notation $A(f) \simeq B(f)$ whenever there exists $c>0$ independent of $f$ such that $c^{-1} A(f) \leq B(f) \leq c A(f)$ and $A(f) \lesssim B(f)$ if at least the first inequality holds.
\\\\
In particular, we have
$$\|f\tilde \|_{\dot B^{-{d/p}}_{\infty,\infty}} = \sup_{\l \in \LL}  2^{jd(\frac{1}{2} - \frac{1}{p})} |c^{(i)}_{j,k}| $$
when $m>d$, which motivates us to define $a^{(i)}_{j,k} := 2^{jd(\frac{1}{2} - \frac{1}{p})} c^{(i)}_{j,k}$ and $\psi_\l$ by $\psi_\l(x) = 2^{jd(\frac{1}{p} - \frac{1}{2})} \varphi_\l (x)$, so that for a given $f$ one may write $f=\sum_\l c_\l \varphi_\l = \sum_\l a_\l \psi_\l$.  With these definitions, we have the equivalence
\begin{equation}\label{c}
\|f\|_{\dot B^s_{a,b}} \simeq \|f\tilde\|_{\dot B^s_{a,b}} = \left\| 2^{j(s + d(\frac{1}{p} - \frac{1}{a}))} \|a^{(i)}_{j,k}\|_{\ell^a_{i,k}} \right\|_{\ell^b_j}
\end{equation}
so that in particular
\begin{equation}\label{d}
\|f\|_{\dot B^{-{d/p}}_{\infty,\infty}} \simeq \|f\tilde \|_{\dot B^{-{d/p}}_{\infty,\infty}}=  \sup_{\l \in \LL} |a^{(i)}_{j,k}|\ ,
\end{equation}
and
\begin{equation}\label{e}
\|f\|_{L^p(\R^d)} \simeq \|f\tilde\|_{L^p(\R^d)} = \left\| \left\{ \sum_{\l \in \LL} |a_{j,k}^{(i)}|^2 2^{\frac{2d}{p}j} \chi_\d(x) \right\}^\frac{1}{2}\right\|_{L^p(\R^d)}\ .
\end{equation}
Note that for any $(i,j,k) \in \LL$, by (\ref{e}) we have
\begin{equation}\label{f}
|a^{(i)}_{j,k}|^p =  \int_{\rd} \left\{ |a_{j,k}^{(i)}|^2 2^{\frac{2d}{p}j} \chi_\d(x) \right\}^\frac{p}{2}\ dx \leq \|f\tilde\|_\lprd^p\ .
\end{equation}
Hence $\lprd \hookrightarrow \dot B^{-{d/p}}_{\infty,\infty}$, and for $\{u_n\} \subset \lprd$ with $u_n = \sum_\l a_{\l,n}\psi_\l$ we see that
\begin{equation}\label{k}
\sup_n \|u_n\tilde \|_\lprd \leq C \quad \Longrightarrow \quad \sup_n \|u_n\tn_\binf = \sup_{\l,n}|a_{\l,n}| \leq C\ ,
\end{equation}
i.e. boundedness in $\lprd$ implies uniform boundedness of the wavelet coefficients by (\ref{d}).
\\\\
We'll need the following two facts regarding the wavelet coefficients of elements of $\lprd$:
\begin{prop}\label{prop1}
Let $\{a_\l\}$ correspond to some $f\in \lprd$.  Then there exists some $\l_0 \in \LL$ such that $\sup_{\l \in \LL} |a_\l| = |a_{\l_0}|$.
\end{prop}
\noindent
The proposition is a simple consequence of (\ref{e}), but for the convenience of the reader we give a proof in the Appendix, see Proposition \ref{prop:a1}.
\\\\
For the second fact, we will need the following definition of ``nonlinear projections":
\\\\
For any $u\in \binf$ and $N \in \N$, we'll say $P_N(u) \in \bar{P}_N(u)$ if
$P_N(u) = \sum_{m=1}^N a_{\l_m} \psi_{\l_m}$ for some ordering $\{\l_m\}_{m=1}^\infty$ of $\LL$ such that
\begin{equation}\label{g}
u = \sum_{m=1}^\infty a_{\l_m} \psi_{\l_m} \qquad \textrm{and} \qquad |a_{\l_m}| \geq |a_{\l_{m+1}}| \ \ \forall m\geq 1\ .
\end{equation}
Note that (\ref{g}) is always possible by Proposition \ref{prop1}.  We'll say that $Q_N(u) \in \bar{Q}_N(u)$ if $Q_N(u) = u-P_N(u)$ for some $P_N(u) \in \bar{P}_N(u)$.  $P_N$ is a ``nonlinear projection'' onto the $N$ ``largest wavelet components'' of a function, and $Q_N$ is the remainder after such a projection.
\begin{lemma}\label{lemma1}
There exists $C=C(p,d) >0$ such that for any $M>0$ and $N\in \N$,
$$\sup_{\scriptsize \begin{array}{c} \|u\|_\lp \leq M,\\ \phantom{sss}Q_N(u) \in \bar{Q}_N(u) \end{array}} \|Q_N(u)\|_\binf \leq C \frac{M}{N^{{1/p}}}\ .$$
\end{lemma}
\noindent
\textbf{Proof:} \quad Fix any $u\in \lprd$ with $\|u\|_\lp \leq M$, fix $\{\l_m\}$ satisfying (\ref{g}) and set $Q_N(u) = \sum_{m=N+1}^\infty a_{\l_m} \psi_{\l_m}$.  Note that, by (\ref{d}) and (\ref{g}),
\begin{equation}\label{h}
\|Q_N(u)\|_\binf \simeq |a_{\l_{N+1}}| \leq \min_{1\leq m \leq N} |a_{\l_m}|\ .
\end{equation}
By (\ref{a1}) of the Appendix we know $\lp \hookrightarrow \dot B^0_{p,p}$, so by (\ref{c}) there exists $C = C(p,d) >0$ such that
\begin{equation}\label{i}
\|a_\l \|_{\ell^p(\LL)} \leq C \|u\|_\lprd
\end{equation}
Now combining (\ref{h}) and (\ref{i}) we see that
$$\left( N \cdot \|Q_N(u)\|^p_\binf \right)^\frac{1}{p} \lesssim \left( \sum_{m=1}^N |a_{\l_m}|^p \right)^\frac{1}{p} \leq
\|a_\l \|_{\ell^p(\LL)}\leq C \|u\|_\lprd \leq CM
$$
and the lemma is proved. \hfill $\Box$
\\\\
We remark here that if one uses the embedding $\dot B^{s_{p,a}}_{a,q} \hookrightarrow \dot B^{s_{p,b}}_{b,b}$ with $b:=\max \{a,q\}$ which is a simple application of Bernstein's inequalities (\ref{aac}) and properties of $\ell^r$ spaces, one easily deduces the similar inequality
\begin{equation}\label{decay2}
\sup_{\scriptsize \begin{array}{c} \|u\|_\besa \leq M,\\ \phantom{sss}Q_N(u) \in \bar{Q}_N(u) \end{array}} \|Q_N(u)\|_\binf \leq C \frac{M}{N^{{1/b}}}\ .
\end{equation}
\section{Main Results}
\noindent
Fix $p,q,r \in \R$ such that $2\leq p < q,r \leq \infty$ and set $s_{p,r}:= d(\frac{1}{r} - \frac{1}{p})$.
Then by Proposition \ref{prop:a3} of the Appendix, there exists $\alpha \in (0,1)$ and $C = C(d,p,q,r) >0$ such that
\begin{equation}\label{j}
\|u\|_\bfin \leq C \|u\|^\alpha_\lprd \|u\|^{1-\alpha}_\binf
\end{equation}
for any $u\in \lprd$.  Note that $\lprd \hookrightarrow \bfin \hookrightarrow \binf$ (see Appendix A).
\\\\
\textbf{Proof of Theorem \ref{thm:a}:}
\\\\
Suppose $\|u_n \|_\lprd \leq \bar{C}$, $\|u_n \tn_\lprd \leq \tilde{C}$, for all $n\in \N$ for some $\bar{C},\tilde{C}>0$, and write ${u_n = \sum_{\l \in \LL} a_{\l,n} \psi_\l}$.  Our first step will be to design an iterative process for describing the sequence $\{u_n\}$ which will allow us to establish (\ref{profiles}) - (\ref{orth4}) of the theorem.
\subsection{Extraction/iteration procedure}
\noindent
\ \\
{({\sc Iterate 0:})
\\\\
Recalling (\ref{k}), if $\lim_{ n\to \infty} \|u_n\|_\binf =0$, then by (\ref{j}) we can write $u_n = r_n^0$ with $\lim_{n\to \infty} \|r_n^0\|_\bfin = 0$, and the statement of the theorem follows.
\\\\
We suppose therefore that $\lim_{n\to\infty}\|u_n\|_\binf \neq 0$, and move to the first iterate:
\\\\
\underline{{\sc Iterate 1:}}
\\\\
By Proposition \ref{prop1}, for each $n\in \N$, there exists $\l_n^1 \in \LL$ such that
$$|a_{\l_n^1}|:=|a_{\l_n^1,n}|  = \max_{\l \in \LL} |a_{\l,n}|\ .$$
By (\ref{d}) and our assumption on $\|u_n\|_\binf$, we may pass to a subsequence such that, for some fixed $i_1 \in \{1, \dots , 2^d -1\}$ and $a_1 \neq 0$,
\begin{equation}\label{l}
\l_n^1 = (i_1, j_n^1,k_n^1) \quad \forall \ n \in \N \qquad \textrm{and} \qquad a_{\l_n^1} \xrightarrow[n\to\infty]{} a_1\ .
\end{equation}
For what follows, let us fix some notation for convenience:
\\\\
If $\tau(x) = ax - b = \frac{x-B}{A}$ for some $a,A \in \R \backslash \{0\}$, $b,B \in \rd$, denote
\begin{equation}\label{m}
(\tau f)(x):=a^{{d/p}} f(\tau(x)) \qquad \textrm{and} \qquad |\tau|:=|\log A| + |B|\ .
\end{equation}
Note that
\begin{equation}\label{n}
\tau_1 (\tau_2 f) = (\tau_2 \circ \tau_1)f
\end{equation}
for any such $\tau_1, \tau_2$.
\\\\
With this notation, we may write
$$u_n = \tau_{1,n} \left(a_{\l_n^1} \psi^{(i_1)} \right) + u_n^1$$
where $\tau_{1,n}(x) := 2^{j_n^1}x - k_n^1$, and $u_n^1$ is just $u_n$ minus one of its wavelet components (i.e., $a_{\l_n^1} \psi_{\l_n^1}$) whose coefficient has the largest possible modulus.  Therefore $\|u_n^1\tn_\binf \leq \|u_n \tn_\binf \leq \tilde{C}$ by (\ref{d}).
\\\\
Suppose $\lim_{n\to\infty}\|u_n^1\|_\binf = 0$.  Then we may write
$$u_n = \tau_{1,n}\left(a_1 \psi^{(i_1)}\right) + r_n^1$$
where $\lim_{n\to\infty}\|r_n^1\|_\bfin = 0$ by (\ref{j}) as follows:
\\\\
Write $r_n^1 = (a_{\l_n^1} - a_1)\psi_{\l_n^1} + u_n^1$, which is just $u_n$ after replacing the coefficient $a_{\l_n^1}$ of $\psi_{\l_n^1}$ by $a_{\l_n^1} - a_1$ which is small for large $n$ by (\ref{l}).  Using (\ref{d}) and (\ref{e}), we therefore see that
$$\|r_n^1 \tn_\binf \leq \|u_n^1 \tn_\binf \xrightarrow[n\to\infty]{} 0$$
and
$$\|r_n^1 \tn_\lprd \leq \|u_n^1 \tn_\lprd \leq \tilde{C}$$
for large $n$.  The statement now follows from (\ref{j}).
\\\\
We have now shown that the statement of the theorem follows when $\lim_{n\to\infty}\|u_n^1\|_\binf = 0$.  We suppose therefore that ${\lim_{n\to\infty}\|u_n^1\|_\binf \neq 0}$, and move to the next iterate:
\\\\
{\sc\underline{Iterate 2:}}
\\\\
By Proposition \ref{prop1} applied to $u_n^1$, for each $n\in \N$, there exists $\l_n^2 \in \LL \backslash \{\l_n^1\}$ such that
$$|a_{\l_n^2}| := |a_{\l_n^2,n}|= \max_{\l \in \LL \backslash \{\l_n^1\}} |a_{\l,n}|\ .$$
By (\ref{d}) and our assumption on $\|u_n^1\|_\binf$, we may pass to a subsequence such that, for some fixed $i_2 \in \{1, \dots , 2^d -1\}$ and $a_2 \neq 0$,
\begin{equation}\label{o}
\l_n^2 = (i_2, j_n^2,k_n^2) \quad \forall \ n \in \N \qquad \textrm{and} \qquad a_{\l_n^2} \xrightarrow[n\to\infty]{} a_2\ .
\end{equation}
Setting $\tau_{2,n}(x):=2^{j_n^2}x - k_n^2$, we can write
$$u_n = \tau_{1,n} \left(a_{\l_n^1} \psi^{(i_1)} \right) + \tau_{2,n} \left(a_{\l_n^2} \psi^{(i_2)} \right) +u_n^2\ .$$
Suppose that $\lim_{n\to\infty}\|u_n^2\|_\binf = 0$.  Let $\tau_n^{(2,1)}:= \tau_{2,n} \circ \tau_{1,n}^{-1}$ and note the following two possibilities (recall (\ref{m})):
\begin{itemize}
\item[(a)] $|\tau_n^{(2,1)}| \to +\infty$ as $n\to\infty$, i.e., ``$\tau_{n,1}$ and $\tau_{n,2}$ are orthogonal'' and
$$u_n = \tau_{1,n} \left(a_1 \psi^{(i_1)} \right) + \tau_{2,n} \left(a_2 \psi^{(i_2)} \right) +r_n^2$$
with $\lim_{n\to\infty}\|r_n^2\|_\bfin = 0$ by (\ref{j}), (\ref{l}) and (\ref{o}) in a way similar to the estimate for $r_n^1$ in {\em Iterate 1}. Note that this means the profiles $a_1 \psi^{(i_1)}$ and  $a_2 \psi^{(i_2)}$ ``live far" (in distance of supports or in scale) from each other after the transformations $\tau_{1,n}$ and $\tau_{2,n}$ respectively, and hence any interactions between the transformed profiles become negligible as $n\to\infty$.
\item[(b)] After possibly passing to a subsequence, $|\tau_n^{(2,1)}| \leq C < \infty$ for all $n$.  We calculate:
$$\tau_n^{(2,1)}(x) = \tau_{2,n}(2^{-j_n^1}(x + k_n^1)) = 2^{j_n^2}(2^{-j_n^1}(x+ k_n^1))-k_n^2 =$$$$= 2^{j_n^2-j_n^1} x - (k_n^2 - 2^{j_n^2-j_n^1} k_n^1) = \frac{x - (2^{j_n^1-j_n^2}k_n^2 - k_n^1)}{2^{j_n^1-j_n^2}}\ ,$$
so (\ref{m}) gives
$$|\log 2^{j_n^1-j_n^2}| + |2^{j_n^1-j_n^2}k_n^2 - k_n^1| \leq C\ .$$
Note therefore that $j_n^2 - j_n^1 \equiv j^{(2,1)}$ for infinitely many $n$ for some $j^{(2,1)} \in \Z$, and then $k_n^2 - 2^{j^{(2,1)}} k_n^1 \equiv b^{(2,1)}$ for infinitely many of those $n$ for some $b^{(2,1)} \in \rd$, due to the fact that we are working on lattices.  We may therefore pass to a subsequence so that
$$\tau_n^{(2,1)} \equiv \tau^{(2,1)}= \textrm{const} \quad \textrm{for all} \ n\in \N, $$
with
$$\tau^{(2,1)}(x) = 2^{j^{(2,1)}}x - b^{(2,1)}\ .$$
Recalling (\ref{n}), we may now write
$$
\begin{array}{rcl}
u_n & = & \tau_{1,n} \left(a_{\l_n^1} \psi^{(i_1)} \right) + \tau_{2,n} \left(a_{\l_n^2} \psi^{(i_2)} \right) +u_n^2\\\\
 & = & \tau_{1,n} \left( a_{\l_n^1} \psi^{(i_1)} + \tau_{1,n}^{-1} (\tau_{2,n} (a_{\l_n^2} \psi^{(i_2)})) \right) +u_n^2 \\\\
 & = & \tau_{1,n} \left( a_{\l_n^1} \psi^{(i_1)} + (\tau_{2,n} \circ \tau_{1,n}^{-1}) (a_{\l_n^2} \psi^{(i_2)}) \right) +u_n^2 \\\\
 & = & \tau_{1,n} \left( a_{\l_n^1} \psi^{(i_1)} + a_{\l_n^2} \tau^{(2,1)}  \psi^{(i_2)} \right) +u_n^2 \\\\
 & =: & \tau_{1,n} \left( a_1 \psi^{(i_1)} + a_2 \tau^{(2,1)}  \psi^{(i_2)} \right) +r_n^2\ .
\end{array}
$$
Comparing the last two lines, we can again show that ${\lim_{n\to\infty} \|r_n^2\|_\bfin = 0}$ in a similar way to {\em Iterate 1} using (\ref{j}), (\ref{l}) and (\ref{o}).
\end{itemize}
This concludes {\em Iterate 2}, and the statement of the theorem follows if ${\lim_{n\to\infty}\|u_n^2\|_\bfin = 0}$.  If not, we would move on to the next iterate.
\\\\
Let us however pause here (before giving the full induction argument for further iterates) to give some geometric meaning to this dichotomy.  For $m=1,2$, note that $|\d_n^m| = (2^{-j_n^m})^d$, and $\d_n^m$ (which corresponds to the location of $\psi_{\l_n^m}$) has $2^{-j_n^m}k_n^m$ as its ``center" (more precisely, the location of the ``corner'' located closest to the origin).
\\\\
Possibility (a) means that either
\begin{itemize}
\item[(i)] the sizes of the cubes $\d_n^1$ and $\d_n^2$ are incomparable as $n\to\infty$; that is,
$$\left|\log \frac{2^{-j_n^2}}{2^{-j_n^1}}\right| = | \log 2^{j_n^1 - j_n^2}| \xrightarrow[n\to\infty]{} \infty\ ,$$
or
\item[(ii)] they have comparable size $2^{-j_n^2} \simeq 2^{-j_n^1}$ for all $n$, and the distance between the cubes {\em relative to their size} becomes infinite as $n\to\infty$.  That is, after dilating space by, say, $2^{j_n^1}$, bringing their sizes to $O(1)$, their new centers ($k_n^1$ and $2^{j_n^1 - j_n^2}k_n^2$) separate:
    $$\left| \frac{2^{-j_n^2}k_n^2 - 2^{-j_n^1}k_n^1}{2^{-j_n^1}} \right| = |2^{j_n^1 - j_n^2}k_n^2 - k_n^1| \xrightarrow[n\to\infty]{} \infty\ .$$
\end{itemize}
In either case, we see that $\psi_{\l_n^1}$ and $\psi_{\l_n^2}$ have negligible interaction for large $n$.
\\\\
Possibility (b) means neither (i) nor (ii) happen, so the cubes have comparable size and {\em relative} (to their size) distance.  Note that $\tau_{m,n}\d_n^m = [0,1)^d$, and we are saying that for a subsequence, after rescaling to $O(1)$ by, say, $\tau_{n,1}$, the relative positions of the cubes are
constant:\footnote{Note that this was stated slightly inaccurately in \cite{jaffard} -- we hope that this explanation clarifies the matter.}}
