TITLE: Why $\{A_n \quad a.a.\}:=\bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k$ is not defined as $\lim_{n\rightarrow\infty} \bigcap_{k=n}^\infty A_k$?
QUESTION [0 upvotes]: Given a probability tripple $(\Omega, F, P)$ and infinite many events $A_1, A_2, \dots, \in F$, a new event "almost always" in $F$ is normally defined as:
$$\{A_n \quad a.a.\}:=\bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k$$
I'm wondering, why it is not defined as
$$\{A_n \quad a.a.\}:=\lim_{n\rightarrow\infty} \bigcap_{k=n}^\infty A_k$$
? The two definitions look equivalent to me and the latter is easier to understand.

REPLY [2 votes]: The first definition is easiest to understand in my opinion. It describes all those sets for which only finitely many $A_n$ are not in $F$.

Regarding your suggestion the set-theoretic
  limit is defined
  as follows. Let the limit infinum and limit supremum be defined as
  \begin{equation*} \liminf_{n\rightarrow \infty}A_n = \bigcup_{n \geq
 1} \bigcap_{j \geq n} A_j \qquad \limsup_{n\rightarrow \infty}A_n =
\bigcap_{n \geq 1} \bigcup_{j \geq n} A_j. \end{equation*} If the two
  coincide the set theoretic limit exists and is equal to the resulting
  set.

You can see that the first definition for "almost always in $F$" is equal to $\liminf_{n\rightarrow \infty}A_n$. As pointed out in the comments $B_n = \bigcap_{j \geq n} A_j$ is an increasing sequence and thus $$\lim_{n\rightarrow \infty} B_n = \liminf_{n\rightarrow \infty}B_n = \bigcup_{n \geq
 1} \bigcap_{j \geq n} B_j = \bigcup_{n \geq
 1} B_n = \bigcup_{n \geq
 1} \bigcap_{j \geq n} A_j= \liminf_{n\rightarrow \infty}A_n.$$


TITLE: Norm, Euclidean Space and Distance
QUESTION [0 upvotes]: To a complete layman, how would you define the following terms intuitively?
$norm$ ,
$euclidean$ $space$ ,
and $euclidean$ $distance$ ?
Note: I have tagged Linear Algebra and Probability Theory because I am trying to read about stochastic convergence which has Euclidean distance mentioned in the source.

REPLY [0 votes]: Norm (I'm assuming Euclidean): length, as you're used to measuring it with a measuring tape.
Euclidean space: a space in which you measure the length in such a way, although it may have more dimensions than you're used to.
Euclidean distance: length of the (shortest) line between the objects in question.


text:
\section{Heterogeneous Treatment Effects}\label{sec:hte}
Now we apply ORF to the problem of estimating \emph{heterogeneous
	treatment effects}.  We will consider the following \emph{extension}
of the \emph{partially linear regression (PLR)} model due to
\citet{Robinson}.
\footnote{The standard PLR model \cite{Robinson} considers solely the case of constant treatment effects, $Y = \ldot{\theta_0}{T} + f_0(X, W) + \eps$, and
the goal is the estimation of the parameter $\theta_0$.} We have $2n$ i.i.d. observations
$D = \{Z_i = (T_i, Y_i, W_i, X_i)\}_{i=1}^{2n}$ such that for each
$i$, $T_i$ represents the treatment applied that can be either
real-valued (in $\mathbb{R}^p$) or discrete (taking values in
$\{0, e_1, \ldots , e_p\}$, where each $e_j$ denotes the standard
basis in $\RR^p$), $Y_i\in \RR$ represents the outcome,
$W_i\in [-1, 1]^{d_\nu}$ represents potential confounding variables
(controls), and $X_i\in \cX = [0, 1]^d$ is the feature vector that
captures the heterogeneity. The set of parameters are related via the
following equations:
\begin{align}
Y &= \ldot{\mu_0(X, W)}{T} + f_0(X, W) + \eps, \label{maineq}\\
T &= g_0(X, W) + \eta, \label{confounding}
\end{align}
where $\eta, \eps$ are bounded unobserved noises such that
$\Ex{}{\eps \mid W, X, T} = 0$ and
$\Ex{}{\eta \mid X, W, \eps} = 0$.\swcomment{added the last $\eps$}
In the main equation \eqref{maineq},
$\mu_0\colon \RR^d\times \RR^{d_\nu}\rightarrow [-1, 1]^p$ represents
the treatment effect function. Our goal is to estimate
\emph{conditional average treatment effect} (CATE) $\theta_0(x)$
conditioned on target feature $x$:
\begin{equation}
\textstyle{\theta_0(x) = \Ex{}{\mu_0(X, W)\mid X=x}}.
\end{equation}

The confounding equation \eqref{confounding} determines the
relationship between treatments variable $T$ and the feature $X$
and confounder $W$.
To create an orthogonal moment for identifying $\theta_0(x)$, we follow the classical \emph{residualization} approach similar to \cite{doubleML}. First, observe
that
\begin{equation*}
Y - \Ex{}{Y \mid X, W} = \ldot{\mu_0(X, W)}{  T - \Ex{}{T\mid X, W}} + \eps
\end{equation*}
Let us define the function $q_0(X, W) = \Ex{}{Y\mid X, W}$, and
consider the residuals
\begin{align*}
\tilde Y &= Y - q_0(X, W)\\
\tilde T &= T - g_0(X, W) = \eta
\end{align*}
 Then we can simplify the equation
as $\tilde Y = \mu_0(X, W) \cdot \tilde T + \eps$. \swedit{As long as
	$\eta$ is independent of $\mu_0(X, W)$ conditioned on $X$ (e.g.
	$\eta$ is independent of $W$ or $\mu_0(X, W)$ does not depend on $W$),
} we also have
$$\Ex{}{\mu_0(X, W) \mid X, \eta} = \Ex{}{\mu_0(X, W)\mid X} = \theta(X).$$
Since $\Ex{}{\eps \mid X, \eta} = \E\bb{\E\bb{\eps\mid X, W, T}\mid X, \eta}=0$, then
$$\Ex{}{\tilde Y \mid X, \tilde T} = \Ex{}{\mu_0(X, W) \mid X}\cdot \tilde T = \theta(X) \cdot \tilde T.$$
This relationship suggests that we can obtain an estimate of
$\theta(x)$ by regressing $\tilde Y$ on $\tilde T$ locally around $X=x$. We can thus define
the \emph{orthogonalized} score function: for any observation
$Z = (T, Y, W, x)$, any parameter $\theta\in \RR^p$, any estimates $q$
and $g$ for functions $q_0$ and $g_0$, the score $\psi(Z; \theta, h(X, W))$ is:
\begin{align*}
\left\{Y - q(X,W) -  \theta \left(T - g(X, W)\rangle\right)  \right\} (T - g(X, W)),
\end{align*}
where $h(X, W) = (q(X, W), g(X, W))$. In the appendix, we show that
this moment condition satisfies local orthogonality, and it identifies
$\theta_0(x)$ as long as as the noise $\eta$ is independent of
$\mu_0(X,W)$ conditioned on $X$ and the expected matrix
$\Ex{}{\eta \eta^\intercal \mid X=x}$ is invertible.  Even though the
approach applies generically, to obtain formal guarantees on the
nuisance estimates via our Forest Lasso, we will restrict their
functional form.

\paragraph{Real-valued treatments.}
Suppose $f_0$ and each coordinate $j$ of $g_0$ and $\mu_0$ are
given by high-dimensional linear functions:
$f_0(X, W) = \langle W, \beta_0(X) \rangle$,
$\mu_0^j(X, W) = \langle W, u_0^j(X) \rangle, \quad g_0^j(X, W) =
\langle W, \gamma_0^j(X)\rangle$, where
$\beta_0(X), \gamma_0^j(X), u_0^j(X)$ are $k$-sparse vectors in
$\RR^{d_\nu}$. Consequently,  $q_0(X, W)$ can be written as

a $k^2$-sparse linear function over degree-2 polynomial
features $\phi_2(W)$ of $W$. Then
as long as
$\gamma_0, \beta_0$ and $\mu_0$ are Lipschitz in $X$ and the
confounders $W$ satisfy $$\Ex{}{\phi_2(W) \phi_2(W)^\intercal\mid X} \succeq \Omega(1) I,$$
then we can use Forest Lasso to estimate both $g_0(x, w)$ and $q_0(x,w)$.
Hence, we can apply the ORF algorithm to get estimation error rates and asymptotic normality results
for $\hat{\theta}$.
\omitarxiv{(see Appendix~\ref{app:hte_cor} for formal statement).}
\omitcm{
\begin{corollary}[Accuracy for real-valued treatments]\label{cor:continuous}
  Suppose that $\beta_0(X)$ and each coorindate
  $u_0^j(X), \gamma_0^j(X)$ are Lipschitz in $X$ and have $\ell_1$
  norms bounded by $O(1)$ for any $X$. Assume that distribution of $X$
  admits a density that is bounded away from zero and infinity.  For
  any feature $X$, the conditional covariance matrices satisfy
  $\Ex{}{\eta \eta^\intercal \mid X} \succeq \Omega(1) I_p$,
  $\Ex{}{W W^\intercal \mid X} \succeq \Omega(1)I_{d_\nu}$ and
  $\Ex{}{\varphi_2(W) \varphi_2(W)^\intercal \mid X} \succeq
  \Omega(1)I_{d_\nu^2 + d_\nu}$, where $\varphi_2(W)$ denotes the
  degree-2 polynomial feature vector of $W$. Then with probability
  $1 - \delta$, ORF returns an estimator $\hat \theta$ such that
  \[
    \textstyle{\|\hat \theta - \theta_0\| \leq O\left(n^{\frac{-1}{2 + 2\alpha d}}\sqrt{\log(n d_{\nu}/\delta)} \right)}
  \]
  as long as the sparsity
  $k \leq O\left(n^{\frac{1}{8 + 8\alpha d}} \right)$ and the
  subsampling rate of ORF $s = \Theta(n^{\alpha d/(1 + \alpha d)})$.
  Moreover, for any $b \in \RR^p$ with $\|b\|\leq 1$, there exists a
  sequence $\sigma_n = \Theta(\sqrt{\polylog(n) n^{-1/(1+\alpha d)}})$
  such that
  \[
    \textstyle{\sigma_n^{-1} \ldot{b}{\hat\theta -\theta} \rightarrow_d \mathcal{N}(0, 1),}
  \]
  as long as the sparsity
  $k=o\left(n^{\frac{1}{8 + 8\alpha d}} \right)$ and the subsampling
  rate of ORF $s = \Theta(n^{\epsilon + \alpha d/(1 + \alpha d)})$ for any $\epsilon>0$.
\end{corollary}
}

\paragraph{Discrete treatments.} We now describe how our theory can be
applied to discrete treatments.  Suppose $f_0$ and each coordinate $j$
of $g_0$ are of the form: $f_0(X, W) = \langle W, \beta_0(X) \rangle$
and $g_0^j(X, W) = \cL(\langle W, \gamma_0^j(X)\rangle)$, where
$\cL(t) = 1/(1 + e^{-t})$ is the logistic function. Note in this case
$\eta$ is not independent of $W$ since
$$\Var(\eta_j) = g_0^j(X,W)(1 - g_0^j(X, W)).$$ To maintain the
conditional independence between $\mu_0(X, W)$ and $\eta$ conditioned
on $X$, we focus on the setting where $\mu_0$ is only a function of
$X$, i.e.  $\mu(X, W) = \theta(X)$ for all $W, X$. In this setting we
can estimate $g_0$ by running a forest logistic lasso for each
treatment $j$. Then we can estimate $q_0(x, W)$ as follows: For each
$t\in \{e_1, \ldots, e_p\}$ estimate the expected counter-factual
outcome function:
$m_0^t(x, W) = \mu_0^t(x, W) + f_0(x, W)$, by running a forest lasso between $Y$ and $X, W$ only among the subset of samples that received treatment $t$. Similarly, estimate $f_0(x, W)$ by running a forest lasso between $Y$ and $X, W$ only among the subset of samples that received treatment $t=0$.
\omitcm{Then observe that:
\begin{equation*}
q_0(x, W) = \sum_{t=1}^p (m_0^t(x, W) - f_0(x, w)) g_0^t(x, W) + f_0(x, W).
\end{equation*}}
\omitarxiv{Then observe that $q_0(x, W)$ can be written as a function of $f_0$, $g_0^t$ and $m_0^t$.}
Thus we can combine these estimates to get an estimate of
$q_0$. Hence, we can obtain a guarantee similar to that of
Corollary~\ref{cor:continuous} (see appendix).

\paragraph{Doubly robust moment for discrete treatments.} In the setting where $\mu$ also depends on $W$ and treatments are discrete, we
can formulate an alternative orthogonal moment that identifies the CATE even when $\eta$ is correlated with $\mu(X, W)$. This moment is based on first constructing unbiased estimates of the counterfactual outcome $m_0^t(X, W) = \mu_0^t(X, W) + f_0(X, W)$ for every observation $X, W$ and for any potential treatment $t$, i.e. even for $t\neq T$. The latter is done by invoking the doubly robust formula \cite{robins1995semiparametric,cassel1976some,kang2007demystifying}:
\begin{equation*}
\textstyle{Y^{(t)} = m_0^t(X, W) + \frac{(Y - m_0^t(X, W))\mathbf{1}\{T=t\}}{g_0^t(X, W)}}
\end{equation*}
with the convention that
\begin{align*}
g_0^0(X, W) &= 1 - \sum_{t\neq 0} g_0^t(X, W)\\
m_0^0(X,W)  &= f_0(X,W)
\end{align*}
Then we can identify the parameter $\theta^t(x)$ using the moment:
\omitarxiv{$\textstyle{\E[Y^{(t)}-Y^{(0)} | X=x] = \theta_t(x).}$}
\omitcm{\begin{equation}
\textstyle{\E[Y^{(t)}-Y^{(0)} | X=x] = \theta_t(x).}
\end{equation}}
One can easily show that this moment satisfies the Neyman
orthogonality condition with respect to the nuisance functions $m$ and
$g$ (see appendix). In fact this property is essentially implied by
the fact that the estimates $Y^{(t)}$ satisfy the double robustness
property, since double robustness is a stronger condition than
orthogonality\omitcm{(see e.g. \cite{Chernozhukov2016locally})}. We will again consider $\mu_0^j(X, W) = \langle W, u_0^j(X)
\rangle$. Then using similar reasoning as in the previous paragraph,
we see that with a combination of forest logistic lasso for $g_0^t$
and forest lasso for $m_0^t$, we can estimate these nuisance functions
at a sufficiently fast rate for our ORF estimator (based on this
doubly robust moment) to be asymptotically normal, assuming they have
locally sparse linear or logistic parameterizations.

\iffalse
\begin{assumpt}[Assumptions for CATE estimation]\quad\label{plrass}
	\begin{enumerate}
		\item $\theta_0\colon \RR^d \rightarrow \RR$ is 1-Lipschitz w.r.t
		$\ell_2$ norm.  The sparse vectors have bounded norms
		$\|\beta_0\|_1, \|\gamma_0\|_1 \leq O(k)$, and the noise terms have
		bounded scales $|\eta_i|, |\eps_i| \leq 1$.

		\item For any feature $X$, the conditional covariance matrix
		$\Ex{}{\eta \eta^\intercal \mid X}$ has eigenvalues bounded away from
		0.
		\item The distribution of $X$ admits a density that is bounded away
		from zero and infinity.  For any feature $X$, the conditional
		covariance matrix has minimum eigenvalue bounded away from 0, that
		is, there is some $\lambda$ with $1/\lambda_0 \leq O(1)$ such that
		$\Ex{}{W_i W_i^\intercal \mid x_i} \succeq \lambda_0 I_p$.
	\end{enumerate}
\end{assumpt}

The first assumption imposes standard Lipchitzness and boundedness
conditions, and the second assumption is useful for the kernel Lasso
analysis.

\begin{corollary}\label{thm:final-hetero-te}
	Under \Cref{plrass}, suppose ORF is instantiated with parameters
	$B\geq n/s$ and
	$s = \left(\frac{n}{k^2\ln{p}}\right)^{\frac{1}{2\tau + 1}}$ with
	$\tau = \frac{1}{2\alpha d}$ and
	$\alpha = \frac{\log(\omega^{-1})}{\pi\log((1 - \omega)^{-1}) }$,
	then the output estimate $\hat \theta$ by the kernel orthogonal
	random forest satisfies
	\[
	\Ex{}{|\hat \theta - \theta_0(x)|} \leq O\left( \left(\frac{\ln
		p}{n} \right)^{\frac{\tau}{{2\tau + 1}}} \right)
	\]
	as long as the support size of the linear functions $\beta_0$ and
	$\gamma_0$ satisfy
	$ k \leq O\left(\left(\frac{n}{\ln p} \right)^{\frac{\tau}{10\tau +
			4}}\right)$.
\end{corollary}
\fi

config: arxiv


TITLE: Partial order given by a stratification on a topological space.
QUESTION [2 upvotes]: A (good) stratification of a topological $X$ is a partition $\{X_i\}_{i\in I}$ ($I$ finite) of $X$, with $X_i$ locally closed, such that $X_i \cap \overline{X_j}\neq \emptyset \Rightarrow X_i \subset \overline{X_j}$. It follows that each $\overline{X_i}$ is the union of some $X_j$. We define the relation $"i \leq j$ if and only if $X_i \subset \overline{X_j}"$ on $I$. Is this a partial order relation? In other words, does $i \leq j$ and $j \leq i$ imply $i=j$? I don't know why, i cannot see it.

REPLY [1 votes]: The key is that if $X$ and $Y$ are (non-empty and) locally closed, and if $X\cap Y=\emptyset$, then it is impossible to have $X\subset \overline Y$ and $Y\subset \overline X$.
Proof:
Write $X=O\cap C$ with $O$ open and $C$ closed.
Since $X\cap Y=\emptyset$ and $Y\subset \overline{X}$, it follows that $Y\subset \left(\overline{X}\setminus X\right)$. But $\left(\overline{X}\setminus X\right)\cap O=\emptyset$ (easy exercise), and so $$Y\cap O=\emptyset$$
Therefore it is impossible for any point in $O$ (and therefore in $X$) to lie in the boundary of $Y$.


TITLE: About $K_{3,3}$ being planar or not
QUESTION [1 upvotes]: I wanted to propose a “proof” that graph $K_{3,3}$ is not planar.
Lets suppose it is planar, then by Euler’s theorem  $V-E+F=2$. Now since $V=6$ and edges $E=9$ , so $F=5$. Now there cannot be a 3-cycle, since in this case two vertices that were not supposed to be would be connected (this is not hard to prove but i will omit). Now $K_{3,3}$ is connected , since we can access to any vertice by jumping around(this is also not hard to prove i think). Now this graph must have cycles, since otherwise graph would be a tree but it isnt since $V-1$ aint $E.$ So there is a cycle , and this cycle must contain 2 vertices that are not connected directly to each other. There can be 3 choose 2 ways to form such cycles whcih means we would have 3 cycles which could produce 4 faces but we needed 5 faces, so that is contradictory.

REPLY [1 votes]: A fast argument to see that there are no $3$-cycles: $K_{3,3,}$ is bipartite, hence every face has an even number of edges, or at least four edges. After that, $F$ faces means at least $4F$ face-edge incidences, which means at least $2F$ edges. This already contradicts  $E=9$ and the  $F=5$ you obtained from Euler.


TITLE: How do you express "a tree is composed of leaves, branches and roots" using mathematical symbols?
QUESTION [1 upvotes]: I am just trying to learn how to express common things using mathematical symbols / equations. I would really love to get an intuitive grasp of how to take thoughts about the world and convert them into mathematical expressions. But I don't know how yet and so am wondering how you approach doing this, and/or what this would look like.
When I briefly think of a tree (like an oak tree), I think it has leaves, branches and roots. So what are the ways you would write that if you wanted to start formalizing it into a mathematical expression?
At first thought, it doesn't seem that "leaves are elements of a tree", like elements of a set, and there is more too the tree than just branches and leaves. So not sure how to begin writing that using mathematical symbols. Any ideas?

REPLY [2 votes]: I don't know if I understood correctly your question intent, but first of all, there isn't a unique way of doing this, probably many equivalent definitions can be formulated.
You may define an undirected graph $G=(V,E)$ with a set of vertices and a set of unordered pairs of vertices (the edges). Then, you may define a tree recursively: A tree is an undirected connected graph with a special vertice called $root$ and the root's adjacent subgraphs are also trees.
Then you may define the leaves as the tree vertices (often called nodes) of degree 1, except maybe the root. Branches are the graph's edges. Graphs and Trees are closely related, and Graphs rely heavly on sets and relations formalism.
But you asked about mathematical models, and while real world trees are conveniently modelled with mathematical trees many times, the model is highly dependent of your goals.
As a cartographer engineer, you may model trees in a terrain as cones or spheres...


TITLE: Double integral problem
QUESTION [1 upvotes]: I'm given that
$$
f(x,y)= \cases{
 \left(y-\tfrac{1}{2}\right)\left(x-\tfrac{1}{2}\right)^{-3} &  \text{if } \left|y-\tfrac{1}{2}\right|<\left|x-\tfrac{1}{2}\right|
\\
0 & \text{otherwise}
}
$$
I want to find the double integrals $$\int_0^1\int_0^1 f(x,y)\,dx\;dy$$ and $$\int_0^1\int_0^1 \left \lvert f(x,y)\right \rvert\,dx\;dy.$$
I've tried splitting each integral up into two integrals where the outer integral goes from $0$ to $\frac{1}{2}$ and from $\frac{1}{2}$ to $1$. But, I am having trouble determining what the inner integral bounds should be.

REPLY [0 votes]: It's not so hard as we're interested only in $\;x,y>0\;$ :
$$\begin{align*}&x,y\ge\frac12\implies y-\frac12<x-\frac12\implies y<x\\{}\\&0<x<\frac12\;,\;\;y\ge\frac12\;\implies\;y-\frac12<\frac12-x\implies y\le-x+1\\{}\\&0<x,y<\frac12\;\implies\;\frac12-y<\frac12-x\implies y>x\\{}\\&x\ge\frac12\,,\,0<y<\frac12\;\implies\;\frac12-y<x-\frac12\implies y>-x+1\end{align*}$$
Draw the above lines that determine the integration domain, and then we get:
$$\int\limits_0^1\int\limits_0^1 f(x,y)\,dxdy=\int\limits_0^{1/2}\int\limits_x^{-x+1} f(x,y) \,dydx+\int\limits_{1/2}^1\int\limits_{-x+1}^x f(x,y)\,dydx$$


TITLE: Functional recursion satisfying $a_{n+1} = p(a_n)$
QUESTION [1 upvotes]: Let $p(x)$ be a polynomial with integer coefficients such that $p(0) = p(1) = 1$. We define the sequence $a_0,a_1,a_2,\ldots,a_n$ that starts with an arbitrary integer $a_0$ and satisfies $a_{n+1} = p(a_n)$ for all $n \in \mathbb{N}$. Prove that the numbers $a_i$, taken two at a time, are relatively prime.

Since $p(0) = p(1) = 1$, we have $p(x) = x(x-1)g(x)+1$ where $g(x)$ is a nonzero polynomial with integer coefficients. How do we continue from here?

REPLY [2 votes]: If $f,g$ are two polynomials, then $f(g(x))= x h(x) + f(g(0))$.
In particular, $p(p(x))=x h(x)+p(p(0))=x h(x) + 1$. By induction, $p^n(x)=xh_n(x)+1$.
Then $a_{k+n}=p^n(a_k)=a_k h_n(a_k)+1$ and so $a_{k+n}$ and $a_k$ are coprime.

REPLY [0 votes]: Let's assume $p(x)=\sum_{k=0}^{n} a_kx^k$ and from $p(0)=a_0=1$ and $p(1)=\sum_{k=0}^{n} a_k=1$. Now $$p(p(x))=\sum_{k=0}^{n} a_kp(x)^k=\sum_{k=0}^{n} a_k \left( \sum_{i=0}^{n} a_ix^i \right)^k=$$
$$\sum_{k=0}^{n} a_k \left(1+ \sum_{i=1}^{n} a_ix^i \right)^k=\sum_{k=0}^{n} a_k \left(1+ xg(x)\right)^k=\sum_{k=0}^{n} a_k \left(1+xG_k(x)\right)=$$
$$\sum_{k=0}^{n} \left(a_k+a_kxG_k(x)\right)=\sum_{k=0}^{n} a_k+x\sum_{k=0}^{n} a_kG_k(x)=\sum_{k=0}^{n} a_k + xT_1(x)=1 + xT_1(x)$$
Now $p(p(p(x)))=p(1 + xT_1(x))=\sum_{k=0}^{n} a_k \left(1 + xT_1(x)\right)^k=1 + xT_2(x)$ similarly and so on by induction.
Then for any $a_i$ and $a_j$ such that $i<j$ we have $a_j=p(p(...p(a_i)...))=1+a_iT_{j-i}(a_i)$


TITLE: How to test the convergence of $\sum^{\infty}_{n=1} \frac{3^n(n-1)(-1)^n}{n^3}$?
QUESTION [2 upvotes]: So $\sum^{\infty}_{n=1} \frac{3^n(n-1)(-1)^n}{n^3}=\sum^{\infty}_{n=1} \frac{(-3)^n(n-1)}{n^3}$. Then I fail to continue. Is there a way to simplify it?

REPLY [1 votes]: The series diverge. It is comparable to sigma 3^n/n^2 and the terms of this approaches infinity as n goes to infinity.


text:
\begin{document}

\maketitle

\begin{abstract}
We show that convex viscosity solutions of the Lagrangian mean curvature equation are regular if the Lagrangian phase has H\"older continuous second derivatives.

\end{abstract}

\section{Introduction}
We establish regularity for convex viscosity solutions of the Lagrangian mean curvature equation
\begin{equation}
    F(D^{2}u)=\sum _{i=1}^{n}\arctan \lambda_{i}=\psi(x) \label{slag}
\end{equation}
under the assumption that $\psi:B_1(0)\to [0,n\pi/2)$ is in $C^{2,\alpha}(B_1(0))$ for some $\alpha\in(0,1)$. Here $\lambda_i$'s are the eigenvalues of the Hessian matrix $D^2u$ and then the phase $\psi$  becomes a potential for the mean curvature of the Lagrangian submanifold  $L=(x,Du(x))\subset \mathbb{C}^n$. The induced Riemannian metric $g$ can be written as
$g=I_n+(D^2u)^2
$ or
\[g_{ij}=\delta_{ij}+u_{ik}\delta^{kl}u_{lk}.
\]
In \cite[(2.19)]{HL}, the mean curvature vector $\vec{H}$ of this Lagrangian submanifold was shown to be
\begin{equation}
\vec{H}=J\nabla_g\psi \label{mean}
\end{equation}
 where $\nabla_g$ is the gradient operator for the metric $g$ and $J$ is the complex structure, or the $\frac{\pi}{2}$ rotation matrix in  $\mathbb{C}^n$. Note that by our assumption on $\psi$, $|H|$ is bounded.

When the phase is constant, denoted by $c$, $u$ solves the special Lagrangian equation
 \begin{equation}
\sum _{i=1}^{n}\arctan \lambda_{i}=c \label{s1}
\end{equation}or equivalently,
\[ \cos c \sum_{1\leq 2k+1\leq n} (-1)^k\sigma_{2k+1}-\sin c \sum_{0\leq 2k\leq n} (-1)^k\sigma_{2k}=0.
\]
 Equation (\ref{s1}) originates in the special Lagrangian geometry of Harvey-
Lawson \cite{HL}. The Lagrangian graph $(x,Du(x)) \subset \mathbb{C}^n$ is called special
when the argument of the complex number $(1+i\lambda_1)...(1+i\lambda_n)$
or the phase $\psi$ is constant, and it is special if and only if $(x,Du(x))$ is a
(volume minimizing) minimal surface in $(\mathbb{C}^n,dx^2+dy^2)$ \cite{HL}.

 A dual form of (\ref{s1}) is the Monge-Amp\`ere equation
\begin{equation}
    \sum_{i=1}^n \ln\lambda_i=c.\label{MA}
\end{equation}
This is the potential equation for special Lagrangian submanifolds in $(\mathbb {C}^n, dxdy)$ as introduced in \cite{Hi}. The gradient graph $(x,Du(x))$ is volume maximizing in this pseudo-Euclidean space as shown by Warren \cite{W}. In the 1980s, Mealy \cite{Me} showed that an equivalent algebraic form of the above equation is the potential equation for his volume maximizing special Lagrangian submanifolds in $(\mathbb C^n, dx^2-dy^2)$.

\smallskip
The regularity of solutions is a fundamental problem for these geometrically and analytically significant equations.  Our main result is
the following:
\begin{theorem}
\label{thm}
Let $u$ be a convex viscosity solution of \eqref{slag} on $B_1(0)\subset\mathbb R^n$ where $\psi\in C^{2,\alpha}(B_1)$.  Then $u\in C^{4,\alpha}(B_1)$, and we have a Hessian bound
\begin{align}
\label{hess_est}
|D^2u(0)|\le C(n, osc_{B_{1/2}}(u), \|\psi\|_{C^{2,\alpha}(B_{1/2})}).
\end{align}
\end{theorem}

\begin{remark}
The case where the phase $\psi$ has a lower regularity is dealt with in \cite{BS3}.
\end{remark}

\smallskip
The convexity of the special Lagrangian equation plays a dominant role
in its regularity theory.

The arctangent operator is concave if $u$ is convex, or if the Hessian has a lower bound $\lambda\ge 0$.  For smooth convex solutions of (\ref{s1}), Hessian estimates have been obtained by Chen-Warren-Yuan \cite{WYJ}. Recently in \cite{CSY}, Chen-Shankar-Yuan confirmed that convex viscosity solutions of \eqref{s1} are smooth. The semiconvex singular solutions constructed by Wang-Yuan \cite{WdY} show that the convexity condition is necessary.  Similarly, if in \eqref{s1} we have critical phase $|c|=(n-2)\pi/2$ or supercritical phase $|c|>(n-2)\pi/2$, then $F(D^2u)$ has convex level sets, but it was shown by Yuan \cite{YY0} that this fails for subcritical phases $|c|<(n-2)\pi/2$.  For critical and supercritical phases, Hessian estimates for \eqref{s1} have been obtained by Warren-Yuan \cite{WY9,WY} and Wang-Yuan \cite{WaY}, see also \cite{L}. Recently in \cite {AB1}, Bhattacharya proved Hessian estimates for solutions of (\ref{slag}) with a $C^{1,1}$ supercritical phase. For subcritical phases, $C^{1,\alpha}$ solutions of \eqref{s1} were constructed by Nadirashvili-Vl\u{a}du\c{t} \cite{NV} and Wang-Yuan \cite{WdY}.  Hessian lower bounds also play a role in the dual equation \eqref{MA}.  The Monge-Amp\'ere type equation $\det D^2u=h(x)$ has Pogorelov-type singular $C^{1-2/n}$ convex viscosity solutions whose graphs contain a line.  However, under the necessary \textit{strict} convexity assumption $\lambda>0$, interior regularity was obtained by Pogorelov \cite{P2} for a smooth enough right hand side $h(x)$, by Urbas \cite{U} if $h(x)$ is Lipschitz and $u(x)\in C^{(1-2/n)^+}$, and by Caffarelli \cite{Caf} if $h(x)$ is merely H\"older.

\smallskip
The convexity
of $u$ creates its own challenges in proving regularity, since it is unstable under smooth approximations of the boundary value.  To use the method of a priori estimates, one would solve the Dirichlet problem for a modified concave equation $\Tilde{F}(D^2u)=\psi(x)$, where $\tilde f(\lambda)=\arctan\lambda$ for $\lambda \geq 0$ and $=\lambda$ for $\lambda<0$.  If we approximate the boundary data and find the smooth solution of this problem, it may no longer be convex, and therefore lacks a connection to the original problem.  Although a priori estimates are available for \eqref{slag} and \eqref{s1}, estimates for $\tilde F$ are unknown.

\smallskip
In showing regularity for convex solutions of \eqref{s1}, the authors of \cite{CSY} managed to avoid a priori estimates altogether.  The basic idea was to change variables using the Lewy-Yuan rotation of the gradient graph, $\bar x+iD\bar u(\bar x)=e^{-i\pi/4}(x+iDu(x))$, such that the Hessian bounds decreased from $0\le \lambda\le+\infty$ to $-1\le\bar\lambda\le 1$. Since minimal graphical tangent cones with such bounds are flat, they were able to deduce regularity in new coordinates using ideas from \cite{Y}. One problem, however, is that defining such a rotation is unclear if convex $u$ is not $C^1$.  To adapt to the lower regularity setting, it was shown in \cite{CSY} that in the smooth case, $\bar u(\bar x)$ can be constructed using the Legendre transform, which is still well defined in the Lipschitz case.

\begin{comment}
\smallskip

\end{comment}

\smallskip
Extending the techniques used for the constant phase (\ref{s1}) to the variable phase equation (\ref{slag}) is difficult because of non-uniform ellipticity.

Interior regularity for solutions of (\ref{slag}) with a $C^{1,1}$ supercritical phase was recently obtained in \cite{AB}, and singular solutions are known if $\psi$ has
H\"older regularity \cite[Remark 1.1]{AB1}.  However, the existence of interior estimates for solutions of (\ref{slag}) with critical and supercritical phase where $\psi\in C^{1,\varepsilon_{0}}$, or even $|\psi|\geq (n-2)\frac{\pi}{2}$ where $\psi\in C^{1,1}$ is still an open question. The difficulty is the lack of a suitable Hessian lower bound $D^2u\ge -I/\delta$, since the linearized operator $DF(\lambda)$ in \eqref{slag} can now have arbitrarily small coefficients.  In the homogeneous case, we can differentiate \eqref{s1} and recover $D^3u$, but upon differentiating \eqref{slag}, we find extra terms involving $D\psi$, which, a priori, could be large compared to the coefficients $DF(\lambda)$ of $D^3u$.  This prevents us from finding Jacobi inequalities such as $\Delta_g\ln\lambda_{max}\ge |\nabla_g\ln\lambda_{max}|^2-C$, which require two differentiations of \eqref{slag}.

\smallskip
In proving Theorem \ref{thm}, we encounter new difficulties involving the new $\psi$ term.

For $\psi$ constant, the approach in \cite{CSY} required showing $\lambda_{max}<\infty$ everywhere by performing a coordinate rotation $\bar z=e^{-i\pi/4}z$ and showing that the new maximum eigenvalue $\bar\lambda_{max}<1$ everywhere.  This was done by showing that $\bar{b}:=\ln\sqrt{1+\bar\lambda_{max}^2}$ is a subsolution of the Laplacian $\Delta_{\bar g}$ on the rotated surface $(\bar x,D\bar u(\bar x))$, to which the strong maximum principle applies.  However, with the presence of the variable phase, extra terms such as $D^2\bar \psi(e_{max},e_{max})$ arise in the expression of $\Delta_g \bar{b}$.  Without close examination, it is very likely that these extra terms destroy the subsolution property of $\bar{b}$.  Remarkably, it turns out that these extra terms vanish at the maximum possible value $\bar\lambda_{max}=1$.  The reason is that the rotation $\bar x=x\cos\alpha+Du(x)\sin\alpha$ is an expanding map if $u$ is strictly convex, with reverse rotation $x(\bar x)$ contracting, so $\bar\psi(\bar x)\sim \psi(x(\bar x))$ changes slowly compared to $\psi(x)$ if $\lambda(D^2u)$ is large.

Since $\bar\lambda_{max}=1$ corresponds to the coordinate blowup $\lambda_{max}=\infty$, the rotated phase turns out to be stationary, and the extra terms vanish.

\begin{comment}
$$
\Delta_{\bar g}b\ge 0,
$$
where $\bar g=I+(D^2\bar u)^2$ is the induced metric on $\mathbb R^n\times\mathbb R^n$ in rotated coordinates $\bar z=e^{-i\pi/4}z$. From here, the strong maximum principle implies that $\bar\lambda_{max}<1$ everywhere.  However, in the variable phase case, it is not at all obvious that this idea still works.

Let $\bar\psi$ be $\psi$ in rotated coordinates.  It was shown in \cite[Proposition 4.1]{AB} that
the variable phase perturbations yield, at Hessian-diagonal points with $\bar\lambda_{max}=\bar \lambda_1=\bar u_{11}$,
\begin{align}
\label{Deltab}
\Delta_{\bar g}b=(\text{constant phase terms})+\frac{\bar \lambda_1}{1+\bar \lambda_1^2}\bar\psi_{\bar 1\bar 1}-\bar\lambda_1\langle \nabla_{\bar g}\bar\psi,\nabla_{\bar g}b\rangle_{\bar g}.
\end{align}
Again we have extra terms, so it is quite conceivable that we can only obtain an inequality of the form
$$
\Delta_{\bar g}b+C|\nabla b|\ge -C,
$$
which is useless for our application.  Quite unexpectedly, we obtain a good inequality:
$$
\Delta_{\bar g}b\ge \langle a(\bar x),\nabla_{\bar g}b\rangle_{\bar g}-c(\bar x)(b_{max}-b)^2,
$$
where $a$ and $c$ are locally bounded coefficients, and $b_{max}=b|_{\bar\lambda=\bar\lambda_{max}}$.  The reason is that rotation is a distance-increasing operation, so the phase $\bar\psi$ in rotated coordinates changes slower compared to its pre-rotated counterpart.  In particular, at a coordinate singularity, it will be stationary, and the bad terms will vanish.

\end{comment}

\smallskip
This paper is divided into the following sections.  In section \ref{sec:rot}, we formulate the Lewy-Yuan rotation for the Lipschitz potential $u(x)$.
In section \ref{sec:reg_baru}, we establish regularity of the rotated potential $\bar u(\bar x)$. In section \ref{sec:reg}, we

deduce regularity of the original potential $u(x)$, and in section \ref{sec:comp}, we prove the Hessian estimate \eqref{hess_est}.

\section{Rotation for Lipschitz potential}
\label{sec:rot}
We formulate Lewy-Yuan rotation for the convex function $u$ solving (\ref{slag}), using the idea introduced in \cite{CSY}; we refer to sections 2.1 and 2.2 in that paper for various details in this section. If $u(x)$ is smooth on $\Omega$, then the gradient graph $z=(x,Du(x))$ is a Lagrangian submanifold of $\mathbb C^n$, and if $u$ is convex, then the downward rotation $\bar z=e^{-i\pi/4}z$ yields another Lagrangian submanifold $(\bar x,D\bar u(\bar x))$ with smooth potential $\bar u(\bar x)$ on $\bar x(\Omega)$.  Because the canonical angles decrease by an angle of $\pi/4$
$$
\arctan\bar\lambda_i(D^2\bar u)=\arctan\lambda_i(D^2u)-\pi/4,
$$
it follows that the rotation sends solutions of \eqref{slag} to solutions $\bar u(\bar x)$ of another Lagrangian mean curvature equation, which now is uniformly elliptic by the Hessian bounds
\begin{align}
\label{bounds}
-I_n\le D^2\bar u\le I_n.
\end{align}
This makes regularity theory tractable in new coordinates using the ideas developed in \cite{Y}.

\smallskip
If we do not assume that $u$ is in $C^1$, then the subdifferential $(x,\pd u(x))$\footnote{
The slopes of tangent planes touching $u$ from below at $x$.  If $u(x)=|x|$, then $\pd u(0)=\bar B_1(0)$.
}
is not a graph over $\Omega$.  Nevertheless, if we rotate the subdifferential $(x,\pd u(x))$ downwards by $\pi/4$, then we still obtain a gradient graph $(\bar x,D\bar u(\bar x))$, where $\bar u\in C^{1,1}$ and is given explicitly by
\begin{equation}
\label{r}
    s\bar{u}(\bar{x})-\frac{c}{2}|\bar x|^2=-\left[su(x)+\frac{c}{2}|x|^2\right]^*(\bar x),\qquad \bar x\in \bar x(\Omega),
\end{equation}
where $c=\cos\pi/4$, $s=\sin\pi/4$, and
$$
f^*(\bar x)=\sup_{x}[x\cdot\bar x-f(x)],\qquad \bar x\in\pd f(\Omega)
$$
is the Legendre transform of convex $f(x)$; see \cite[Proposition 2.1]{CSY}.  The image domain $\bar\Omega=\bar x(\Omega)=\pd \tilde u(\Omega)$ is open and connected by \cite[Lemma 2.1]{CSY},
which comes from the fact that
\begin{equation}
\tilde u(x):=su+\frac{c}{2}|x|^2 \label{u1}
\end{equation}
is strictly convex.

\medskip
The transform \eqref{r} is order preserving, $f\le g\to \bar f\le \bar g$, hence, it preserves uniform limits.  It follows that $\bar u(\bar x)$ satisfies the same Hessian bounds \eqref{bounds} as in the smooth case.  We could then use interpolation to see that $D\bar u_n\to D\bar u$ locally uniformly provided $u_n\to u$ uniformly, but this also follows from strict convexity, as in \cite[Proposition 2.3]{CSY}, i.e. if $\bar x\in \pd \tilde u(a)\cap\pd \tilde v(b)$ for $u,v$ convex, then
\begin{align}
\label{xsmall}
|b-a|^2\le 2\sqrt 2(|\tilde u-\tilde v|(a)+|\tilde u-\tilde v|(b)).
\end{align}
The smallness of $|D\bar u-D\bar v|$ would then follow from reverse rotation $a=-s\bar x+cD\bar u$.

\smallskip
Following \cite[Propositions 2.2,2.3]{CSY}, we show that viscosity solutions are preserved under this rotation.

\begin{proposition}
Let $u(x)$ be a convex viscosity solution of (\ref{slag}) in $B_{1.2}(0)$. Then the $\pi/4$-rotation $\bar u$ in (\ref{r}) is a corresponding viscosity solution of
\begin{equation}
    \bar F(D^2\bar{u})=\sum_{i=1}^n \arctan \bar{\lambda}_i(D^2\bar{u})=\bar\psi(\bar x,D\bar u)=\psi(x)-n\pi/4 \label{rs}
\end{equation}
in open $\bar{\Omega}=\partial \Tilde{u}(B_1(0))$, where $\tilde{u}$ is as defined in (\ref{u1}).
\end{proposition}

\begin{proof} We first prove the following claim.

\begin{claim} \label{1}
We show that $\bar u$ is a supersolution of (\ref{rs}) in $\bar{\Omega}$.
\end{claim}
Let $\bar{P}$ be a quadratic polynomial touching $\bar{u}$ from below locally
somewhere on the open set $\bar{\Omega},$ say at the origin
$\bar{0}\in\partial\tilde{u}(  0).$ Since at $\bar{0}$, $D^2\bar{P}\leq D^2\bar{u}\leq I_n$, we assume $D^{2}\bar{P}<I_n$, by subtracting $\varepsilon|\bar x|^2$ from $\bar{P}$, and then letting $\varepsilon\rightarrow 0$. This
guarantees the existence of its pre-rotated quadratic polynomial $P,$ and also confirms that
$\bar{P}$ touches $\bar{u}$ from below in an open neighborhood of the closed set $\partial\tilde{u}(  0).$ Using the order preservation
of $\alpha$-rotation that is also valid for reverse rotation, and continuity of the gradient mapping $\partial\tilde{u}$ in \cite[Corollary 24.5.1]{R}, we see that the pre-rotated quadratic polynomial $P$ touches $u$ from below
somewhere in an open neighborhood of $0\in B_{1}(  0).$ Using the fact that $u$ is a supersolution of (\ref{slag}) we get
\[\sum_{i=1}^n\arctan\lambda_i(D^2P)\leq \psi(0),
\]which in turn implies
\[\bar F(\bar D^2\bar P)=\sum_{i=1}^n\arctan\bar \lambda_i(D^2\bar P)\leq \psi(0)-n\pi/4=\bar\psi(\bar 0,D\bar P(\bar 0)).
\]
Therefore, the claim holds good.
\begin{claim}\label{2}
We show that $\bar u$ is a subsolution of (\ref{rs}) in $\bar{\Omega}$.
\end{claim}

The first part of this proof is the same as \cite[Prop 2.3, Step1]{CSY} where a convolution $u_\varepsilon$ of $u$ is considered and the smooth $\alpha$-rotation $\bar{u}_\varepsilon$ is shown to be well defined on $\overline{\Omega}=\partial \tilde{u}(B_1(0))\subset D\tilde{u}_\varepsilon (B_{1.1}(0))$ for $\varepsilon>0$ small enough.\\
Next, observe that (\ref{slag}) is concave when $u$ is a convex function. Applying \cite[Theorem 5.8]{CC}, we see that the solid convex average $u_\varepsilon=u*\rho_\varepsilon$ is a subsolution of
\[F(D^2u_\varepsilon)\ge\psi_\varepsilon(x)=\psi*\rho_\varepsilon(x)
\] for $\varepsilon>0$ small enough. Combining \cite[Prop 2.1]{CSY} with the first part of this proof, we see that the smooth $\alpha$-rotation $\bar{u}_\varepsilon$ is a subsolution of
\begin{equation*}
    \bar{F}(D^2\bar{u}_\varepsilon)=\sum_{i=1}^n \arctan \bar{\lambda}_i(D^2\bar{u}_\varepsilon)\ge \bar\psi_\varepsilon(\bar{x}_{\varepsilon},D\bar u_{\varepsilon})=\psi_\varepsilon(x_\e)-n\alpha,
\end{equation*}
in $\overline{\Omega}$, where $x_\e=c\bar x-s D\bar u_\e(\bar x)$.  By \eqref{xsmall}, $x_\e\to x=c\bar x-s\,D\bar u(\bar x)$ uniformly on $\bar\Omega$, so $\psi_\e(x_\e(\bar x))\to\psi(x(\bar x))$ uniformly on $\bar\Omega$.   Since $\bar u_\e\to \bar u$ uniformly on $\bar\Omega$, and they are viscosity subsolutions of locally uniformly convergent equations $\bar F(D^2\bar u_\e)=f_\e(\bar x)$, it follows from \cite[Proposition 2.9]{CC} that $\bar u$ is a subsolution of the limiting equation \eqref{rs}.

\smallskip
Therefore, from claims \ref{1} and \ref{2}, we see that $\bar u$ is a viscosity solution of (\ref{rs}) in $\bar{\Omega}$.
\end{proof}

\section{Regularity of the rotated potential}
\label{sec:reg_baru}
We first show that $\bar u\in VMO(B_{1/2})$ followed by $\bar u\in C^{2,\alpha}(B_{1/2})$.
We recall the notion of VMO functions.
\begin{definition}[Vanishing mean oscillation]
Let $\Omega\subset\mathbb{R}^n$. A locally integrable function $v$ is in $VMO(\Omega)$ with modulus $\omega(r,\Omega)$ if
\[\omega(r,\Omega)=\sup_{x_0\in \Omega,0<r\leq R}\frac{1}{|{B_r(x_0)\cap \Omega}|} \int_{B_r(x_0)\cap \Omega} |v(x)-v_{x_0,r}|\rightarrow 0, \text{ as $r\rightarrow 0$}
\]
where $v_{x_0,r}$ is the average of $v$ over $B_r(x_0)\cap \Omega.$
\end{definition}

\begin{proposition}[VMO Estimates]
\label{prop:vmo}
Let $\bar u$ be a $C^{1,1}$ viscosity solution of (\ref{rs}) in $B_{1}(0)\subset \mathbb R^n$, where $|D^2\bar u|\leq 1$ and $ \bar\psi$ is Lipschitz continuous. Then $D^2 \bar u \in VMO(B_{1/2})$ and the $VMO$ modulus of $\bar u$, denoted by $\omega(r)\rightarrow 0$ as $r\rightarrow 0$.
\end{proposition}
\begin{proof}
Suppose the contrary is true. Then we can find $\varepsilon>0$ and sequences $\{\bar x_k\rightarrow0\}\subset B_{1/2}$, $\{r_k \rightarrow 0\}$, and a family of $C^{1,1}$ viscosity solutions $\{\bar u_k\}$ of (\ref{rs}) with $|D^2\bar u_k|\leq 1$, such that
\[
\frac{1}{|B_{r_k}|}\int_{B_{r_k}}|D^2\bar u_k-(D^2\bar u_k)_{\bar x_k,r_k}|\geq \varepsilon.
\]
Next, we blow up $\{\bar u_k\}$. For $|\bar y|\leq \frac{1}{r_k}$, we set
\[\bar v_k(y)=\frac{\bar u_k(\bar x_k+r_k\bar y)-\nabla \bar u_k(\bar x_k). r_k \bar y-\bar u_k(\bar x_k)}{r_k^2}.
\]
Note that $||\bar v_k||_{C^{1,1}(B_r)}\leq C(r)$ and $\bar v_k$ is a viscosity solution of (\ref{rs}). Then we can find a subsequence, still denoted by $\{ \bar v_k\}$ and a function $\bar v_r\in C^{1,1}(B_r)$ such that $\bar v_k\rightarrow \bar v_r$ in $C^{1,\alpha}(B_r)$ as $k\rightarrow\infty$. Note that $|D^2\bar v_r|\leq 1.$
Viscosity solutions are closed under $C^0$ uniform limits, so $\bar v_r$ is a viscosity solution of $\bar F(D^2 \bar v)= \bar \psi(0)$ in $B_r$. Now applying the $W^{2,\delta}$ estimate from \cite[Prop 7.4]{CC}, we get
\[ ||D^2\bar v_k-D^2\bar v_r||_{L^{\delta}(B_{r/2})}\leq C(r)||\bar v_k-\bar v_r||_{L^{\infty}(B_r)}
\]
where the RHS goes to $0$ as $k\rightarrow \infty$. Using the fact that $|D^2\bar v_k|, |D^2\bar v_r|\leq 1$, we see that for any $p>0$,  $||D^2\bar v_k-D^2\bar v_r||_{L^{p}(B_{r/2})}$ approaches $0$ as $k\rightarrow\infty$.

\smallskip
Next, by the diagonalization method, we can find another subsequence, again denoted by $\{\bar v_k\}$ and $\bar v\in C^{1,1}(\mathbb {R}^n)$, such that as $k\rightarrow \infty$, we have  $\bar v_k\rightarrow \bar v$ in $W^{2,p}_{loc}(\mathbb {R}^n)$ and $|D^2\bar v|\leq 1$.
Note that $\bar v$ will be a viscosity solution of $\bar F(D^2\bar v)=\bar \psi(0)$ in $\mathbb{R}^n$.
From \cite[(2.9)]{HL} we see that the mean curvature vector $\vec{H}$ of the Lagrangian submanifold $(x,Du(x))$
is given by (\ref{mean}). This shows that $L_{\bar v}=\{(\bar x,D \bar v(\bar x))|\bar x\in \mathbb {R}^n\}$ is a minimal surface in $\mathbb R^{2n}$. Since $\bar v_k$ converges to $\bar v$ in $W^{2,n}_{loc}(\mathbb{R}^n)$, we apply the monotonicity formula \cite[Pg 84]{sim} and \cite[Theorem 19.3]{sim}, to conclude that $L_{\bar v}$ is a minimal cone.

\smallskip
Finally, we claim that $L_{\bar v}$ is smooth away from the origin (vertex). Let $\bar p$ away from the origin be a singular point of $L_{\bar v}$. We blow up $L_{\bar v}$ at $\bar p$ to get a tangent cone, which is a lower dimensional special Lagrangian cone times a line. Now we
repeat the same process if the resulting cone still has a singularity away from the origin. At the end we get a special Lagrangian cone which is smooth away from the origin and the absolute value of the eigenvalues of the Hessian of the potential function are still bounded by 1. Then by \cite[Prop 2.2]{YY} the cone is flat. This contradicts Allard’s regularity result \cite[Theorem 24.2]{sim}. So, now applying \cite[Prop 2.2]{YY}, we see that $L_{\bar v}$ is flat, i.e. $\bar v$ is a quadratic polynomial. Therefore we have
\begin{align*}
    0=\frac{1}{|B_1|}\int_{B_1}|D^2\bar v-(D^2\bar v)_{0,1}|=\lim_{k\rightarrow \infty}\frac{1}{|B_1|}\int_{B_1}|D^2\bar v_k-(D^2\bar v_k)_{0,1}|\\
    =\lim_{k\rightarrow \infty}\frac{1}{|B_{r_k}|}\int_{B_{r_k}}|D^2\bar u_k-(D^2\bar u_k)_{\bar x_k,r_k}|\ge \varepsilon
\end{align*}
which is a contradiction.
\end{proof}

By translation invariance of the VMO seminorms, $D^2\bar u(\bar x+h)$ will also be in $VMO(B_{1/2})$ if $h$ is small.  This means we can conclude the following but for the sake of completeness, we provide a proof similar to Proposition \ref{prop:vmo}.
\begin{corollary}
\label{cor:vmo}
Let $h\in \mathbb R^n$ be sufficiently small, and $\bar u$ and $\bar\psi$ be as defined in Proposition \ref{prop:vmo}.  Then a continuous function $K(.,.)$ of $D^2\bar u(\bar x)$ and $D^2\bar u(\bar x+h)$ is in $VMO(B_{1/2})$.
\end{corollary}

\begin{proof}
Letting $\bar u^h_k(\bar x)=\bar u_k(\bar x+ h)$, we repeat the proof of Proposition \ref{prop:vmo}, assuming instead that
$$
\frac{1}{|B_{r_k}|}\int_{B_{r_k}}|K(D^2\bar u_k,D^2\bar u^{h_k}_k)-(K(D^2\bar u_k,D^2\bar u^{h_k}_k))_{\bar x_k,r_k}|\ge\e.
$$
Rescaling as before via $\bar v_k$ and $\bar v_k^{h_k}$, we take subsequences and send $k\to\infty$ to obtain limits $\bar v$ and $\bar v^h$.  The convergence is in $W^{2,p}_{loc}(\mathbb R^n)$, so after a subsequence, we can assume $D^2\bar v_k$ and $D^2\bar v^{h_k}_k$ converge almost everywhere.  By the dominated convergence theorem and $|D^2\bar v_k|\le 1$, it follows that $K(D^2\bar v_k,D^2\bar v_k^{h_k})$ converges in $L^1_{loc}(\mathbb R^n)$.  In fact, $\bar v$ and $\bar v^h$ are quadratic polynomials, so $K(D^2\bar v,D^2\bar v^h)$ is a constant. Therefore we have
\begin{align*}
0=\frac{1}{|B_1|}\int_{B_1}|K-(K)_{0,1}|&=\lim_{k\to\infty}\frac{1}{|B_{1}|}\int_{B_{1}}|K(D^2\bar v_k,D^2\bar v_k^{h_k})-(K(D^2\bar v_k,D^2\bar v_k^{h_k}))_{0,1}|\\
&=\lim_{k\to\infty}\frac{1}{|B_{r_k}|}\int_{B_{r_k}}|K(D^2\bar u_k,D^2\bar u^{h_k}_k)-(K(D^2\bar u_k,D^2\bar u_k^{h_k}))_{\bar x_k,r_k}|\\
&\ge\e,
\end{align*}
which is a contradiction.
\end{proof}

Note that the $VMO(B_{1/2})$ seminorms of $K(D^2\bar u,D^2\bar u^h)$ are independent of $h$ if $h$ is small.  The point of $K$ is we need to take a difference quotient of \eqref{rs}.

\begin{proposition}[$C^{2,\alpha}$ Estimates]
\label{prop:C2a}
Let $\bar u$ be a $C^{1,1}$ viscosity solution of (\ref{rs}) in $B_{1}(0)\subset \mathbb R^n$, where $|D^2\bar u|\leq 1$ and $ \bar\psi$ is Lipschitz continuous. Then $\bar u \in C^{2,\alpha}(B_{1/2})$ for all $\alpha\in(0,1)$.
\end{proposition}

\begin{proof}
Letting $v^h=[\bar u(\bar x+h)-\bar u(\bar x)]/|h|$, we obtain the linearized equation
\begin{align}
\label{slag:lin}
F^h_{ij}v^h_{ij}=\bar\psi^h(\bar x),
\end{align}
where
$$
F^h=\int_0^1\frac{1}{1+x^2}((1-t)D^2\bar u+tD^2\bar u^h)dt,\qquad \bar\psi^h(\bar x)=\frac{\bar\psi(\bar x+h)-\bar\psi(\bar x)}{|h|}.
$$
For small $h$, equation \eqref{slag:lin} holds
pointwise\footnote{In fact, $\bar u(\bar x)$ is twice differentiable everywhere, which is easily shown using the rescaling procedure in Proposition \ref{prop:vmo}.}
on $B_{1/2}$, and can be thought of as a linear equation for $v^h$ in nondivergence form, with, by Corollary \ref{cor:vmo}, $VMO_{loc}$ coefficients $F^h_{ij}$ and $L^\infty_{loc}$ right hand side $\bar \psi^h$, each of whose seminorms are independent of $h$.  Recalling the interior $W^{2,p}$ estimates due to \cite{CFL}, see also \cite[Theorem 2.1]{V}\footnote{Although these estimates assume solutions are in $W^{2,p}_0(\Omega)$, adding a cutoff function as in \cite[Proof of Theorem 9.11]{GT} yields standard interior estimates.},

we deduce that $\forall p$ large, $v^h\in W^{2,p}_{loc}$ inside a slightly smaller domain in $\bar x(B_1)$ with local estimates independent of $h$.  Since for difference quotient $v^h$ we have $D^2v^h\in L^p_{loc}$, it follows that $D^2\bar u\in W^{1,p}_{loc}$ for all large $p$, hence by the Sobolev Embedding Theorem $D^2\bar u\in C^{\alpha}(B_{1/2})$ for all $\alpha\in(0,1)$.
\end{proof}

Now from Schauder theory, the rotated potential $\bar u(\bar x)$ is as regular as the equation allows.

\begin{corollary}
\label{cor:higher}
Let $u$ be a convex viscosity solution of \eqref{slag} on $B_1(0)$, where $\psi(x)$ is Lipschitz.  Then rotated potential $\bar u(\bar x)$ is a $C^{2,\alpha}$ solution of \eqref{rs} on $\bar x(B_1(0))$ for all $\alpha\in(0,1)$.  If $\psi\in C^{2,\alpha}$ for some $\alpha\in (0,1)$, then $\bar u\in C^{4,\alpha}$.
\end{corollary}

Indeed, since $\bar u\in C^{1,1}$, we have $D\bar u\in C^{0,1}$ and $\bar\psi\in C^{0,1}$, and Proposition \ref{prop:C2a} applies, so $D\bar u\in C^{1,\alpha}$.  If we also know $\psi\in C^{2,\alpha}$, then this means $\bar\psi\in C^{1,\alpha}$. Recalling equation \eqref{slag:lin} for the difference quotient of $\bar u$, it follows from Schauder estimates that $D\bar u\in C^{2,\alpha}$, so $\bar\psi\in C^{2,\alpha}$.  Taking two difference quotients in a similar way, we deduce $D^2\bar u\in C^{2,\alpha}$.

\begin{comment}
\begin{remark}
For use in section \ref{sec:comp}, we mention that the Hessian bound $|D^2\bar u|\le 1$ ensures that the $C^{2,\alpha}$ seminorms of $\bar u(\bar x)$ on compactly contained subdomains of $\bar x(B_1(0))$ depend universally: only on the dimension, $\alpha$, the Lipschitz seminorms of $\psi(x)$, and the subdomains.  Indeed, in Proposition \ref{prop:vmo}, the VMO seminorms have universal dependence, since they were obtained by a compactness argument, and similarly with Corollary \ref{cor:vmo}.  And in Proposition \ref{prop:C2a}, the local $W^{2,p}$ estimates depend only on the VMO seminorms, the Lipschitz seminorms of $\psi$, the subdomains, and $p$, so the same will be true of the corresponding local $C^{2,\alpha}$ estimates.
\end{remark}
\end{comment}

\section{Proof of Theorem \ref{thm}}

\label{sec:reg}

The final step is to show that $\bar\lambda_{max}<1$ on $\bar x(B_1)$.  Indeed, once we prove this, we can compare with quadratics to convert to original variables.  Let $\bar Q(\bar x)$ touch $\bar u$ from above near $\bar x$ and suppose, after lowering, that $D^2\bar Q<I_n$.  By the order preservation of rotations, we see that $Q(x)$ touches $u$ from above near $x$.  Moreover, $D^2Q<\infty$ by the transformation law
$$
D^2Q(x)=\left(\frac{1+t}{1-t}\right)\,(D^2\bar Q(\bar x)).
$$
Thus $\lambda_{max}<\infty$ on $B_1$, i.e. $u\in C^{1,1}(B_1)$.  This means $\bar x(x)=cx+sDu(x)$ is Lipschitz, so the above formula with $Q$ replaced by $u$ implies $u\in C^{2,1}(B_1)$. This means $\bar x\in C^{1,1}$, and as before, we deduce $u\in C^{3,1}(B_1)$.  Since $D^2\bar u\in C^{2,\alpha}(B_1)$ and $\bar x\in C^{2,1}$, we conclude that $u\in C^{4,\alpha}(B_1)$.
We now establish the desired inequality.
\begin{remark}
We need $\psi \in C^{2,\alpha}$ only for the following proposition.

\end{remark}

\begin{proposition}
\label{prop:max}
Let $u$ be a convex viscosity solution of \eqref{slag} on $B_1(0)$, with $\psi(x)\in C^{2,\alpha}$.  Then $\bar\lambda_{max}<1$ on $\bar x(B_1(0))$.
\end{proposition}

\begin{proof}
Suppose $\bar\lambda_{max}=1$ at a point $\bar x$, or more generally that $1=\bar\lambda_{1}=\cdots=\bar\lambda_m>\bar\lambda_{m+1}\ge \cdots\ge \bar\lambda_n$ at $\bar x$.  We also assume $D^2\bar u$ is diagonal at $\bar x$. From \cite[Lemma 4.2]{AB} we get the following for $\bar b_m:=\frac{1}{m}\sum_1^m\ln \sqrt{1+\bar\lambda_k^2}$ when the phase is variable:
$$
m\Delta_{\bar g}\bar b_m=(\text{constant phase terms})+\sum_{i=1}^m\frac{\bar\lambda_i}{1+\bar\lambda_i^2}\bar\psi_{\bar i\bar i}-m\sum_{i=1}^m\bar \lambda_i\bar g^{\bar{i}\bar i}\bar\psi_{\bar i}\pd_{\bar i}\bar b_m.
$$
Note that $\bar b_m$ is $C^{2,\alpha}$ in a neighborhood $U$ of $\bar x$.  The constant phase terms are nonnegative if the Hessian bounds \eqref{bounds} are true, as in \cite[Section 3]{CSY}.  The term involving $\nabla \bar b_m$ is harmless, so it suffices to lower bound the second term, which will have two contributions:
$$
\bar\psi_{\bar{ii}}=\frac{\pd^2}{\pd \bar x_i^2}\psi\left(\frac{\bar x-D\bar u(\bar x)}{\sqrt 2}\right)=-\frac{1}{\sqrt 2}\psi_a\partial_{\bar a}\bar\lambda_i+\frac{1}{2}\psi_{ii}(1-\bar\lambda_i)^2.
$$
The first term yields a harmless contribution to the maximum principle:
$$
\sum_{a=1}^n\psi_a\sum_{i=1}^m\frac{\bar\lambda_i}{1+\bar\lambda_i^2}\partial_{\bar a}\bar\lambda_i=m\sum_{a=1}^n\psi_a\partial_{\bar a}\bar b_m.
$$
For the second term, we start with
$$
\frac{1}{2}\sum_{i=1}^m\frac{\bar\lambda_i}{1+\bar\lambda_i^2}\psi_{ii}(1-\bar\lambda_i)^2\ge -c(\bar x)(1-\bar\lambda_m)^2,
$$
for some locally bounded $c(\bar x)$.  Next, we shrink $U$ so that $\bar\lambda_m> 0$ in $U$.  By the convexity of $\bar b(t):=\ln\sqrt{1+t^2}$ in $[0,1]$,
$$
0<\frac{\bar b(1)-\bar b(0)}{1-0}\le \frac{\bar b(1)-\bar b(t)}{1-t},
$$

so putting $t=\bar\lambda_m(\bar y)$ for $\bar y\in U$ yields
$$
(1-\bar \lambda_m)^2\le C(\bar b(1)-\bar b(\bar\lambda_m))^2\le C(m\bar b(1)-\sum_{i=1}^{m} \bar b(\bar\lambda_i))^2.
$$
We thus conclude that in a sufficiently small neighborhood $U$ of $\bar x$,
$$
\Delta_{\bar g}\bar b_m\ge \langle a(\bar x),\nabla_{\bar g}\bar b_m\rangle_{\bar g}-c(\bar x)(\bar b_{m,max}-\bar b_m)^2
$$
for some
bounded continuous $a,c$, where $\bar b_{m,max}=\bar b_m|_{\bar\lambda_1=\cdots=\bar\lambda_m=1}$.  The strong maximum principle still holds because the right side vanishes at an interior maximum, according to \cite[Lemma 3.4 and Theorem 3.5]{GT}.  It follows that $\bar b_{m}\equiv \bar b_{m,max}$ and $\bar\lambda_{max}\equiv 1$ on an open set containing $\bar x$.  Since $\bar x(B_1)$ is connected, we conclude this is true everywhere:
$\bar\lambda_{max}\equiv 1$.  However, because $u(x)$ is bounded, we can touch it from above somewhere in $B_1(0)$ by a sufficiently tall quadratic $Q$.  The rotation $\bar Q$ then touches $\bar u$ from above somewhere in $\bar x(B_1)$.  But $D^2Q<\infty$ corresponds\footnote{See \cite[End of section 3]{CSY}.} to $D^2\bar Q<I_n$, a contradiction.
\end{proof}

