added more pdes

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}
@@ -666,7 +666,7 @@
     \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}
@@ -865,11 +865,11 @@
     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*}
@@ -1068,14 +1068,14 @@
       \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.
@@ -1134,7 +1134,7 @@
     We may suppose $\varepsilon<1$. Let $v\in E$ such that $d(v,F)=\delta>0$. Then, consider $u=\frac{v-x_0}{\norm{v-x_0}}$, where $x_0\in F$ satisfies $\delta\leq \norm{v-x_0}\leq \frac{\delta}{1-\varepsilon}$. Finally, $\forall x\in F$:
     $$\norm{u-x}=\frac{\norm{v-(x_0+\norm{v-x_0}x)}}{\norm{v-x_0}}\geq \frac{\delta}{\norm{v-x_0}}\geq 1-\varepsilon$$
   \end{proof}
-  \begin{theorem}[Riesz's theorem]
+  \begin{theorem}[Riesz's theorem]\label{RFA:riesz_ball}
     Let $E$ be a normed vector space. If the unit closed sphere $\{x\in E:\norm{x}=1\}$ is compact, then $\dim E<\infty$.
   \end{theorem}
   \begin{proof}
@@ -1579,7 +1579,7 @@
     \end{align*}
   \end{proposition}
   \subsubsection{Duality and adjoint operator}
-  \begin{theorem}[Riesz representation theorem]
+  \begin{theorem}[Riesz representation theorem]\label{RFA:riesz_rep}
     Let $(H,\dotp{\cdot}{\cdot})$ be a Hilbert space. The map $$\function{}{H}{H^*}{x}{\dotp{\cdot}{x}}$$ is semilinear, bijective and isometric.
   \end{theorem}
   \begin{corollary}

Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex

@@ -184,11 +184,12 @@
   \begin{definition}
     Let $E$, $F$ be Banach. We say that $F$ is \emph{embedded} in $E$ if $F\subseteq E$ and the inclusion map $i:F\hookrightarrow  E$ is continuous. We say that $F$ is \emph{compactly embedded} in $E$ if $F\subseteq E$ and the inclusion map $i:F\hookrightarrow  E$ is compact.
   \end{definition}
-  \begin{theorem}[Gagliardo, Nirengerg and Sobolev's inequality]
+  \begin{theorem}[Gagliardo, Nirengerg and Sobolev's inequality]\label{ATFAPDE:gagliardo_nirengerg_sobolev}
     For all $1\leq p\leq\frac{d}{m}$, there is a constant $C=C(p,m,d)$ so that $\forall u\in \mathcal{C}_0^\infty(\RR^d)$ we have:
     $$
       \norm{u}_{q}\leq C\sum_{\abs{\alpha}= m}{\norm{\partial^\alpha u}_p}$$
     where $\displaystyle\frac{1}{q}=\frac{1}{p}-\frac{m}{d}$.
+    That is, we have the continuous embedding $W^{m,p}(\RR^d)\hookrightarrow L^q(\RR^d)$. In particular for $m=1$, we have $W^{1,p}(\RR^d)\hookrightarrow L^{p^*}(\RR^d)$ with $\displaystyle\frac{1}{p^*}=\frac{1}{p}-\frac{1}{d}$.
   \end{theorem}
   \begin{proof}
     By induction, it suffices to prove the result only for $m=1$. We will prove only the case $d=2$. We start with $p=1$. Let $u\in \mathcal{C}_0^\infty(\RR^2)$. We have: