ended pde ICT

Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex

@@ -28,7 +28,7 @@
     Let $\Sigma$ be a $\sigma$-algebra over a set $\Omega$. A \emph{measure} over $\Omega$ is any function $$\mu:\Sigma\longrightarrow[0,\infty]$$ satisfying the following properties:
     \begin{enumerate}[ref = $\sigma$-additivity]
       \item $\mu(\varnothing)=0$.
-            \item\label{RFA:sigmaadditivity} \emph{$\sigma$-additivity}: If $(A_n)\in\Sigma$ are pairwise disjoint, then: $$\mu\left(\bigsqcup_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty \mu(A_n)$$
+      \item\label{RFA:sigmaadditivity} \emph{$\sigma$-additivity}: If $(A_n)\in\Sigma$ are pairwise disjoint, then: $$\mu\left(\bigsqcup_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty \mu(A_n)$$
     \end{enumerate}
   \end{definition}
   \begin{definition}
@@ -39,8 +39,8 @@
     \begin{enumerate}
       \item If $A\subseteq B$, then $\mu(B\setminus A)=\mu(B)-\mu(A)$.
       \item If $A\subseteq B$, then $\mu(A)\leq\mu(B)$.
-            \item\label{RFA:incresingseq} If $A_n\nearrow A$, then $\displaystyle\mu(A)=\lim_{n\to\infty} \mu(A_n)$.
-            \item\label{RFA:decresingseq} If $A_n\searrow A$ and $\mu(A_1)<\infty$, then $\displaystyle\mu(A)=\lim_{n\to\infty} \mu(A_n)$.
+      \item\label{RFA:incresingseq} If $A_n\nearrow A$, then $\displaystyle\mu(A)=\lim_{n\to\infty} \mu(A_n)$.
+      \item\label{RFA:decresingseq} If $A_n\searrow A$ and $\mu(A_1)<\infty$, then $\displaystyle\mu(A)=\lim_{n\to\infty} \mu(A_n)$.
     \end{enumerate}
   \end{proposition}
   \begin{sproof}
@@ -113,12 +113,12 @@
     The outer measure has the following properties:
     \begin{enumerate}
       \item $\om{\varnothing}=0$.
-            \item\label{RFA:measureB} If $A\subseteq B\subseteq\RR^n$, then $\om{A}\leq \om{B}$.
+      \item\label{RFA:measureB} If $A\subseteq B\subseteq\RR^n$, then $\om{A}\leq \om{B}$.
       \item \label{RFA:measureC} If $(A_k)\subseteq \RR^n$, then: $$\om{\bigcup_{k=1}^\infty A_k}\leq \sum_{k=1}^\infty \om{A_k}$$
       \item \label{RFA:measureD}If $I\subseteq \RR^n$ is an open interval and $I\subseteq A\subseteq \cl{I}$, then $\om{A}=\vol(I)$.
-            \item\label{RFA:measureE} If $I_1,\ldots,I_N\subseteq \RR^n$ are disjoint intervals, then: $$\displaystyle \om{\bigsqcup_{k=1}^N I_k}= \sum_{k=1}^N \vol(I_k)$$
+      \item\label{RFA:measureE} If $I_1,\ldots,I_N\subseteq \RR^n$ are disjoint intervals, then: $$\displaystyle \om{\bigsqcup_{k=1}^N I_k}= \sum_{k=1}^N \vol(I_k)$$
       \item If $A,B\subseteq \RR^n$ and $d(A,B):=\inf\{d(a,b):a\in A,b\in B\}>0$, then $\om{A\sqcup B}=\om{A}+\om{B}$.
-            \item\label{RFA:measureG} If $A\subseteq\RR^n$ and $x\in\RR^n$, then $\om{A+x}=\om{-A}=\om{A}$\footnote{Here $A+x:=\{a+x:a\in A\}$ and $-A:=\{-a:a\in A\}$}.
+      \item\label{RFA:measureG} If $A\subseteq\RR^n$ and $x\in\RR^n$, then $\om{A+x}=\om{-A}=\om{A}$\footnote{Here $A+x:=\{a+x:a\in A\}$ and $-A:=\{-a:a\in A\}$}.
     \end{enumerate}
   \end{theorem}
   \begin{sproof}
@@ -658,7 +658,7 @@
     \begin{enumerate}[ref = Triangular inequality]
       \item $\|\vf{u}\|=0\iff \vf{u}=0$
       \item $\|\lambda \vf{u}\|=|\lambda|\|\vf{u}\|$
-            \item\label{RFA:triangularineq} $\|\vf{u}+\vf{v}\|\leq \|\vf{u}\|+\|\vf{v}\|\quad$(\emph{triangular inequality})
+      \item\label{RFA:triangularineq} $\|\vf{u}+\vf{v}\|\leq \|\vf{u}\|+\|\vf{v}\|\quad$(\emph{triangular inequality})
     \end{enumerate}
     We define a \emph{normed vector space} as a pair $(E,\|\cdot\|)$ that satisfy the previous properties.
   \end{definition}
@@ -857,11 +857,11 @@
     The case of $L^\infty(E)$ is easy and the first two properties for $L^p(E)$, $p\geq 1$, too (remember \mcref{RFA:postmonotoneE}). It's missing to prove the \mref{RFA:triangularineq} (also called \emph{Minkowski inequality} in this case): $$\norm{f+g}_p\leq \norm{f}_p+\norm{g}_p$$
     We have that:
     \begin{align*}
-      {\norm{f+g}_p}^p & =\int_E\abs{f+g}\abs{f+g}^{p-1}                                            \\
-                       & \leq \int_E\abs{f}\abs{f+g}^{p-1}+\int_E\abs{g}\abs{f+g}^{p-1}             \\
+      {\norm{f+g}_p}^p & =\int_E\abs{f+g}\abs{f+g}^{p-1}                                                  \\
+                       & \leq \int_E\abs{f}\abs{f+g}^{p-1}+\int_E\abs{g}\abs{f+g}^{p-1}                   \\
       \begin{split}
-        & \leq \left[{\left(\int_E\abs{f}^p\right)}^{1/p}+{\left(\int_E\abs{g}^p\right)}^{1/p}\right]\cdot \\
-        & \hspace{2.5cm}\cdot{\left(\int_E\abs{f+g}^{(p-1)\frac{p}{p-1}}\right)}^{1-1/p}
+         & \leq \left[{\left(\int_E\abs{f}^p\right)}^{1/p}+{\left(\int_E\abs{g}^p\right)}^{1/p}\right]\cdot \\
+         & \hspace{2.5cm}\cdot{\left(\int_E\abs{f+g}^{(p-1)\frac{p}{p-1}}\right)}^{1-1/p}
       \end{split} \\
                        & =(\norm{f}_p+\norm{g}_p)\frac{{\norm{f+g}_p}^p }{{\norm{f+g}_p}}
     \end{align*}
@@ -1056,14 +1056,14 @@
       \item\label{RFA:TcontinuousC} $T(B_E)$ is bounded on $F$, where $B_E:=\{x\in E:\norm{x}_E\leq 1\}$.
       \item\label{RFA:TcontinuousD} $\norm{T}<\infty$.
       \item\label{RFA:TcontinuousE} $\exists C\geq 0$ such that $\forall x\in E$ we have: $$\norm{Tx}_F\leq C\norm{x}_E$$
-      If, moreover, $T$ is continuous, $\norm{T}$ is the least of such constants $C$.
+            If, moreover, $T$ is continuous, $\norm{T}$ is the least of such constants $C$.
     \end{enumerate}
   \end{theorem}
   \begin{sproof}
     \begin{enumerate}[leftmargin=1.5cm]
       \item[\mref{RFA:TcontinuousA}$\implies$\mref{RFA:TcontinuousB}:] Let $x\in E$ and $(x_n)\in E$ such that $\displaystyle \lim_{n\to\infty}x_n=x$. Then $\displaystyle \lim_{n\to\infty}(x_n - x)=0$ and the continuity and linearity imply $\displaystyle \lim  _{n\to\infty}(Tx_n-Tx)=0$.
       \item[\mref{RFA:TcontinuousB}$\implies$\mref{RFA:TcontinuousC}:] The continuity at the origin of $T$ implies that given $\varepsilon =1$, $\exists \delta>0$ such that: $$T(B_E(0,\delta))\subseteq B_F(0,1)$$
-        The linearity of $T$ implies that $T(B_E(0,1))\subseteq B_F(0,1/\delta)$.
+            The linearity of $T$ implies that $T(B_E(0,1))\subseteq B_F(0,1/\delta)$.
       \item[\mref{RFA:TcontinuousC}$\implies$\mref{RFA:TcontinuousD}:] Consequence of \mcref{RFA:normT}.
       \item[\mref{RFA:TcontinuousD}$\implies$\mref{RFA:TcontinuousE}:] By the definition of supremum we have: $$\norm{T\left(\frac{x}{\norm{x}_E}\right)}_F\leq \norm{T}$$ And so $\norm{Tx}_F\leq \norm{T}\norm{x}_E$.
       \item[\mref{RFA:TcontinuousE}$\implies$\mref{RFA:TcontinuousA}:] Evident.
@@ -1354,7 +1354,7 @@
   \begin{theorem}[Open mapping theorem]
     Let $E$, $F$ be a Banach spaces and $T:E\rightarrow F$ be a surjective bounded operator. Then, $T$ is open.
   \end{theorem}
-  \begin{theorem}[Closed graph theorem]
+  \begin{theorem}[Closed graph theorem]\label{RFA:closedgraph}
     Let $E$, $F$ be a Banach spaces and $T:E\rightarrow F$ be an operator. Consider the graph of $T$: $$\graph(T)=\{(x,y)\in E\times F:y=Tx\}$$
     Then, $T$ is bounded if and only if $\graph(T)$ is a closed set on $E\times F$.
   \end{theorem}