\section{Hessian estimate by compactness}
\label{sec:comp}
We now prove \eqref{hess_est}.  If the estimate fails, then there is a sequence of $C^{4,\alpha}$ convex solutions $u_k$ and $C^{2,\alpha}$ phases $\psi_k$ to \eqref{slag} with
$$
\lambda_{max}[D^2u_k(0)]\to \infty,\qquad u_k(0)=Du_k(0)=0,\qquad  \|u_k\|_{C^1(B_{1/2}(0))}+\|\psi_k\|_{C^{2,\alpha}(B_{1/2}(0))}\le C.
$$
Since $\|u_k\|_{C^1}$ is bounded, we can pass to a subsequence and assume $u_k$ converges uniformly to $u$ on $B_{1/2}(0)$.  Note that $u$ is necessarily convex, so $\tilde u:=\frac{1}{\sqrt 2}(u+\frac{1}{2}|x|^2)$, as in Section 2, is uniformly convex.  This means $\partial\tilde u(B_{r}(0))$ is open and contains $B_{r/\sqrt 2}(0)$.  Moreover, if we shrink slightly and suppose $\bar x\in \partial\tilde u(B_{1/2-\delta}(0))$, then $\bar x\in\partial\tilde u_k(B_{1/2}(0))$ for large enough $k$, since if $\bar x\in \partial\tilde u_k(x_k)$, then uniform convexity yields, via \eqref{xsmall},
$$
|x-x_k|^2\le C\|\tilde u-\tilde u_k\|_{L^\infty(B_{1/2}(0)}\to 0.
$$
So the rotated sequence $\bar u_{k}$ is defined on arbitrarily large subsets of $\partial\tilde u(B_{1/2}(0))$, converging uniformly to $\bar u$ thereabouts by the order preservation of rotation.  It follows that $\bar u$ is the locally uniform limit on $\partial\tilde u(B_{1/2}(0))$ of smooth rotations $\bar u_k$.

\medskip
The smooth rotations $\bar u_k$ have eigenvalues which blowup: $\bar\lambda_{max,k}(0)\to 1$.  To see this for $\bar u$, we use the $C^{2,\alpha}$ estimates for $\bar u_k$, noting that $\psi_k$ converges to some $\psi\in C^{2,\alpha}(B_{1/2}(0))$ in the norm of $C^{2,\alpha/2}(B_{1/2}(0))$ after taking a subsequence.  Along a subsequence, it follows that $\bar u_k$ eventually converges locally in $C^{4,\alpha/2}$ to $\bar u$ in $\partial \tilde u(B_{1/2}(0))$, so $\bar\lambda(0)=1$.  Moreover, $\bar u$ is a $C^{4,\alpha}$ solution of the rotated equation $\bar F(D^2u)=\bar\psi$, so $\bar\lambda_{max}\equiv 1$ by the strong maximum principle arguments in Proposition \ref{prop:max}.  This is a contradiction, since $\bar\lambda_{max}<1$ somewhere on $\partial\tilde u(B_{1/2}(0))$ for bounded convex $u$.

\begin{remark}
In fact, a stronger Hessian estimate than \eqref{hess_est} holds:
\begin{equation}
    |D^2u(0)|\leq C_1\exp [C_2\, (osc_{B_{1/2}}u) ^{2n-2}]
\end{equation}
where $C_1$ and $C_2$ are positive constants depending on $||\psi||_{C^{1,1}(B_{1/2})}$ and $n$.  This result follows from the methods in \cite{AB}.  The proof in \cite{AB} goes through if the supercriticality condition $|\psi|\ge (n-2)\pi/2+\delta$ is replaced by the convexity condition $D^2u\ge 0$.  See also \cite{BS2}, where such an estimate is generalized to the case of $\psi=\psi(x,u)$.
\end{remark}

\textbf{Acknowledgments.} The authors are grateful to Professor Yu Yuan for his guidance, support, and helpful discussions. RS was partially supported by the NSF Graduate Research Fellowship Program under grant No. DGE-1762114.

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\end{document}


TITLE: How to subtract a fraction where a fraction is part of the denominator already?
QUESTION [0 upvotes]: I am to subtract two fractions:
$\frac{1}{\frac{1}{x}+2}-\frac{2}{1}$
I can see the answer in my text book being just x, the only in between line I'm shown is:
$\frac{1}{\frac{1}{x}+2}-\frac{2}{1}$ =
$x + 2 - 2$ = x
I'm trying to arrive at this myself and got lost:
$\frac{1}{\frac{1}{x}+2}-\frac{2}{1}$
Least common denominator (lcd) is the product of the two denominators, in this case $1(\frac{1}{x}+2)$
So, the left side is already using the lcd so I only need to transform the right side by multiplying by $\frac{1}{x}+2$. I'll do this to both the numerator and denominator to keep them in proportion:
$\frac{1}{\frac{1}{x}+2}-\frac{2}{1}$ =
$\frac{1}{\frac{1}{x+2}} - \frac{-2(\frac{1}{x+2})}{\frac{1}{x+2}}$
Now they both have the same denominator so I can subtract one from the other:
$\frac{1-2(\frac{1}{x}+2)}{\frac{1}{x+2}}$
This is as far as I can take it.
How can I arrive at $x + 2 - 2$ = x?
Seeking baby, more granular steps if possible.
Screen shot of the question in full in case I've missed anything or made typos:

REPLY [1 votes]: Method 1:
Whenever I see "small denominators" (nested fractions), I try to clear the small denominators.
\begin{align*}
\frac{1}{\frac{1}{x}+2} - \frac{2}{1}
&= \frac{1}{\frac{1}{x}+2} \cdot \underbrace{\frac{x}{x}}_{1} - \frac{2}{1}  \\
&= \frac{x}{1+2x} - \frac{2}{1}  \\
&= \frac{x}{1+2x} - \frac{2}{1} \cdot \underbrace{\frac{1+2x}{1+2x}}_{1} \\
&= \frac{x}{1+2x} - \frac{2+4x}{1+2x} \\
&= \frac{x - (2+4x)}{1+2x}  \\
&= \frac{-2-3x}{1+2x}  \text{.}
\end{align*}
Method 2:
Don't try to be tricky with the common denominator.
\begin{align*}
\frac{1}{\frac{1}{x}+2} - \frac{2}{1}
&= \frac{1}{\frac{1}{x}+2} - \frac{2}{1} \cdot \underbrace{\frac{\frac{1}{x}+2}{\frac{1}{x}+2}}_{1}  \\
&= \frac{1 - 2(\frac{1}{x}+2)}{\frac{1}{x}+2}  \\
&= \frac{-3 - \frac{2}{x}}{\frac{1}{x}+2} \cdot \underbrace{\frac{x}{x}}_{1} \\
&= \frac{-3x - 2}{1+2x}  \text{.}
\end{align*}
I notice that neither of these is "$x$".  I suspect there is a typo' in the presented problem.  Let's start from $\frac{1}{\frac{1}{x+2}}$ instead.
\begin{align*}
\frac{1}{\frac{1}{x+2}} - \frac{2}{1}
&= \frac{1}{\frac{1}{x+2}} \cdot \underbrace{\frac{x+2}{x+2}}_{1} - \frac{2}{1}  \\
&= \frac{x+2}{1} - \frac{2}{1}  \\
&= \frac{x+2 - 2}{1}  \\
&= \frac{x}{1}  \\
&= x  \text{.}
\end{align*}


TITLE: Outer measure of dirac measure
QUESTION [1 upvotes]: Let $\delta_x$ be the dirac measure and $\delta_x^*$ the corresponding outer measure. Show that for all $A \subset (0,1]$ the equation
$\delta_x^*(A):=\cases{1, x \in A \\ 0, x \notin A}$
holds.
Note that the outer measure is definded as
$\mu^∗
(A) = \inf \{\sum_{j=1}^\infty
\mu(A_j ) : A_j ∈ B, j ∈ \mathbb{N},
\bigcup_{j=1}^\infty
A_j ⊃ A\}$
In this case B is  the ring of all finite unions of left-half-open intervals.
What I've tried:
Define the set $\{x:x \in A\}$ and show that the outer measure on it is $1$. Let $(A_n)$ be a sequence in $B$. Then  this set is a subset of $\bigcup_{n \in \mathbb{N}}A_n.$ Hence there has to be a $m \in \mathbb{N}$ so that $x \in A_m$.
I'm stuck on this point. I also need to prove that all subsets $A$ of $(0,1]$ are $\delta_x^*$-measurable. Any help is appreciated!

REPLY [1 votes]: Let $A\subseteq(0,1]$. It is easy to see that
$$0\leq\delta_x^*(A)\leq1.$$
Suppose first that $x\in A$. Then for any sequence $\{A_j\}_{j\in\mathbb{N}}$ in $B$ with $A\subseteq\bigcup_{j\in\mathbb{N}}A_j$ we have that $x\in A_k$ for some $k\in\mathbb{N}$. In particular then
$$\sum_{j=0}^\infty\delta_x(A_j)\geq\delta_x(A_k)=1.$$
Taking the infimum over all such sequences $\{A_j\}_{j\in\mathbb{N}}$ we get that $\delta_x^*(A)\geq1$, and so $\delta_x^*(A)=1$.
Suppose now that $x\notin A$. Choose a sequence $\{A_j\}_{j\in\mathbb{N}}$ in $B$ such that $A\subseteq\bigcup_{j\in\mathbb{N}}A_j=(0,1]\setminus\{x\}$ (this can easily be done explicitly if you're unsure about why such a sequence exists). This means that $\delta_x(A_j)=0$ for all $j\in\mathbb{N}$, and so
$$0\leq\delta_x^*(A)\leq\sum_{j=0}^\infty\delta_x(A_j)=0,$$
from which it follows that $\delta_x^*(A)=0$.
So we have now proven than $\delta_x^*(A)=\delta_x(A)$ for any $A\subseteq(0,1]$. I'll leave the measurability part of the question to you, as that should be a very short and straightforward argument.