$$\tau^{(2,1)}(\tau_{1,n} \d_n^2) = \tau_{1,n}\d_n^1\ .$$
This gives $\tau^{(2,1)} \tau_{1,n} = \tau_{2,n} \iff \tau^{(2,1)} = \tau_{2,n}\tau_{1,n}^{-1}$, as we had previously designated due to {\em algebraic} considerations.
\\\\
We will now continue to inductively define all possible iterates.  From now on, when we pass to a subsequence at the $N^{\textrm{th}}$ iterate, we will leave the first $N$ sequence elements unchanged, obtaining a ``diagonal'' subsequence which will work for all iterates.
\\\\
Suppose, after $N$ iterations, $\lim_{n\to\infty} \|u_n^{N'}\|_\binf \neq 0$ for all $N' < N$, and the conclusion of the $N^{\textrm{th}}$ iterate is as follows (and holds for all previous iterates):
\\\\
{\sc \underline{Iterate N:}}
\begin{equation}\label{p}
u_n = \sum_{l=1}^{L_N} \tau_{m_1(l),n} \left( \sum_{\mu = 1}^{M_N(l)} a_{\l_n^{m_\mu(l)}} \tau^{(m_\mu(l),m_1(l))} \psi^{(i_{m_\mu(l)})}  \right) + u_n^N
\end{equation}
where $\sum_{l=1}^{L_N} M_N(l) = N$, $m_\mu(l)$ are never repeated (i.e., $m_\mu(l) = m_{\mu'}(l') \Longrightarrow {\mu = \mu', l=l'}$), $i_{m_\mu(l)} \in \{1,\dots, 2^d -1\}$,
$$\tau_{m_\mu(l),n}(x) := 2^{j_n^{m_\mu(l)}}x - k_n^{m_\mu(l)}$$
and
$$\tau^{(m_\mu(l),m_1(l))}(x): = 2^{j^{(m_\mu(l),m_1(l))}}x - b^{(m_\mu(l),m_1(l))}$$
for some $j_n^{m_\mu(l)},j^{(m_\mu(l),m_1(l))} \in \Z$, $k_n^{m_\mu(l)}\in \Z^d$ and $b^{(m_\mu(l),m_1(l))} \in \rd$ such that the following hold:
\begin{enumerate}
\item For each $\mu$ and $l$,
$$\l_n^{m_\mu(l)} = (i_{m_\mu(l)}, j_n^{m_\mu(l)},k_n^{m_\mu(l)}) \quad \forall \ n \in \N$$
\item $j_n^{m_\mu(l)} - j_n^{m_1(l)} \equiv j^{(m_\mu(l),m_1(l))}= \textrm{const}$ for all $n\in \N$,
\item $k_n^{m_\mu(l)} - 2^{j^{(m_\mu(l),m_1(l))}}k_n^{m_1(l)} \equiv b^{(m_\mu(l),m_1(l))}= \textrm{const}$ for all $n\in \N$,
\item $\tau_{m_\mu(l),n} \tau_{m_1(l),n}^{-1} \equiv \tau^{(m_\mu(l),m_1(l))} = \textrm{const}$ for all $n\in \N$,\\\\ (4. is a consequence of 2. and 3.)
\item for any $l\neq l'$,
\begin{equation}\label{six}
|\tau_{m_1(l),n}\tau_{m_1(l'),n}^{-1}|  \xrightarrow[n\to\infty]{} +\infty\ ,
\end{equation}
\item $\displaystyle{|a_{\l_n^m}|:= |a_{\l_n^m,n}| = \max_{\l \in \LL \backslash \{\l_n^{m'}\}_{m'=1}^{m-1}} |a_{\l,n}|}$ for any $m \in \N$, and
\item there exists $a_{m_\mu(l)} \in \R \backslash \{0\}$ for each $\mu$ and $l$ such that
\begin{equation}\label{q}
a_{\l_n^{m_\mu(l)}} \xrightarrow[n\to\infty]{} a_{m_\mu(l)}\ .
\end{equation}
\end{enumerate}
Note that in {\em Iterate 2}, case (a) we had the above with $L_2 =2$, $M_2(1) = M_2(2) =1$, $m_1(1)=1$ and $m_1(2) =2$, and {\em Iterate 2} case (b) was $L_2 =1$, $M_2(1) =2$, $m_1(1) =1$, $m_2(1) =2$.
\\\\
If $\lim_{n\to\infty}\|u_n^N\|_\binf =0$, then the statement of the theorem follows as before by writing
\begin{equation}\label{r}
u_n = \sum_{l=1}^{L_N} \tau_{m_1(l),n} \left( \sum_{\mu = 1}^{M_N(l)} a_{{m_\mu(l)}} \tau^{(m_\mu(l),m_1(l))} \psi^{(i_{m_\mu(l)})}  \right) + r_n^N\ ,
\end{equation}
comparing (\ref{r}) with (\ref{p}) and estimating $r_n^N$ as in {\em Iterate 1} and {\em Iterate 2} using (\ref{j}) and (\ref{q}).
\\\\
Therefore we assume again that $\lim_{n\to\infty}\|u_n^N\|_\binf \neq 0$, and move to the next iterate:
\\\\
{\sc \underline{Iterate N+1:}}
\\\\
We pass to a subsequence so that
$$a_{\l_n^{N+1}} \xto a_{N+1} \neq 0$$
where $|a_{\l_n^{N+1}}|:=|a_{\l_n^{N+1},n}| = \|u_n^N \tn_\binf$ for each $n$ and
$$\l^{N+1}_n = (i_{N+1},j^{N+1}_n,k_n^{N+1})\in \LL$$
for all $n$ for some fixed $i_{N+1} \in \{1,\dots,2^d -1\}$.
\\\\
Let $\tau_{N+1,n}(x):= 2^{j^{N+1}_n}x-k_n^{N+1}$ and define $\tau_n^{(N+1,m_1(l))}:=\tau_{{N+1},n}\tau_{m_1(l),n}^{-1}$ for $l = 1, \dots, L_N$.  We then have the following two possibilities:
\begin{itemize}
\item[(a)]  $|\tau_n^{(N+1,m_1(l))}| \xto +\infty$ for all $l\in \{1,\dots,L_N\}$.  In this case, we set
$$L_{N+1} = L_N + 1, \quad M_{N+1}(L_{N+1}) = 1 \quad \textrm{and} \quad m_1(L_{N+1}) = N+1\ $$
which gives (\ref{p}) with $N$ replaced by $N+1$ with all the required properties.  That is, we start a new function in slot $L_{N}+1$ so that $u_n$ is a composed of rescalings of $L_N + 1$ functions (as opposed to $L_N$ functions as in {\em Iterate N}), the new function being $a_{\l_n^{N+1}} \psi^{(i_{N+1})}$.  (These functions become constant in $n$ after writing this in the form of (\ref{r}), the new function being $a_{N+1} \psi^{(i_{N+1})}$.)
\item[(b)]  There is some $\bar{l} \in \{1,\dots,L_N\}$ and $C>0$ such that, for some subsequence,
$$|\tau_n^{(N+1,m_1(\bar{l}))}| \leq C \quad \textrm{for all}\ n\in \N\ .$$
In this case, we pass to a subsequence so that
$$\tau_n^{(N+1,m_1(\bar{l}))} \equiv \tau^{(N+1,m_1(\bar{l}))} = \textrm{const} \quad \textrm{for all}\ n\in\N$$
with $\tau^{(N+1,m_1(\bar{l}))}$ defined in a way analogous to 2., 3. and 4. of ${\textit{Iterate N}}$ with $m_\mu(l)$ replaced by $N+1$ and $l$ replaced by $\bar{l}$ everywhere else.  We then set
$$L_{N+1} = L_N, \quad M_{N+1}(\bar{l}) = M_N(\bar{l}) + 1 \quad \textrm{and} \quad m_{M_{N+1}(\bar{l})} = N+1\ .$$
That is, we still have re-scalings of $L_N$ functions as in {\em Iterate N}, and we just add the new component, $a_{\l_n^{N+1}} \psi_{\l_n^{N+1}}$, to the $\bar{l}^{\textrm{th}}$ one so that the $\bar{l}^{\textrm{th}}$ term in the sum in (\ref{p}) becomes
$$\tau_{m_1(\bar{l}),n} \left( \left( \sum_{\mu = 1}^{M_N(\bar{l})} a_{\l_n^{m_\mu(\bar{l})}} \tau^{(m_\mu(\bar{l}),m_1(\bar{l}))} \psi^{(i_{m_\mu(\bar{l})})} \right)+ a_{\l_n^{N+1}} \tau^{(N+1,m_1(\bar{l}))} \psi^{(i_{N+1})} \right)\ ,$$
and all others remain the same.
\end{itemize}
In either case, if $\lim_{n\to\infty}\|u_n^{N+1}\|_\binf = 0$, the statement of the theorem follows as in {\em Iterate N}.  If $\lim_{n\to\infty}\|u_n^{N+1}\|_\binf \neq 0$ then we move to {\em Iterate N+2} and so on, so we have now inductively defined all possible iterates.
\subsection{Proof of properties (\ref{profiles}) - (\ref{orth2}) and (\ref{orth4})}
If $\lim_{n\to\infty}\|u_n^{\bar{N}}\|_\binf = 0$ for some $\bar{N} \in \N$ then we stop the iterations at ${\mathit{Iterate \bar{N}}}$ and the main statement of the theorem follows (with $r_n^L \equiv r_n^{\bar{N}}$, etc., for large $L$).
\\\\
Therefore we suppose now that $\lim_{n\to\infty}\|u_n^N\|_\binf \neq 0$ for all $N\in \N$, and we pass to a diagonal subsequence so that the statements in {\em Iterate N} hold for all $N\in \N$ (without changing subsequences).
\\\\
Let $L_\infty := \lim_{N\to \infty}L_N$ and $M_\infty(l):= \lim_{N\to \infty}M_N(l)$ for any $l\in [1,L_\infty] \cap \N$.  Then for any $L\in [1,L_\infty] \cap \N$ we write
$$u_n = \sum_{l=1}^{L} \tau_{m_1(l),n} \left( \sum_{\mu = 1}^{M_\infty(l)} a_{{m_\mu(l)}} \tau^{(m_\mu(l),m_1(l))} \psi^{(i_{m_\mu(l)})}  \right) + r_n^L
$$
\begin{equation}\label{s}
\qquad \qquad  = \sum_{l=1}^{L}  \left( \sum_{\mu = 1}^{M_\infty(l)} a_{{m_\mu(l)}} \tau_{m_\mu(l),n} \psi^{(i_{m_\mu(l)})}  \right) + r_n^L  \ .
\end{equation}
If $M_\infty(l) < \infty$ for some $l$, the term in the parentheses in (\ref{s}) is a well-defined element of $\lprd$, since it is a wavelet expansion with only finitely many components.  Hence we have also
\begin{equation}\label{aa}
\phi_l := \sum_{\mu = 1}^{M_\infty(l)} a_{{m_\mu(l)}} \tau^{(m_\mu(l),m_1(l))} \psi^{(i_{m_\mu(l)})} \in \lprd
\end{equation}
by the invariance of the usual $\lprd$ norm under $ \tau_{m_1(l),n}$.
\\\\
If $M_\infty(l) = +\infty$ then (\ref{e}) and Fatou's lemma (along with the boundedness of $\{u_n\}$) imply $\phi_l \in \lprd$ as well, a fact we leave to the Appendix.  Indeed, apply Lemma \ref{lemma:a4} of the Appendix with $a_n^\mu = a_{\l_n^{m_\mu(l)}}$, $\l_n^\mu = \l_n^{m_\mu(l)}$ and $C=\bar{C}$.
\\\\
Property (\ref{orth1}) is clear from our construction, so we must now show that
$$\lim_{L \to L_\infty} \left( \varlimsup_{n\to\infty} \|r_n^L\|_\bfin \right) =0$$
which will establish (\ref{orth2}), for which we turn to Lemma \ref{lemma1}.
\\\\
Let us first assume that $L_\infty = +\infty$ (the case $L_\infty < +\infty$ is even easier).
\\\\
Note first by standard embeddings (see the Appendix) that we have
$$\sum_{m=1}^\infty |a_{\l_n^m}|^p = \|u_n\tn_{\dot B^0_{p,p}}^p \lesssim \|u_n\|_\lprd^p \leq \bar{C}^p \quad \textrm{for all}\ n\in\N\ ,$$
so by Fatou's lemma we have
$$\sum_{m=1}^\infty |a_m|^p = \sum_{m=1}^\infty \varliminf_{n\to\infty}|a_{\l_n^m}|^p \leq \varliminf_{n\to\infty}\sum_{m=1}^\infty |a_{\l_n^m}|^p \lesssim \bar{C}^p < \infty\ ,$$
hence
\begin{equation}\label{t}
\lim_{m\to\infty}|a_m| =0\ .
\end{equation}
We will need this fact in what follows.
\\\\
Fix any $L\in \N$.  For any $N$ large enough that $L_N \geq L$ (recall we are assuming $L_\infty = +\infty$), by (\ref{p}) we can write
$$u_n = \sum_{l=1}^{L_N}  \left( \sum_{\mu = 1}^{M_N(l)} a_{\l_n^{m_\mu(l)}} \tau_{m_\mu(l),n} \psi^{(i_{m_\mu(l)})}  \right) + u_n^N$$
$$= \sum_{l=1}^{L}  \left( \sum_{\mu = 1}^{M_\infty(l)} a_{m_\mu(l)} \tau_{m_\mu(l),n} \psi^{(i_{m_\mu(l)})}  \right) + r_n^L$$
which gives
\begin{equation}\label{u}
\begin{array}{rcl}
r_n^L & = & \displaystyle{\sum_{l=1}^{L}  \left( \sum_{\mu = 1}^{M_N(l)} (a_{\l_n^{m_\mu(l)}} - a_{m_\mu(l)}) \tau_{m_\mu(l),n} \psi^{(i_{m_\mu(l)})}  \right)}\\\\
&  & \quad - \ \displaystyle{ \sum_{l=1}^{L} \left( \sum_{\mu = M_N(l)+1}^{M_\infty(l)} a_{m_\mu(l)} \tau_{m_\mu(l),n} \psi^{(i_{m_\mu(l)})}  \right)      }\\\\
&  & \quad + \ \displaystyle{ \sum_{l=L+1}^{L_N} \left( \sum_{\mu = 1}^{M_N(l)} a_{\l_n^{m_\mu(l)}} \tau_{m_\mu(l),n} \psi^{(i_{m_\mu(l)})}  \right) \ +\  u_n^N     }\\\\
& =: & I - II + III + IV
\end{array}
\end{equation}
(where if $L=L_N$ or $M_N(l) = M_\infty(l)$, we replace the appropriate sum by zero).  We'll estimate, using the embedding $\lprd \hookrightarrow \bfin$,
\begin{equation}\label{v}
\|r_n^L\|_\bfin \lesssim \|I\|_\lprd + \|II\|_\lprd + \|III\|_\bfin + \|IV\|_\bfin\ .
\end{equation}
We claim, in fact, that this leads to an estimate of the form
\begin{equation}\label{w}
\|r_n^L\|_\bfin \leq \e_{N,L}(n) + \e_L^1(N) + \e_L^2(n) + \frac{C}{N^\frac{1}{p}}\ ,
\end{equation}
where
\begin{equation}\label{x}
\begin{array}{c}
\displaystyle{\lim_{n\to\infty} \e_{N,L}(n) = 0 \ \ \forall\ N,L \in \N\ , \quad  \lim_{N\to\infty} \e_L^1(N) =0 \ \ \forall\ L \in \N\ ,}\\\\
\displaystyle{\textrm{and}\qquad \lim_{L\to\infty}\e_L^2=0 \quad \textrm{where} \quad \e_L^2:=\lim_{n\to\infty} \e_L^2(n)\ .}
\end{array}
\end{equation}
In this case, the desired estimate on $r_n^L$ is proved as follows:  For any $\bar{\e}>0$, there exists $\bar{L}\in \N$ such that $\e^2_{L_0} < {\bar{\e}/3}$ for any $L_0 \geq \bar{L}$.  For any such $L_0$, there exists $N_0 = N_0(L_0)\in \N$ large enough that $L_{N_0} \geq L_0$ and ${\max \{\e_{L_0}^1(N_0), {{C N_0^{-{1/p}}}} \} < {\bar{\e}/3}}$.  Hence by (\ref{w})-(\ref{x}), for any $\bar{\e}>0$ there exists $\bar{L}\in \N$ such that, for all $L_0 \geq \bar{L}$,
$$\varlimsup_{n\to\infty} \|r_n^{L_0}\|_\bfin \leq \lim_{n\to\infty} \left( \e_{N_0,L_0}(n)\ +\ \e_{L_0}^1(N_0)\ +\ \e_{L_0}^2(n)\ +\ \frac{C}{N_0^\frac{1}{p}}\right) =$$
$$\qquad \qquad \qquad \qquad
\qquad = \qquad \  0\quad  +\  \e_{L_0}^1(N_0) \ +\  \e_{L_0}^2\  +\  \frac{C}{N_0^\frac{1}{p}}\qquad   <\qquad  \bar{\e}\ ,
$$
which is exactly (\ref{orth2}).
\\\\
Let us pause here to address (\ref{orth4}) as well:  It is clear that $III+IV$ is just $u_n$ with some of its wavelet components removed.  Therefore, using (\ref{e}) and (\ref{u}) - (\ref{w}), we can estimate
$$\|r_n^L\tn_\lprd \leq \e_{N,L}(n) + \e_L^1(N) + \|u_n\tn_\lprd
$$
and hence for any $\e>0$, by taking first $N$ large and then $n$ large, we see by (\ref{x}) that for any $L\in \N$ we have
$$\left| \|r_n^L\tn_\lprd - \|u_n\tn_\lprd  \right| < \e $$
for $n$ sufficiently large which establishes (\ref{orth4}).
\\\\
Turning back to the proof of (\ref{orth2}), we'll now estimate individually each term of (\ref{v}).  Note first the following important change of variable formula:
\\\\
For any $l$ and $M$ and any numbers $a_\mu (n)$ possibly depending on $n$, we may write
\begin{equation}\label{y}
\sum_{\mu=1}^M |a_\mu(n)|^2 2^{\frac{2d}{p} j_n^{m_\mu(l)}} \chi_{\d_n^{m_\mu(l)}}(x) = 2^{\frac{2d}{p}j_n^{m_1(l)}} \tilde{\chi}_l (2^{j_n^{m_1(l)}}x - k_n^{m_1(l)} )
\end{equation}
where
\begin{equation}\label{z}
\tilde{\chi}_l(y) = \sum_{\mu=1}^M |a_\mu(n)|^2 2^{\frac{2d}{p}j^{(m_\mu(l),m_\mu(1))}} \chi_{\d^{(m_\mu(l),m_\mu(1))}}(y)
\end{equation}
and
$$\d^{(m_\mu(l),m_\mu(1))} = (\tau^{(m_\mu(l),m_\mu(1))})^{-1} ([0,1)^d)$$
(see the proof of Lemma \ref{lemma:a4}).  Recall that $\chi_\d$ is the characteristic function of the cube $\d$, and note that $\tilde \chi_l$ is bounded with compact support for each $l$, and of course is independent of $n$.
\pagebreak
\\\\
\underline{{\sc Estimate of I:}}
\\\\
Using (\ref{e}) and (\ref{y}), for any $l\in \{1,\dots,L\}$ the change of variable $y=2^{j_n^{m_1(l)}}x - k_n^{m_1(l)}$ gives
$$\Bigg\| \sum_{\mu=1}^{M_N(l)} (a_{\l_n^{m_\mu(l)}} - a_{m_\mu(l)}) \tau_{m_\mu(l),n} \psi^{(i_{m_\mu(l)})}
\tilde{\Bigg\|}^p_\lprd =$$
$$=
\int \left\{ \sum_{\mu=1}^{M_N(l)}|a_{\l_n^{m_\mu(l)}} - a_{m_\mu(l)}|^2 2^{\frac{2d}{p}j^{(m_\mu(l),m_\mu(1))}} \chi_{\d^{(m_\mu(l),m_\mu(1))}}(y) \right\}^\frac{p}{2}dy
$$
$$
\leq 2 \int \left\{ \sum_{\mu=1}^{M_N(l)}|a_{m_\mu(l)}|^2 2^{\frac{2d}{p}j^{(m_\mu(l),m_\mu(1))}} \chi_{\d^{(m_\mu(l),m_\mu(1))}}(y) \right\}^\frac{p}{2}dy \leq \tilde{C}^p < \infty
$$
for all $n$ sufficiently large, depending on $N$, and the last inequality is clear from the proof of Lemma \ref{lemma:a4}.  Lebesgue's Dominated Convergence Theorem now gives that the integral on the left tends to zero as $n\to \infty$ for fixed $N$, hence, after summing from $1$ to $L$ ($L<\infty$), we have
$$\|I\|_\lprd \leq \e_{N,L}(n) \qquad \textrm{with} \qquad \lim_{n\to\infty} \e_{N,L}(n) = 0$$
as desired.
\\\\
\underline{{\sc Estimate of II:}}
\\\\
Similarly, for $l \in \{1,\dots,L\}$, we have
$$\Bigg\| \sum_{\mu=M_N(l)+1}^{M_\infty(l)} a_{m_\mu(l)} \tau_{m_\mu(l),n} \psi^{(i_{m_\mu(l)})}
\tilde{\Bigg\|}^p_\lprd =$$
$$=
\int \left\{ \sum_{\mu=M_N(l)+1}^{M_\infty(l)} |a_{m_\mu(l)}|^2 2^{\frac{2d}{p}j^{(m_\mu(l),m_\mu(1))}} \chi_{\d^{(m_\mu(l),m_\mu(1))}}(y) \right\}^\frac{p}{2}dy
$$
$$
\leq  \int \left\{ \sum_{\mu=1}^{M_\infty(l)}|a_{m_\mu(l)}|^2 2^{\frac{2d}{p}j^{(m_\mu(l),m_\mu(1))}} \chi_{\d^{(m_\mu(l),m_\mu(1))}}(y) \right\}^\frac{p}{2}dy \leq \tilde{C}^p < \infty
$$
uniformly in $N$ by Lemma \ref{lemma:a4}.  Now, again by Lebesgue's Dominated Convergence Theorem, the left-hand side tends to zero as $N\to \infty$.  Summing from $1$ to $L$, we have
$$\|II\|_\lprd \leq \e_L(N) \qquad \textrm{where} \qquad \lim_{N\to\infty} \e_L(N) =0$$
as desired.
\\\\
\underline{{\sc Estimate of III:}}
\\\\
Since $III$ is just $u_n$ after replacing some of its wavelet coefficient by zero, (\ref{e}) gives
\begin{equation}\label{three}
\|III\tn_\lprd \leq \|u_n \tn_\lprd \leq \tilde{C}\ .
\end{equation}
Note that, by our construction, $m_1(l)$ is increasing with $l$ and $m_\mu(l)$ is increasing with $\mu$ for each fixed $l$.  Therefore, since also $|a_{\l_n^m}|$ is monotonically decreasing in $m$ for each fixed $n$, we see that
$$\|III\tn_\binf = |a_{\l_n^{m_1(L+1)}}|$$
which is independent of $N$.  Since $m_1(L+1) \to +\infty$ as $L\to \infty$, (\ref{t}) gives
\begin{equation}\label{four}
\|III\tn_\binf \leq \e_L(n)\ ,
\end{equation}
where
\begin{equation}\label{five}
\lim_{L\to\infty} \left(\lim_{n\to\infty} \e_L(n) \right) = \lim_{L\to\infty} |a_{m_1(L+1)}| = 0\ .
\end{equation}
The desired estimate now follows from (\ref{three}) - (\ref{five}) and the interpolation inequality (\ref{j}).
\\\\
\underline{{\sc Estimate of IV:}}
\\\\
Similarly, $\|u_n^N\tn_\lprd \leq \|u_n \tn_\lprd \leq \tilde{C}$, and since $u_n^N \in \bar{Q}_N(u_n)$, Lemma \ref{lemma1} gives $\|u_n^N\|_\binf \leq CN^{-{{1/p}}}$ for some $C>0$, independently of $n$.  The desired estimate now follows from (\ref{j}).
\\\\
Estimate (\ref{w}) is now proved, and defining $\phi_l$ as in (\ref{aa}), we have
$$u_n = \sum_{l=1}^L \tau_{m_1(l),n} \phi_l + r_n^L, \qquad \textrm{with} \qquad \lim_{L\to\infty} \left( \varlimsup_{n\to\infty} \|r_n^L\|_\bfin \right) =0$$
thus establishing (\ref{profiles}) - (\ref{orth2}) of the theorem in the case $L_\infty = \infty$.  The case $L_\infty < \infty$ is even easier; starting with $L=L_\infty$ from the beginning, the term $III$ completely disappears from (\ref{u}) so that $\lim_{n\to\infty} \|r_n^{L_\infty}\|_\bfin = 0$ by the considerations above.
\subsection{Proof of (\ref{orth3})}
We now turn to the proof of (\ref{orth3}):
\\\\
For convenience, let's define
$$\l_n^0 = (0,0,0)\ , \quad a_{\l_n^0} = a_0 = \psi^{(0)} = j^{(0,m_1(l))} = b^{(0,m_1(l))} =0\ ,$$
and replace both $L_\infty$ and $M_\infty(l)$ by $\infty$ by making the convention that in case either is finite, we set $m_\mu(l) = 0$ whenever $l> L_\infty$ or $\mu > M_\infty(l)$.
\\\\
With this convention, we can write
\begin{equation}\label{ab}
u_n = \sum_{l=1}^{\infty}  \sum_{\mu = 1}^{\infty} a_{\l_n^{m_\mu(l)}} \tau_{m_\mu(l),n} \psi^{(i_{m_\mu(l)})}  \ ,
\end{equation}
which is just a particular ordering (depending on $n$), of the wavelet components of $u_n$.  Therefore (\ref{e}) gives
\begin{equation}\label{ac}
\int \left\{ \sum_{l=1}^{\infty}  \sum_{\mu = 1}^{\infty} |a_{\l_n^{m_\mu(l)}}|^2 2^{\frac{2d}{p}j_n^{m_\mu(l)}} \chi_{\d_n^{m_\mu(l)}} (x) \right\}^\frac{p}{2} dx = \|u_n \tn_\lprd^p \leq \tilde{C}^p
\end{equation}
uniformly in $n$.  A change of variables as in (\ref{y}) and an application of Lemma \ref{lemma:a4} shows that for any $l\in \N$ (recall (\ref{aa})),
\begin{equation}\label{za}
\begin{array}{c}
\displaystyle{\|\tau_{m_1(l),n}\phi_l \tn_p^p = \left\{   \sum_{\mu = 1}^{\infty} |a_{m_\mu(l)}|^2 2^{\frac{2d}{p}j_n^{m_\mu(l)}} \chi_{\d_n^{m_\mu(l)}} (x) \right\}^\frac{p}{2} dx  = \qquad } \\\\
\phantom{f}  \qquad \displaystyle{=\left\{   \sum_{\mu = 1}^{\infty} |a_{m_\mu(l)}|^2 2^{\frac{2d}{p}j^{(m_\mu(l),m_\mu(1))}} \chi_{\d^{(m_\mu(l),m_\mu(1))}}(y) \right\}^\frac{p}{2}dy \leq   \tilde{C}^p}
\end{array}
\end{equation}
for any $n\in\N$ -- that is, within each profile one can replace the wavelet coefficients by their pointwise limits.  What we want to show now is that
$$\sum_{l=1}^\infty \|\tau_{m_1(l),n}\phi_l \tn_p^p \leq \liminf_{n\to\infty} \|u_n\tn_p^p, $$
which is (\ref{orth3}), and at this point we have established that all quantities are well-defined.
\\\\
To simplify notation, let's denote
$$\eta^{\mu,l}_n(x):= 2^{\frac{2d}{p}j_n^{m_\mu(l)}} \chi_{\d_n^{m_\mu(l)}} (x) \quad \textrm{and} \quad \tilde{\eta}^{\mu,l}(y):= 2^{\frac{2d}{p}j^{(m_\mu(l),m_\mu(1))}} \chi_{\d^{(m_\mu(l),m_\mu(1))}}(y)\ .$$
Truncating (\ref{ac}) and (\ref{za}) to finite sums, Fatou's lemma applied as in Lemma \ref{lemma:a4} yields
$$\sum_{l=1}^L \int \left\{   \sum_{\mu = 1}^{M} |a_{m_\mu(l)}|^2 \tilde{\eta}^{\mu,l}(y) \right\}^\frac{p}{2}dy
\leq \liminf_{n\to\infty} \sum_{l=1}^L \int \left\{   \sum_{\mu = 1}^{M} |a_{\l_n^{m_\mu(l)}}|^2 \tilde{\eta}^{\mu,l}(y) \right\}^\frac{p}{2}dy .
$$
We claim moreover that
$$\sum_{l=1}^L \int \left\{   \sum_{\mu = 1}^{M} |a_{\l_n^{m_\mu(l)}}|^2 \eta^{\mu,l}_n(x) \right\}^\frac{p}{2}dx
= \int \left\{  \sum_{l=1}^L \sum_{\mu = 1}^{M} |a_{\l_n^{m_\mu(l)}}|^2 \eta^{\mu,l}_n(x) \right\}^\frac{p}{2}dx  +\e_1(n)\ ,
$$
where $\lim_{n\to\infty}\e_1(n) =0$.  (This is simply the quantity on the right of the previous line after a change of variable.)  Let us postpone this claim for the moment.
\\\\
Using (\ref{ac}) to bound the last term above by $\|u_n\tn_p^p + \e_1(n)$ and inserting this into the previous inequality we have
$$\sum_{l=1}^L \int \left\{   \sum_{\mu = 1}^{M} |a_{m_\mu(l)}|^2 \tilde{\eta}^{\mu,l}(y) \right\}^\frac{p}{2}dy \leq \liminf_{n\to\infty}\|u_n\tn_p^p\ .$$
Now that both sides are independent of n, using Fatou's lemma to let $M\to\infty$ and then letting $L\to\infty$ and using identity (\ref{za}) yields (\ref{orth3}) as desired.
\\\\
We are reduced therefore to showing that
\begin{equation}\label{ag}
\begin{array}{c}
    \displaystyle{ \lim_{n\to\infty} I_{L,M}(n) := }\\\\
 \displaystyle{ \lim_{n\to\infty}\Bigg| \int \left\{ \sum_{l=1}^{L}  \sum_{\mu = 1}^{M} |a_{\l_n^{m_\mu(l)}}|^2 2^{\frac{2d}{p}j_n^{m_\mu(l)}} \chi_{\d_n^{m_\mu(l)}} (x) \right\}^\frac{p}{2} dx } \qquad  \qquad  \qquad \\\\
 \qquad  \qquad  \qquad   \displaystyle{ -\ \sum_{l=1}^{L} \int \left\{  \sum_{\mu = 1}^{M} |a_{\l_n^{m_\mu(l)}}|^2 2^{\frac{2d}{p}j_n^{m_\mu(l)}} \chi_{\d_n^{m_\mu(l)}} (x) \right\}^\frac{p}{2} dx} \Bigg| =0
\end{array}
\end{equation}
Note first that if $p=2$, (\ref{ag}) is trivial so we may assume that $p>2$.
\\\\
We'll use the elementary inequality (see, e.g., equation (1.10) in \cite{gerard}) that for any $L\in \N$ there exists a constant $C_L=C_L(p) >0$ such that for any $\{A_l\} \subset [0,\infty)$,
\begin{equation}\label{aj}
\left| \left( \sum_{l=1}^L A_l \right)^\frac{p}{2} - \sum_{l=1}^L (A_l)^\frac{p}{2} \right| \leq \ \ C_L \sum_{l\neq l'} A_l (A_{l'})^{\frac{p}{2}-1}\ .
\end{equation}
We can therefore estimate the function in (\ref{ag}) using (\ref{aj}) and the notation of (\ref{y}) and (\ref{z}) as
$$I_{L,M}(n) \leq C_L \sum_{l\neq l'} I^{l,l'}(n)$$
with
$$I^{l,l'}(n):= \int \left(2^{dj_n^{m_1(l)}}\right)^{(\frac{2}{p})} \tilde \chi_l (2^{j_n^{m_1(l)}}x - k_n^{m_1(l)})
\left(2^{dj_n^{m_1(l')}}\right)^{(1-\frac{2}{p})} \tilde \chi_{l'}^{(\frac{p}{2}-1)} (2^{j_n^{m_1(l')}}x - k_n^{m_1(l')})\ dx\ .
$$
(We have absorbed the coefficients $|a_{\l_n^{m_\mu(l)}}|^2$ into the constant since they converge and hence are uniformly bounded.) Fix $l$ and $l'$, and set the notation
\begin{equation}\label{zzz}
x_n^{l_1,l_2}(w):=2^{j_n^{m_1(l_2)}-j_n^{m_1(l_1)}}w + \left[2^{j_n^{m_1(l_2)}-j_n^{m_1(l_1)}}k_n^{m_1(l_1)}- k_n^{m_1(l_2)}\right]
\end{equation}
for $w\in \rd$ and any $l_1,l_2\in \N$.  Then by appropriate changes of variables, we see
\begin{equation}\label{ak}
\begin{array}{rcl}
I^{l,l'}(n) & = &\displaystyle{ \left(2^{j_n^{m_1(l')}-j_n^{m_1(l)}}\right)^{(1-\frac{2}{p})d} \int  \tilde \chi_l (y)
\ \tilde \chi_{l'}^{(\frac{p}{2}-1)} (x_n^{l,l'}(y))\ dy} \\\\
 & = &\displaystyle{ \left(2^{j_n^{m_1(l)}-j_n^{m_1(l')}}\right)^{(\frac{2}{p})d} \int  \tilde \chi_l (x_n^{l',l}(z))
\ \tilde \chi_{l'}^{(\frac{p}{2}-1)} (z)\ dz}\ .
\end{array}
\end{equation}
By orthogonality of $\tau_{m_1(l),n}$ and $\tau_{m_1(l'),n}$ (i.e. (\ref{six}), recall (\ref{m})), there are two possibilities:
\begin{enumerate}
\item One of the pre-factors in (\ref{ak}) tends to zero in which case $I^{l,l'}(n) \to 0$ as $n\to \infty$ since the functions in the integrands are bounded with compact support.
\item Both pre-factors are bounded, hence
$$2^{j_n^{m_1(l')}-j_n^{m_1(l)}} \in (c_1,c_2) \qquad \forall \ n$$
for some $c_2>c_1 >0$, and
$$\left|2^{j_n^{m_1(l')}-j_n^{m_1(l)}}k_n^{m_1(l)} - k_n^{m_1(l')}\right| \xto +\infty\ .$$
In this case, in light of (\ref{zzz}) we see that $I^{l,l'}(n) \equiv 0$ for sufficiently large $n$, since the supports of the two functions in the integrand in (\ref{ak}) become disjoint.
\end{enumerate}
Summing over a finite number of $l$ and $l'$ concludes the proof of (\ref{ag}) and of the theorem. \hfill $\Box$
\\\\
Note that we did not need to use the compact support of the $\psi^{(i)}$ (as was emphasized in \cite{jaffard}), only the compact support of the functions $\chi_\d$ (in fact, $\chi_\d \in \lprd$ would have been sufficient) and the relation (\ref{e}).  Hence, any well-localized basis functions $\psi^{(i)}$ would be sufficient for our purposes (e.g., the divergence-free wavelet bases of \cite{bf} or \cite{lr}).
\section{Other applications of the method}
\noindent
We can similarly (and in fact, more easily) deduce the following as well:
\\\\
Fix any $p \in [-\infty,+\infty]\backslash \{0\}$, and for any $a\geq 1$ set $s_{p,a}:= \frac{d}{a} - \frac{d}{p}$.  Bernstein's inequalities (\ref{aac}) immediately give us $\dot B^{s_{p,a}}_{a,q} \hookrightarrow \dot B^{s_{p,b}}_{b,r}$ whenever $1\leq a < b\leq +\infty$ and $1\leq q \leq r \leq +\infty$.  (The choice of $p$ dictates the common scaling of these spaces, which coincides with that of $L^p(\mathbb{R}^d)$ when $p>0$.)  If, moreover, we have the slight improvement that $r\geq \frac{b}{a} q$, the methods used to prove Theorem \ref{thm:a} easily enable us to prove the following similar result:
\begin{thm}\label{thm:aa}
Let $\{u_n\}_{n=1}^\infty$ be a bounded sequence in $\besa$.  There exists a sequence $\{\phi_l\}_{l=1}^\infty \subset \besa$ and for each $l\in \N$ a sequence $\{(j_n^l,k_n^l)\}_{n=1}^\infty \subset \Z \times \Z^d$, both depending on $\{u_n\}$, such that, after possibly passing to a subsequence in $n$,
\begin{equation}\label{profilesa}
u_n(x) = \sum_{l=1}^L (2^{j^l_n})^{{d/p}} \phi_l(2^{j^l_n}x - k^l_n) + r_n^L(x)
\end{equation}
for any $L \in \N$ where the following properties hold:
\begin{enumerate}
\item For $l\neq l'$, the sequences $\{(j_n^l,k_n^l)\}$ and $\{(j_n^{l'},k_n^{l'})\}$ are orthogonal in the following sense:
\begin{equation}\label{orth1a}
\left|\log \left(2^{(j_n^l - j_n^{l'})}\right)\right| + \left|2^{(j_n^l - j_n^{l'})}k_n^{l'} - k_n^l\right|  \xrightarrow[n\to\infty]{} +\infty
\end{equation}
\item The remainder $r_n^L$ satisfies the following smallness condition:
\begin{equation}\label{orth2a}
\lim_{L\to\infty} \left(\varlimsup_{n\to \infty} \|r_n^L\|_\besb \right) = 0
\end{equation}
\item There is a norm $\| \cdot \tilde \|_\besa$ which is equivalent to $\|\cdot \|_\besa$ such that for each $n \in \N$,
\begin{equation}\label{orth3a}
\left\| \left(\|(2^{j^l_n})^{{d/p}} \phi_l(2^{j^l_n}x - k^l_n) \tn_\besa\right)_{l=1}^\infty \right\|_{\ell^\tau} \leq \varliminf_{n'\to\infty} \|u_{n'}\tn_\besa
\end{equation}
where $\tau := \max \{a,q\}$, and for any $L \in \N$,
\begin{equation}\label{orth4a}
 \|r_n^L \tn_\besa \leq  \|u_{n}\tn_\besa + \circ(1) \quad \textrm{as} \quad n\to\infty\ .
\end{equation}
\end{enumerate}
\end{thm}
\noindent
The proof of Theorem \ref{thm:a} depended in part on the ``improved Sobolev-type'' embedding inequality\footnote{We believe moreover that in many settings of this type one can complete the proofs without explicitly using such an improved embedding inequality.  This will be addressed soon in \cite{ck}.} (\ref{j}).  The slight improvement on the distance between $r$ and $q$ is to allow us the following inequality, proved in Proposition \ref{prop:a3}, which is the analogue of (\ref{j}):
\begin{equation}\label{eq:sobc}
\|u\|_\besb \leq \|u\|^\alpha_\besa \|u\|^{1-\alpha}_\binf \ .
\end{equation}
The proofs of statements (\ref{profilesa}), (\ref{orth1a}), (\ref{orth2a}) and (\ref{orth4a}) are similar to (and, in fact, simpler than) the proofs of the analogous statements in Theorem \ref{thm:a} so we will neglect the details (see also \cite{ck}).  We will just mention that inequality (\ref{eq:sobc}) will replace inequality (\ref{j}), and one can replace the inequality in Lemma \ref{lemma1} by (\ref{decay2}).  This takes care of the key ingredients of the improved Sobolev-type inequality and the decay of non-linear wavelet projections.
\\\\
As a technical point, we also briefly mention the following:  Whereas in the proof of Theorem \ref{thm:a} one often needed to use Fatou's Lemma twice, once for an $L^q$-type norm and once for an $\ell^q$-type norm (for example in the proof of Lemma \ref{lemma:a4}), in the case of Besov spaces one proceeds in the same way but now both are $\ell^q$-type norms due to (\ref{c}), and so one no longer has to be cautious about pointwise limits in $\R^d$.  In this case, the ``change of variables" which allows the use of Fatou's lemma comes from the fact that within each profile, one can ``shift the labels" on the coefficients in a way depending on $n$ so that when one takes the Besov norm the only dependence on $n$ is in the coefficients (and not on the order in which they are summed).
\\\\
We turn therefore to the proof of (\ref{orth3a}), the proof of which is necessarily different since one cannot use the orthogonality of the scales and shifts which appear in the profile decomposition because our norms do not involve a space variable, only wavelet coefficients.  This will be compensated by the natural orthogonalities in the wavelet basis in spaces of the type given by (\ref{c}).
\\\\
The proof is due to the following basic inequality:
\\\\
Suppose $\LL$ is decomposed into a disjoint union of sets $E_i$:  $\LL = \cup_{i=1}^\infty E_i$, and suppose $f= \sum f_\l \psi_\l \in \besa$.  Then one has
$$
\left\| \left( \left\| \sum_{\l \in E_i} f_\l \psi_\l \right\|_X \right)_{i=1}^\infty \right\|_{\ell^\tau} \leq \left\| \sum_{\l \in \LL} f_\l \psi_\l \right\|_X
$$
where $\|\cdot \|_X := \| \cdot \tn_\besa$ and $\tau:= \max \{a,q\}$.  The inequality is a simple consequence of properties of $\ell^r$ spaces and definition (\ref{c}).
\\\\
Letting $E_i$ correspond to the wavelet components of $u_n$ which contribute to the rescaling of the $i$'th profile $\phi_i$ at a certain step of the iteration process, and bounding the term on the right by the full expansion of $u_n$, taking limits and using Fatou's lemma gives the desired result.  We refer the interested reader to the upcoming work \cite{ck} for more details.
\appendix
\section{Appendix}
\noindent
In this appendix we collect some known propositions regarding wavelets and embeddings of Besov and Lebesgue spaces, as well as a lemma (Lemma \ref{lemma:a4}) which is necessary for the proof of our main theorem.
\begin{prop}\label{prop:a1}
Let $\{a_\l\}$ correspond to some $f\in \lprd$, and for $A>0$ denote $\LL_A := \{ \l \in \LL \ | \ |a_\l| \geq A\ \}$.  Then $\#(\LL_A) < \infty$ for any $A>0$.  (In particular, $\exists \l_0 \in \LL$ such that $\sup_{\l \in \LL} |a_\l| = |a_{\l_0}|$.)
\end{prop}
\noindent
(For a countable set $\Omega$ we are denoting by $\#(\Omega)$ the number of elements in $\Omega$.)
\\\\
\textbf{Proof:}
\\\\
Fix $A>0$, and suppose to the contrary that $\#(\LL_A) = +\infty$.
\\\\
Suppose that there exists $\{\l_m = (i_m,j_m,k_m) \} \subset \LL_A$ such that $j_m \to -\infty$ as $m \to\infty$.  Then $|\d_m| = |\d(\l_m)| \to +\infty$ (recall that $\d(\l)$ is the cube associated to $j$ and $k$ if $\l = (i,j,k)$ and has volume $2^{-dj}$) and we can pass to a subsequence so that $|\d(\l_m)|$ increases so quickly that
\begin{equation}\label{aaa}
|\d^{(m)}| \geq \tfrac{1}{2} |\d(\l_m)| = \tfrac{1}{2}\cdot 2^{-dj_m}\ ,
\end{equation}
where we define $\d^{(m)} := \d(\l_m) \backslash \cup_{n=1}^{m-1} \d(\l_n)$.  Then for any $N\geq 2$,
\begin{equation}\label{aab}
\begin{array}{c}
\displaystyle{\|f\tn_\lprd^p   \geq \int_\rd \left\{ \sum_{m=1}^\infty |a_{\l_m}|^2 2^{\frac{2d}{p}j_m} \chi_{\d_m} \right\}^\frac{p}{2} \geq
\sum_{M=2}^{N+1} \int_{\d^{(m)}} \left\{ \sum_{m=1}^\infty |a_{\l_m}|^2 2^{\frac{2d}{p}j_m} \chi_{\d_m} \right\}^\frac{p}{2}} \\\\
\displaystyle{ \geq
\sum_{M=2}^{N+1} \int_{\d^{(m)}} \left\{ |a_{\l_M}|^2 2^{\frac{2d}{p}j_M} \chi_{\d_M} \right\}^\frac{p}{2} \geq \sum_{M=2}^{N+1} \frac{A^p}{2} = N \frac{A^p}{2}\ ,  }
\end{array}
\end{equation}
which implies (letting $N \to \infty$) that $\|f\|_\lprd = +\infty$ contrary to assumption.
\\\\
Similarly, if there exists $\{\l_m  \} \subset \LL_A$ such that $j_m \to +\infty$, for fixed $N \geq 2$ we can pass to a subsequence (depending on $N$) so that $|\d(\l_m)| \to 0$ rapidly enough that (\ref{aaa}) holds for $2\leq m \leq N+1$ after reversing the order of the first $N+2$ elements of the sequence.  By (\ref{aab}) therefore we have $\|f\tn_\lprd^p \geq N \frac{A^p}{2}$ which is false for a suitably large $N$ depending on $f$.
\\\\
Hence $i$ and $j$ can take only finitely-many values for $\l \in \LL_A$, so $\#(\LL_A) =+\infty$ implies that there exists $\{\l_m\} \subset \LL_A$ such that $|k_m| \to +\infty$.  Passing to a subsequence so that $j_m \equiv \bar{j} = \textrm{const}$ and $|k_m| \to +\infty$ sufficiently rapidly we can ensure that the $\d_m$ are mutually disjoint, so $\d^{(m)} = \d_m$ for all $m$.  Equations (\ref{aaa}) and (\ref{aab}) then again contradict the assumption that $f\in \lprd$ and the proposition is proved. \hfill $\Box$
\\\\
For the following two propositions, we will use the Littlewood-Paley description of Besov and Triebel-Lizorkin spaces in terms of the frequency localization operators $\D_j$, where $\widehat{\D_j f}$ is supported in a neighborhood of $\{|\xi| = 2^j\}$ for $j\in \Z$.  For example, one defines the Besov spaces by $\|f\|_{\dot B^s_{r,q}} = \left\| 2^{sj}\|\D_j f\|_{L^r(\rd)} \right\|_{\ell^q(\Z)}$.
\begin{prop}\label{prop:a2}
Fix $p,q,r \in \R$ such that $2\leq p \leq q,r \leq \infty$, and set ${s_{p,r}:= d(\tfrac{1}{r} - \tfrac{1}{p})}$ (note $s_{p,r} < 0$).  Then
$$\lprd \hookrightarrow \brq\ .$$
\end{prop}
\pagebreak
\noindent
\textbf{Proof:}
\\\\
Note first by Bernstein's inequalities that
\begin{equation}\label{aac}
2^{jd(\frac{1}{r}-\frac{1}{p})}\|\D_jf\|_{L^r(\rd)} \lesssim \|\D_jf\|_\lprd
\end{equation}
for
$$1\leq p \leq r \leq +\infty\ .$$
Taking $\ell^q$ norms of both sides of (\ref{aac}) gives $\dot B^0_{p,q} \hookrightarrow \bfin$.
\\\\
Note now that $\|f\|_\lprd \simeq \|f\|_{F^0_{p,2}} = \|\{\sum_j |\D_jf|^2\}^\frac{1}{2}\|_\lprd$.  Using the simple fact that $\|c_j\|_{\ell^\infty} \leq \|c_j\|_{\ell^\alpha}$ for any $\alpha >0$ implies\footnote{Indeed, $a>0$ and $\frac{a}{b} \in [0,1]$ gives
$\|c_j\|_{\ell^b} = \||c_j|^{1-\frac{a}{b}}|c_j|^\frac{a}{b}\|_{\ell^b} \leq  \||c_j|^{1-\frac{a}{b}}\|_{l^\infty}\||c_j|^\frac{a}{b}\|_{\ell^b}
\leq  \||c_j|^{1-\frac{a}{b}}\|_{l^{\frac{a}{1-\frac{a}{b}}}}\||c_j|^\frac{a}{b}\|_{\ell^b} = \|c_j\|_{\ell^a}$.} that $\|c_j\|_{\ell^b} \leq   \|c_j\|_{\ell^a}$ for $0< a \leq b \leq \infty$, we have
$$  \Big\{\sum_j |\D_jf|^p\Big\}^\frac{1}{p}(x) \leq \Big\{ \sum_j |\D_jf|^2\Big\}^\frac{1}{2}(x)$$
for any $x\in \rd$, as long as $p\geq 2$.  Taking $\lprd$ norms of both sides, Fubini's theorem now gives
\begin{equation}\label{a1}
\|f\|_{\dot B^0_{p,p}}= \left\|\Big\{\sum_j |\D_jf|^p\Big\}^\frac{1}{p}\right\|_\lp \leq \|f\|_{F_{p,2}^{0}} \lesssim \|f\|_\lp\ .
\end{equation}
Since $0<p\leq q \leq \infty$, we see therefore that
$$\lprd \hookrightarrow \dot B^0_{p,p} \hookrightarrow \dot B^0_{p,q} \hookrightarrow \bfin$$
and the proposition is proved. \hfill $\Box$
\begin{prop}\label{prop:a3}
Suppose $2\leq p < q,r \leq \infty$.  Then for any $\alpha \in (\max \{\frac{p}{r},\frac{p}{q}\},1)$, there exists $C=C(d,p,q,r,\alpha)$ such that
$$\|f\|_\brq \leq C \|f\|^\alpha_\lprd \|f\|^{1-\alpha}_\binf\ .$$
\end{prop}
\noindent
\textbf{Proof:}
\\\\
Fix such an $\alpha$ and set $a:=r\alpha$, $b:=q\alpha$.  Since
$$\frac{p}{r} < \alpha < 1 \qquad \Longrightarrow \qquad 1<a<r\ ,$$
H\" older's inequality gives
$$2^{jd(\frac{1}{r} - \frac{1}{p})} \|\D_jf\|_r \leq \left[2^{jd(\frac{1}{a} - \frac{1}{p})} \|\D_jf\|_a\right]^\alpha \left[ 2^{-j\frac{d}{p}} \|\D_jf\|_\infty\right]^{1-\alpha} \ .$$
Writing this as $\e_{r,j} \leq \e_{a,j}^\alpha \e_{\infty,j}^{1-\alpha}$, we have
$$\|\e_{r,j}\|_{\ell^q} \leq \|\e_{a,j}^\alpha \|_{\ell^q} \|\e_{\infty,j}^{1-\alpha}\|_{l^\infty} = \|\e_{a,j}\|_{\ell^b}^\alpha  \|\e_{\infty,j}\|_{l^\infty}^{1-\alpha}$$
which means that
$$\|f\|_\bfin \leq \|f\|_{\dot B^{s_{p,a}}_{a,b}}^\alpha \|f\|^{1-\alpha}_\binf \ .$$
But by Proposition \ref{prop:a2}, $\|f\|_{\dot B^{s_{p,a}}_{a,b}} \lesssim \|f\|_\lprd$ provided only that $2\leq p \leq a,b \leq +\infty$.  This is exactly guaranteed by our assumptions on $\alpha$, and the proposition follows. \hfill $\Box$
\\\\
Finally, we prove the following crucial lemma:
\begin{lemma}\label{lemma:a4}
For each $n$, define $\phi_n \in L^p(\R^d)$ by
$$\phi_n = \sum_{\mu \in \N} a_n^\mu \psi_{\l_n^\mu}$$ and suppose the following are satisfied:
\begin{enumerate}
\item There exists $C>0$ such that $\|\phi_n\|_p \leq C$ for all $n \in \N$,
\item for each $\mu\in \N$, there exists $a^\mu \in \R$ such that $a_n^\mu \to a^\mu$ as $n\to \infty$, and
\item for each $\mu\in \N$, there exists $j^{(\mu,1)} \in \Z$ and $b^{(\mu,1)} \in \R^d$ such that
$$j_n^\mu - j_n^1 \equiv j^{(\mu,1)} \qquad \textrm{and} \qquad k_n^\mu - 2^{j^{(\mu,1)}} k_n^1 \equiv b^{(\mu,1)} \qquad \textrm{for all}\ n\in \N\ ,$$
and $\l_n^\mu = (i_\mu,j_n^\mu, k_n^\mu)$ for some $i_\mu \in \{1,\dots,2^d-1\}$ for all $\mu,n\in\N$.
\end{enumerate}
Then $\tilde\phi_n := \sum_{\mu\in \N} a^\mu \psi_{\l_n^\mu} \in L^p(\R^d)$ for any $n\in \N$ and there exists $\phi \in L^p(\R^d)$ such that $\tilde\phi_n \equiv \tau_{n,1}\phi$ for all $n\in \N$.
\end{lemma}
\ \\
\noindent
\textbf{Proof:}
\\\\
Note that
$$
\begin{array}{rcl}
\d_n^\mu & = & \{2^{j_n^\mu}x - k_n^\mu \in [0,1)^d\ \} \\\\
& = & \{2^{j_n^1 + j^{(\mu,1)}}x - 2^{j^{(\mu,1)}}k_n^1 - b^{(\mu,1)} \in [0,1)^d\ \} \\\\
& = & \{2^{j^{(\mu,1)}}( 2^{j_n^1}x -k_n^1) - b^{(\mu,1)} \in [0,1)^d\ \} \\\\
& = & \{\tau^{(\mu,1)}(\tau_{n,1}(x)) \in [0,1)^d\ \}\\\\
& = & (\tau_{n,1})^{-1}[ (\tau^{(\mu,1)})^{-1}([0,1)^d)]
\end{array}
$$
where $\tau^{(\mu,1)}(x) := 2^{j^{(\mu,1)}}x - b^{(\mu,1)}$.  Defining $\d^{(\mu,1)} := (\tau^{(\mu,1)})^{-1}([0,1)^d)$, we can now easily see that
$$\chi_{\d_n^\mu}(x) = \chi_{\d^{(\mu,1)}}(\tau_{n,1}(x))\ .$$
\pagebreak
Using now the change of variables $y = \tau_{n,1}(x) = 2^{j_n^1}x - k_n^1$, we see that
$$I_n := \int \left\{ \sum_\mu |a_n^\mu|^2 2^{\frac{2d}{p}j_n^\mu} \chi_{\d_n^\mu}(x)   \right\}^{{p/2}} dx =
\int \left\{ \sum_\mu |a_n^\mu|^2 2^{\frac{2d}{p}j_n^\mu} \chi_{\d^{(\mu,1)}}(\tau_{n,1}(x))   \right\}^{{p/2}} dx =
$$
$$=\int \left\{ \sum_\mu |a_n^\mu|^2 2^{\frac{2d}{p}j^{(\mu,1)}} \chi_{\d^{(\mu,1)}}(y)   \right\}^{{p/2}} dy
$$
Note now that the only dependence on $n$ in the last integral is in the terms $a_n^\mu$.
\\\\
Fix first any $y\in \R^d$ such that the following is finite for every $n$ (it is true almost everywhere because $\phi_n \in L^p$ implies $I_n$ is finite for each $n$, and a countable union of sets of measure zero has measure zero), and write
$$\sum_\mu |a_n^\mu|^2 2^{\frac{2d}{p}j^{(\mu,1)}} \chi_{\d^{(\mu,1)}}(y) = \int_{\N} |a_n^\mu|^2 2^{\frac{2d}{p}j^{(\mu,1)}} \chi_{\d^{(\mu,1)}}(y) d\alpha (\mu)$$
where $\alpha$ is the discrete measure.  By Fatou's Lemma, we have
$$\int_{\N} \left\{\liminf_{n\to\infty} |a_n^\mu|^2 2^{\frac{2d}{p}j^{(\mu,1)}} \chi_{\d^{(\mu,1)}}(y)\right\} d\alpha (\mu) \leq \liminf_{n\to\infty} \int_{\N} |a_n^\mu|^2 2^{\frac{2d}{p}j^{(\mu,1)}} \chi_{\d^{(\mu,1)}}(y) d\alpha (\mu)\ ,$$
hence
$$\left\{ \sum_m |a^\mu|^2 2^{\frac{2d}{p}j^{(\mu,1)}} \chi_{\d^{(\mu,1)}}(y)\right\}^\frac{p}{2} \leq \left\{ \liminf_{n\to\infty} \sum |a_n^\mu|^2 2^{\frac{2d}{p}j^{(\mu,1)}} \chi_{\d^{(\mu,1)}}(y)\right\}^\frac{p}{2} =$$$$=  \liminf_{n\to\infty} \left\{ \sum |a_n^\mu|^2 2^{\frac{2d}{p}j^{(\mu,1)}} \chi_{\d^{(\mu,1)}}(y)\right\}^\frac{p}{2} \ .$$
Integrating now on both sides of the previous inequality in $y$ and applying Fatou's lemma again with respect to the usual Lebesgue measure on $\R^d$, we have
$$I := \int \left\{ \sum_\mu |a^\mu|^2 2^{\frac{2d}{p}j^{(\mu,1)}} \chi_{\d^{(\mu,1)}}(y)   \right\}^{{p/2}} dy \leq \liminf_{n\to\infty} I_n$$
and we know by equivalence of the wavelet norms and the usual Lebesgue norms that $I_n \lesssim \|\phi_n\|_p^p \leq C$ hence ${I \lesssim C <\infty}$.  Changing back to the variable $x$ in $I$ (which gives the same formula as the original definition of $I_n$ as an integral in $x$ only with $a_n^\mu$ replaced by $a^\mu$), we see again by the equivalence of norms that
$\|\tilde{\phi}_n\|_p^p \lesssim I < \infty$.  (If one is uncomfortable considering the wavelet expansion of a potentially non-existence element of $\lprd$, one can alternatively show that the relevant sequences of partial sums are Cauchy by similar calculations to obtain the existence of $\tilde \phi_n$.)
\\\\
It is obvious from the setup that $\tau_{n,1}^{-1} \tilde{\phi}_n$ is independent of $n$, hence the last statement in the claim.  Indeed, we must have
$$\tau_{n,1}^{-1} \tilde{\phi}_n = \sum_\mu a^\mu \tau^{(\mu,1)}\psi^{(i_\mu)}$$
and the lemma is proved.
\hfill $\Box$

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TITLE: Prove a Sequence is Cauchy
QUESTION [4 upvotes]: Prove that sequence {(2n+1)/n} is Cauchy.
I understand the definition of a Cauchy sequence; however, I'm not sure how to find the necessary value of N to satisfy the prove.
I know that you can simply proof that the sequence is Cauchy by stating that it converges to 2. But, for this specific problem we are asked to use strictly the definition of a Cauchy sequence in writing the proof.
Thanks for the help in advance.

REPLY [3 votes]: There is a theorem"a sequence is convergent iff it is a Cauchy sequence"
so it needs to prove that the given sequence is convergent which emphasises that it is Cauchy.$$a_n=2+\frac{1}{n}\to 2$$i.e. the sequence is convergent and hence it is Cauchy sequence


TITLE: using trig sub vs u-sub with $ \int_0^1 x^3 \sqrt{1 - x^2} \, dx $?
QUESTION [0 upvotes]: I'm having a hard time understanding when I should use $u$-sub rather than trig sub. I have the problem
$$
\int_0^1 x^3 \sqrt{1 - x^2} \, dx
$$
My initial attempt was to use $x = \sin u$ as my trig sub but this was wrong. I didn't think to use $u$-sub as the $dx$ and $du$ wouldn't cancel out nicely. But the answer says to just use the $u$ equation and solve for $x$ then plug that into the $x^3$.
This brings me to my next question. If I have an square root integral and both $a$ and $x$ are $1$, do I need to use $u$-sub rather than trig sub?

REPLY [1 votes]: Letting $u=x^2$ transforms the integral into $$
\begin{aligned}
\int_0^1 x^3 \sqrt{1-x^2} d x &=\int_0^1 x^2 \sqrt{1-x^2} \cdot x d x \\
&=\frac{1}{2} \int_0^1 u \sqrt{1-u} d u
\end{aligned}
$$
Decomposing the integrand into two yields
$$
\begin{aligned}
I &=\frac{1}{2} \int_0^1[1-(1-u)] \sqrt{1-u} d u \\
&=\frac{1}{2} \int_0^1 \sqrt{1-u} d u-\int_0^1(1-u)^{\frac{3}{2}} d u \\
&=\frac{1}{2}\left[-\frac{2}{3}(1-u)^{\frac{3}{2}}+\frac{2}{5}(1-u)^{\frac{2}{2}}\right]_0^1\\&=\frac{2}{15}
\end{aligned}
$$

For fun, we can do it by integration by parts.
$$
\begin{aligned}
I &=-\frac{1}{3} \int_0^1 x^2 d\left(1-x^2\right)^{\frac{1}{2}} \\
&=-\left[\frac{x^2\left(1-x^2\right)^{\frac{3}{2}}}{3}\right]_0^1+\frac{2}{3} \int_0^1 x\left(1-x^2\right)^{\frac{3}{2}} d x \\
&=-\frac{2}{3}\left[\frac{1}{5}\left(1-x^2\right)^{\frac{5}{2}}\right]_0^1=\frac{2}{15}
\end{aligned}
$$


TITLE: Topology on cartesian product and product topology
QUESTION [0 upvotes]: Let X and Y be sets and let $\tau$ be topology on X$\times$Y. Then there are $\tau_1$ and $\tau_2$ are topologies on X and Y , respectively such that $\mathcal B$={U$\times$V : U$\in$$\tau_1$ and V$\in$$\tau_2$} is basis for $\tau$ or not ? If it is true then I want to proof. But if not then I want to example.

REPLY [2 votes]: First of all.If you have topologies $T_1,T_2$ of $X,Y$ respectively then the $\mathcal B$={U$\times$V : U$\in$$T_1$ and V$\in$$T_2$} is a basis for a topology in $X \times Y$ which is called product topology.
You ask the other direction.Just have in mind that for every basis there is a unique topology and for a topology there isn't a unique basis.So what you ask doesn't really stand.
For example if $X=Y=\Bbb R$ then $X \times Y=\Bbb R^2$ which is homeomorphic to $\Bbb C$ and the open sets there are the open disks and with your sayings you make it more difficult because then the disk would be described as the union of two product open intervals of $\Bbb R$,which is more complicated.
I mean that if you have something more complicated than $\Bbb C$ how would you break it in a product of Spaces?


TITLE: if $\int _a^{\infty }\left|f\left(x\right)\right|\:dx$ converges then $\int _a^{\infty }f\left(x\right)\:dx$ also converge.?
QUESTION [2 upvotes]: if $\int _a^{\infty }\left|f\left(x\right)\right|\:dx$ converges then $\int _a^{\infty }f\left(x\right)\:dx$ also converge.
I wonder what i missing. we proved in our course thar for every $g$ >$f$>0 if $\int _a^{\infty }f\left(x\right)\:dx$ converges than $\int _a^{\infty }g \left(x\right)\:dx$ also converges.
i can't just say $\left|f\right|\:>\:f$ and use this theorem?

REPLY [1 votes]: Since $f$ is not necessarily non-negative you cannot use this theorem. However, notice that for any $A\lt B$,
$$\left|\int_a^A f(x)dx-\int_a^Bf(x)dx\right|\leqslant \int_A^B|f(x)|dx$$
and by the convergence of $\int_a^{+\infty}|f(x)|dx$, we conclude that the limit $\lim_{A\to +\infty}\int_a^Af(x)dx$ exists.
If you want to use the result $0\leqslant f\leqslant g$, notice that $0\leqslant f^+\leqslant f$ and the same with $f^-$. We deduce that $\int_a^{+\infty}f^+(x)dx$ and $\int_a^{+\infty}f^-(x)dx$ are convergent and since $f=f^+-f^-$ that $\int_a^{+\infty}f(x)dx$ is convergent.


TITLE: Why does the co-efficent of friction between plane and body does change depending on the angle of the plane?
QUESTION [1 upvotes]: First of all I know there is a similar question to this one. However, it didn't help me resolve my problem.
Let's say there is a body on a angled plane and the system is in equilibrium (the body doesn't move). The forces on the body are friction, normal force and weight.
$\mu_s$ and $\mu_k$ are the same in my case (and in the question that I try to solve).
I chose that the axis $x$ is parallel to the plane.
Thus from First Law of Newton,
$$\Sigma F_x = mg\sin\alpha - f = 0$$
$$\Sigma F_y = N - mg\cos\alpha = 0$$
From here I get that
$$N = mg\cos\alpha$$
$$f = \mu N = mg\sin\alpha$$
I put what I found about $N$:
$$\mu mg\cos\alpha = mg\sin\alpha$$
From here I can say that
$$\mu = \frac{\sin\alpha}{\cos\alpha} =\tan\alpha$$
I found that $\mu$ depends in the angle of the plane. Nevertheless, I read that $\mu$, as friction co-efficent, will never change.
What are my mistakes or misunderstandings?

REPLY [1 votes]: There are two mistakes in your thinking.  Considering static friction first, $\alpha$ is the minimum angle at which the body first starts to slide; if $\tan\alpha$ is less than $\mu_s$, the body won't slide at all.  Considering kinetic friction, when $\tan\alpha=\mu_k$, the body can be moving at constant velocity; at larger angles, it will be accelerating.


TITLE: Classes of finite non-abelian groups with non-trivial center .
QUESTION [1 upvotes]: I'm looking for an interesting class of finite non-abelian groups with non-trivial center whose representation theory is well-known. In particular, I'm interested in such groups with the additional condition that there is a linear character that is non-trivial on the center of the group.
For example: Consider $\mathcal{D}_{2n}=\left\langle a,b\mid a^n=b^2=1, bab=a^{n-1}\right\rangle$ where $n$ is divisible by $2$ once, then $\ker(\mathcal{D}_{2n})=\left\{1,a^{\frac{n}{2}}\right\}$. There are four linear characters determined by $a\mapsto \pm 1$ and $b\mapsto \pm 1$. The two characters that map $a$ to $-1$ have the value $-1$ on $a^{\frac{n}{2}}$ since $\frac{n}{2}$ is odd. Thus this class satisfies all conditions.
If some other class springs to mind, I would be very interested. I'm not interested in examples found by taking direct sums of easier classes.

REPLY [2 votes]: It is a fact that all the (complex valued) linear characters of a finite group $G$ are trivial exactly in the commutator subgroup $G':=[G,G]$. Because the target group $GL_1(\Bbb{C})=\Bbb{C}^*$ is abelian, the elements of $G'$
are in the kernel of all the 1-dimensional characters. The reverse inclusion follows from the fact that the quotient $G/G'$ is abelian, and a finite abelian group has enough 1-dimensional characters that only $1_G$ is in the intersection of their kernels.
So you want groups such that $Z(G)$ is not contained in the derived group $G'$. If $A$ is an abelian group, then this holds for groups of the form $G=H\times A$ with $H$ an arbitrary group. Here $G'=H'\times\{1_A\}$ but
$\{1_H\}\times A\subseteq Z(G)$, so the criterion is satisfied. Admittedly this is not a very exciting family of examples.

Another family of examples are the groups $G=GL_n(K)$, where $K$ is a finite field. Here $Z(G)$ consists of scalar matrices, and $G'\le SL_n(K)$ (more often than not there is equality). Unless $|K|-1$ is a factor of $n$ there will be scalar matrices with determinants $\neq1$.


TITLE: Finding primitive roots modulo $x$
QUESTION [0 upvotes]: I'm just starting with number theory and by now I know how to test whether a given number $\alpha$ is a primitive root mod $p$ or not. But I'm not sure yet how it works if $\alpha$ isn't given. I saw a small example where they ask you to find a primitive modulo $18$. So what was done is the following: $\phi(18) = 6$, so we look for an element of order 6. There are four possibilities: $5$, $7$, $11$ and
$13$. Can someone explain to me what $\phi(18) = 6$ means and how $5$, $7$, $11$ and $13$ are of order $6$?
Edit: So when I know these $4$ possibilities, I can easily test them but I'm not sure how to arrive at these possibilities.

REPLY [1 votes]: $\phi (18)=6$ means that there are six integers $n$ with $1\le n \le 17$ (we can automatically exclude $0\equiv 18\mod 18$) which are coprime to $18$ - they have no common factor with $18$.
These $6$ numbers have multiplicative inverses modulo $18$ and therefore form an abelian group - the multiplicative group of units modulo $18$. If this group is cyclic, then a primitive root modulo $18$ is a generator of this cyclic group.
[Note that if $n$ is a prime the group is always cyclic - for primes the order is $n-1$ - and there is always a primitive root. But if $n=8$ or $n=15$ for example the group is not cyclic.]
Now the cyclic group of order $6$ has just two generators. Here the six elements are $1,5,7,11,13,17$. We have, modulo $18$
$1=1$ has order $1$
$5^2=25\equiv 7; 5^3\equiv 5\times 7\equiv 35\equiv 17\equiv -1$
Note that it can be very useful to introduce negative numbers into the calculations to keep arithmetic simple. Now the order of an element of a group of order $6$ must divide the order of the group, so must be $1,2,3,6$ and for the element $5$ we have excluded $1,2,3$ and it follows that $5$ must have order $6$ and therefore is a primitive root. We have $5^4\equiv 5\times -1=-5\equiv 13, 5^5\equiv 5\times 13\equiv 65\equiv 11, 5^6\equiv 5\times 11=55\equiv 1$.
$7^2=49\equiv 13; 7^3\equiv 7\times 13\equiv 91\equiv 1$ so that $7$ has order $3$ and is not a primitive root.
$11^2=121\equiv 13; 11^3\equiv 143\equiv -1$ so that $11$ is a primitive root for the same reasons as $5$.
$13^2=(-5)^2\equiv 25\equiv 7; 13^3\equiv 13\times 7\equiv 1$ so $13$ has order $3$ and is not a primitive root (note how using negative numbers made this easier)
$17^2\equiv (-1)^2\equiv 1$ so $17$ has order $2$ and is not a primitive root.
We could have used information about the structure of the cyclic group to make this easier. In general finding primitive roots is not easy - there are methods which will help, but no ways to guarantee a quick result.
For example, here we know that $17\equiv -1$ has order $2$, and if we find an element $a$ of order $3$ before we find an element of order $6$ we know that multiplying an element of order $3$ by an element of order $2$ in an abelian group will give us an element of order $6$, so we identify $-a$ as a primitive root.
Here $13$ has order $3$ and $-13\equiv 5$ has order $6$. Also $7$ has order $3$ so that $-7\equiv 11$ has order $6$.


TITLE: Wave function of free particle
QUESTION [2 upvotes]: According to wiki, the wave function for a free particle is:
$$ \psi(\mathbf{r}, t) = Ae^{i(\mathbf{k}\cdot\mathbf{r}-\omega t)} $$
that with the necessary restrictions and uni-dimensional became:
$$ \psi(\mathbf{r}, t) = Ae^{i k \left( x-\frac{h}{2m\lambda} t \right) } $$

this wave function must fulfill normalization, thus
$$|A|^2 x \rvert_{-\infty}^\infty = 1 $$
something impossible for any $A $?

to which kind of particles is this wave function applicable ?

In case it is applicable to photons, we could compare with the electromagnetic waves:

$$ Ae^{i k \left( x-vt \right) } $$
(with $v=c$ in empty space)
we can see that in newtonian $v$ is a term related to the medium, being a universal constant in the case of empty space, while in quantum it is $\frac{h}{2m\lambda}$, something that depends only in particle characteristics $m$ and $\lambda$. How to match both facts?

REPLY [1 votes]: About your first question, the wave function of a free particle does not belong to the vector space $L_2(\mathbb{R})$ of the functions having the square modulus integrable in $\mathbb{R}$. Formally it lives in a generalized vector space, whose elements are known as"distributions" in mathematical literature. I don't remember the mathematical details.
However, physically this reflects the Heisemberg's principle. In this representation you know exactly the momentum of the particle, which is $\hbar k$ (simply apply the momentum operator $-i\hbar \partial/\partial x$ to the function and check what you get); so Heisemberg's principle states that you have total uncertainty on the position, and indeed the probability distribution in space is a constant $|A|^2$. The non integrability is a pathology in the formalism due to the fact that you don't have any uncertainty in the momentum value.
Coming to the second question: this can be applied to any free particle if you take the general form $Ae^{i(kx-\omega_k t)}$, where $\omega_k$ is the dispersion relation, which depends on the specific problem. For example this is a solution of Schrodinger equation with no external potential if $\omega_k = \hbar k^2/2m$.
The problem with photons is that, being fully relativistic ($m=0$) objects they are not correctly described by the Schrodinger equation. In fact the correct dispersion relation for phonons is NOT $\omega_k=\hbar k^2/2m$ clearly, because this would not make sense given $m=0$. Instead it is $\omega_k = c k $, as we already know for electromagnetic waves; but this would be consistent with a fully relativistic generalization of Schrodinger equation. (And that was question 3 :) Hope this helped!)


TITLE: Doubling metrics, doubling measures, Lebesgue density
QUESTION [2 upvotes]: As stated in this question,
Lebesgue differentiation theorem holds on locally doubling space?
and proved here,
http://www.math.uiuc.edu/~tyson/595f15lecture2.pdf
the Lebesgue differentiation theorem (LDT) holds in doubling measure spaces.
It is also known that every complete doubling metric supports a doubling measure:
http://www.ams.org/journals/proc/1998-126-02/S0002-9939-98-04201-4/S0002-9939-98-04201-4.pdf
[From this I conclude that if $\mu$ is a probability measure on a complete doubling metric space $(X,d)$ and $\mu\ll\nu$, where $\nu$ is a doubling measure (whose existence is guaranteed), then $\mu$ is itself a doubling measure and hence satisfies the LDT. EDIT: this is false, ignore.]
Does the LDT hold for probability measures on complete doubling metric spaces? (Can additionally assume $(X,d)$ is compact if needed.)

REPLY [3 votes]: A version of Lebesgue differentiation for non-doubling measures on doubling metric space appears in section 3 of:
T. Hytonen, A FRAMEWORK FOR NON-HOMOGENEOUS ANALYSIS ON METRIC SPACES, AND THE RBMO SPACE OF TOLSA
http://www.raco.cat/index.php/PublicacionsMatematiques/article/download/191394/387513
I leave it to your judgment whether this is of use to you.


TITLE: Existence of a cubic planar graph with one hexagonal face and four square faces.
QUESTION [3 upvotes]: I'm currently playing around with Euler's Formula and found the following for cubic planar graphs:
$$
\sum_{k=1}^F f_k = 6F-12,
$$
where $f_k$ is the degree of the $k$th face. I tried to apply this formula on graphs without triangles. The smallest example I found is:
$\hskip2.5in$
where $\sum_1^6 4=24=6\cdot 6 -12$. Adding one more face gives $30$ and now I'm trying for several hours (CPU time) to draw a graph $G$ with six faces of degree $4$ and one with $6$. An example $G'$ with five of degree $4$ and two of $5$ was right at hand, by extending the central square and the outer face to a pentagon.
$G'$ and $G$ both have (or should have) $F=7,\; E=15$ and $V=10$ due to Euler's formula. $G$ further is bipartite since it has only even degree faces. This means that the set of vertices splits into two sets of $5$ vertices.

Is it impossible to draw such a graph, and if so: Why?

REPLY [4 votes]: Let $G$ be a cubic planar graph with one hexagonal face and four square faces. By Euler's formula
$$V-E+F=2,$$
where $F=7$ and $E=\tfrac32V$, and hence $V=10$. Let $G_1$ and $G_2$ be the induced subgraphs on the vertices of the hexagonal face, and on the remaining $4$ vertices respectively. Without loss of generality $G_2$ lies inside the hexagon $G_1$. Because $G$ is cubic there are precisely $6$ edges going out from $G_1$, hence going into $G_2$. It follows that there are precisely $3$ edges in $G_2$. As $G$ does not contain triangular faces $G_2$ is either a claw or a path.
If $G_2$ is a claw then each of the three leaves is adjacent to two vertices in $G_1$. Then removing the root of the claw from $G$ we can smooth out the remaining three leaves of $G_2$ to get a new planar graph on the vertices of $G_1$. Because $G$ is triangle-free this smoothing out produces three distinct edges that are not edges of $G_1$. But there is no such planar graph, a contradiction.
If $G_2$ is a path then contracting the edges adjacent to the terminal vertices yields two vertices that are each adjacent to three distinct vertices of $G_1$. As this graph is planar, it is the following graph:

Because $G$ is triangle-free the two terminal vertices of $G_2$ are adjacent to non-adjacent vertices in $G_1$, i.e. to the vertices on the left and right in the picture. Then the two middle vertices of $G_2$ are adjacent to the top and bottom vertices of $G_1$ in the picture. This is impossible as $G$ is planar.
Ths shows that there is no cubic planar graph with one hexagonal face and four square faces.


TITLE: When is internal energy proportional to temperature?
QUESTION [1 upvotes]: A supervisor of mine claims that the following formula holds
$$
U = \rho V C T,
$$
where $U$ is internal energy, $\rho$ is density, $V$ is volume, $C$ is specific heat capacity and $T$ is temperature. I have tried to find this formula in the literature, but all I can find is the well known equation
$$
dU = \rho V C dT.
$$
I have mentioned this to my supervisor, and he replied that the latter equation can be derived from the first by assuming $(\rho V C)$ to be constant, for then $dU = d(\rho V C T) = \rho V C dT$ according to him. However, I am a bit alarmed by the fact that I am still unable to find the aforementioned equation for $U$ in any of the literature I have read so far. Could anyone here please settle the debate about whether or not the first equation above is generally valid? Also, if it is not generally valid, I would be interested in knowing if it is valid under any other assumptions than $(\rho V C)$ = constant. Thanks in advance!

REPLY [0 votes]: In classical systems, the internal kinetic energy is proportional to temperature. And also quadratic terms in the internal potential energy, according to the equipartition principle. This gives the Dulong & Petit law for the heat capacity of solids.
So that sounds general, but it cannot be true at phase transitions. It also breaks down at low temperatures when quantum effects are important, where the heat capacity must go to zero according to the Third law (Nernst's law).


TITLE: If $A$ is bounded, then $\partial A$ is bounded.
QUESTION [0 upvotes]: Let $A \subset \mathbb{R}^2 $, $A \neq \emptyset$.
If $A$ is bounded, then $\partial A$ is bounded.?

REPLY [0 votes]: In every neighbourhood of a boundary point there are points that belong to the set. If the elements of the set are all concentrated in a certain ball centered in $(0,0)$, the points in the boundary cannot be outside the closure of that ball.


TITLE: Photoemission - A surface process?
QUESTION [1 upvotes]: The square law of Fowler for photoemission into a vacuum is based on the assumption that the photoelectron current is proportional to the number of electrons available in the metal per unit volume with the velocity component normal to the metal surface.
With this volume assumption Fowler gives the expression for the photoelectron current as:
$$
I \propto \int_{u_0}^{\infty} \ln \left[1+e^{\left(E_{\mathrm{f}}-\frac{1}{2} m u^2\right) / k_{\mathrm{B}} T}\right] \mathrm{d} u
$$
and $E_{\mathrm{f}}$ is the Fermi energy of metal, $\Phi$ the work function of metal, $h v$ the radiation energy, $u$ the velocity of electron perpendicular to the surface, $m$ the mass of electron, $k_{\mathrm{B}}$ Boltzmann's constant.
The expression converted to a function of energy is:
$$
I_{\mathrm{p}} \propto \int_{E_{\mathrm{f}}+\Phi-N_{\mathrm{A}} h \nu}^{\infty} \ln \left[1+e^{\left(E_{\mathrm{f}}-E\right) / R T}\right] E^{-1 / 2} \mathrm{~d} E
$$
where $N_{\mathrm{A}}$ is Avogadro's number.
Now, this paper suggests that rather than considering the number of available electrons per unit volume as proportional to photoelectron current the number of electrons available per unit surface area. This may be done by multiplying the integrand in the first expression by $u$.
I do not understand why multiplying by a velocity allows us to consider the surface area rather than the volume process.

REPLY [1 votes]: The Matthews & Khan paper says the photoelectric current is proportional to the number of electrons per unit area per unit time, which are the correct units if a density per unit volume is multiplied by a velocity.
Photoelectric emission from surfaces is a surface effect, so it makes sense that the photoelectric current (charge per unit time) per unit area will depend on the number of electrons arriving at the surface per unit area per unit time.  The "surface" is the region where light can penetrate and photoelectric electrons can escape without scattering so much that they no longer have enough energy to leave the material.


TITLE: definition of metric space
QUESTION [1 upvotes]: from the actual definition of metric space ,we know that
metric space - a set of points such that for every pair of points there is a nonnegative real number called their distance that is symmetric and satisfies the triangle inequality
i am interested  what is a  symmetric distance?i know  triangle equality,something sum of two length is more then third one,but what about  symmetric distance?thanks in advance

REPLY [6 votes]: Saying that the metric (or distance) is symmetric just means that the distance from $x$ to $y$ is always the same as the distance from $y$ to $x$. In symbols, for all $x,y\in X$ we have $$d(x,y)=d(y,x)\;.$$


TITLE: Can we use a magnetic vector potential in the case of time varying $E$-fields?
QUESTION [0 upvotes]: Most discussions of the magnetic vector potential defined through $\mathbf{B}=\nabla\times\mathbf{A}$ are only for working with static electric fields (for example, Griffiths:

If we instead require that
$$\nabla\times\mathbf{B}=\nabla\times(\nabla\times\mathbf{A}) = \nabla(\nabla\cdot\mathbf{A})\ -\nabla^2\mathbf{A} = \mu_0\mathbf{J} +\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}$$
using the full form of Ampere's law, then we can still have the freedom to choose an $\mathbf{A}$ with zero divergence, giving us a form of Poisson's equation:
$$\nabla^2\mathbf{A} = -\mu_0\mathbf{J} -\mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}$$
Which can be solved in principle using analogies to the electric scalar potential $V$.
Why is this form of the magnetic vector potential in its full generality seldom seen? Is there a contradiction which prevents its use?

REPLY [1 votes]: Yes it can be used,
Given,$ \nabla × A = B$
Then
$\nabla ×E = -\frac{\partial B}{\partial t}$
$\nabla × E = -\frac{\partial (\nabla × A)  }{\partial t}$
$\nabla × E = -\nabla × \frac{\partial  A }{\partial t}$
$\nabla × E  + \nabla × \frac{\partial  A }{\partial t} = 0$
$\nabla × (E  +  \frac{\partial  A }{\partial t})= 0$
Because the curl of this quantity is zero, it can be written as the gradient of a scalar function ( or the negative of a gradient of a scalar function, which is used to match the definition of electrostatic potential)
$E  +  \frac{\partial  A }{\partial t} = -\nabla V$
$E = -\nabla V  -\frac{\partial  A }{\partial t} $
From here we can substitute the definitions of E into the other maxwell equations
This will obtain 2 equations that interlink A and V in terms of the source terms $\rho$ and J.
using the "lorenz" gauge choice ( not lorentz) we can decouple these and solve easier( or using the coulomb gauge)
This is called the potential formulation of maxwells equations


TITLE: prove Bézout's Theorem: few axioms, restricted language
QUESTION [3 upvotes]: I take Bézout's theorem to be the statement that given two different irreducible algebraic curves in two-dimensional projective space over an arbitrary field $K$, of degrees $m$ and $n$, the number of points of intersection does not exceed $mn$.
The only proof I know assumes that $K$ is infinite, so that via a projective transformation you can prevent $[1,0,0]$ from sharing the same line with any pair of distinct points of intersection.
Could I prove the same theorem in a more restricted language: perhaps one in which the infinite size of $K$ or the possibility of field extensions cannot be expressed?

REPLY [2 votes]: I'm confused about what you're looking for. You state a version of Bézout's theorem in a way that applies to an arbitrary field:

Given two different irreducible algebraic curves in two-dimensional projective space over an arbitrary field $K$, of degrees $m$ and $n$, the number of points of intersection does not exceed $mn$.

For clarity, I'll call this statement Bézout's theorem. You say that you only know how to prove this for infinite fields. That's fine. Here's where I get confused:

Could I prove the same theorem in a more restricted language: perhaps one in which the infinite size of  or the possibility of field extensions cannot be expressed?

It doesn't really matter whether a language can express the infinite size of $K$, does it? What matters is whether your proof uses the assumption that $K$ is infinite. And since the proof you have in mind uses this assumption, it's not a proof of the full statement of Bézout's theorem, and it can't be used to "automatically" give a proof.
Of course, Bézout's theorem is true for arbitrary fields. Here's a proof that bootstraps from your proof. Fix two irreducible algebraic curves in two-dimensional projective space over an arbitrary field $K$. Embed $K$ in its algebraic closure $\overline{K}$. Now $\overline{K}$ is infinite, so by your proof, the number of points of intersection in $\mathbb{P}^2(\overline{K})$ is at most $mn$. But every point of intersection in $\mathbb{P}^2(K)$ is also a point of intersection in $\mathbb{P}^2(\overline{K})$, so $mn$ is also an upper bound over $K$.
It seems like you were looking for some kind of abstract "logical" approach to transferring the result from infinite fields to arbitrary fields... one way to view the proof I gave above is to observe that Bézout's theorem can be stated as a family of universal sentences (sentences of the form $\forall \overline{x}\, \varphi(\overline{x})$, where $\varphi(\overline{x})$ is quantifier-free) in the first-order language of fields, one sentence for each pair of degrees $(m,n)$. Now universal sentences are preserved under substructure: e.g. if a universal sentence is true in some field, it's automatically true in any subfield. Since every field is a subfield of an infinite field, the statement of Bézout's theorem transfers from infinite fields to arbitrary fields.