From c867a1389de10222c3f61087f2763042137293b5 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?V=C3=ADctor?= Date: Sun, 17 Sep 2023 18:19:14 +0200 Subject: [PATCH] updated all analysis --- .../Real_and_functional_analysis.tex | 8 +- ...topics_in_functional_analysis_and_PDEs.tex | 161 ++++++++++++++++++ main_math.tex | 2 +- 3 files changed, 166 insertions(+), 5 deletions(-) diff --git a/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex b/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex index 5c2286f..a045837 100644 --- a/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex +++ b/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex @@ -976,7 +976,7 @@ We say that $F$ is \emph{pointwise equicontinuous} if it is equicontinuous at each point of $X$. Finally, we say that $F$ is \emph{uniformly equicontinuous} if $\forall \varepsilon>0$ $\exists \delta>0$ such that $\forall x,y\in X$ with $d(x,y)<\delta$ we have: $$d_Y(f(x),f(y))<\varepsilon\quad\forall f\in F$$ \end{definition} \begin{definition} - Let $(X,d)$ be a metric space and $F\subseteq X$. We say that $F$ is relatively compact on $X$ if $\overline{F}$ is compact on $X$. + Let $(X,d)$ be a metric space and $F\subseteq X$. We say that $F$ is \emph{relatively compact} on $X$ if $\overline{F}$ is compact on $X$. \end{definition} % \begin{lemma} % Let $(X,d)$ be a metric space and $F\subseteq X$. $F$ is relatively compact on $X$ if and only if any bounded sequence $(x_n)\in F$ has a partial convergent subsequence. @@ -1621,16 +1621,16 @@ \end{proposition} \subsubsection{Lax-Milgram theorem} \begin{definition} - Let $H$ be a Hilbert space and $a:H\times H\rightarrow\CC$ be a bilinear map. We say that $a$ is \emph{continuous} if $\exists C>0$ such that $\forall u,v\in H$ we have: $$\abs{a(u,v)}\leq C\norm{u}\norm{v}$$ + Let $H$ be a Hilbert space and $a:H\times H\rightarrow\RR$ be a bilinear map. We say that $a$ is \emph{continuous} if $\exists C>0$ such that $\forall u,v\in H$ we have: $$\abs{a(u,v)}\leq C\norm{u}\norm{v}$$ \end{definition} \begin{definition} - Let $H$ be a Hilbert space and $a:H\times H\rightarrow\CC$ be a bilinear map. We say that $a$ is \emph{coercive} if $\exists\alpha>0$ such that $\forall u\in H$ we have: $$a(u,u)\geq\alpha\norm{u}^2$$ + Let $H$ be a Hilbert space and $a:H\times H\rightarrow\RR$ be a bilinear map. We say that $a$ is \emph{coercive} if $\exists\alpha>0$ such that $\forall u\in H$ we have: $$a(u,u)\geq\alpha\norm{u}^2$$ \end{definition} \begin{definition} Let $H$ be a Hilbert space and $a:H\times H\rightarrow\CC$ be a bilinear map. We say that $a$ is \emph{symmetric} if $\forall u,v\in H$ we have: $$a(u,v)=\overline{a(v,u)}$$ \end{definition} \begin{theorem}[Lax-Milgram theorem] - Let $H$ be a Hilbert space and $a:H\times H\rightarrow\CC$ be a continuous and coercive bilinear map. Then, $\forall L\in H^*$ there exists a unique $u\in H$ such that: $$a(u,v)=L(v)\quad \forall v\in H$$ + Let $H$ be a Hilbert space and $a:H\times H\rightarrow\RR$ be a continuous and coercive bilinear map. Then, $\forall L\in H^*$ there exists a unique $u\in H$ such that: $$a(u,v)=L(v)\quad \forall v\in H$$ In addition, if $\mathcal{H}$ is a real Hilbert space and $a$ is symmetric, then $u$ is the unique minimizer of: $$\min_{v\in H}\left\{\frac{1}{2}a(v,v)-L(v)\right\}$$ \end{theorem} diff --git a/Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex b/Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex index 08db5f8..88a5f14 100644 --- a/Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex +++ b/Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex @@ -90,6 +90,7 @@ where the convergence of the integral is due to the weakly convergence of $f_n$ to $f$. \end{proof} \subsection{Sobolev spaces} + \subsubsection{Basic definitions} \begin{definition}[Sobolev spaces] Let $\Omega\subseteq \RR^d$ be an open set, $m\in\NN$ and $1\leq p\leq \infty$. We define the \emph{Sobolev spaces} $W^{m,p}$ as: $$ @@ -179,5 +180,165 @@ \norm{u-\fint_\Omega u}_p\leq C\norm{\grad u}_p $$ \end{theorem} + \subsubsection{Sobolev embeddings} + \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] + 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 $\displaystyle\frac{1}{q}=\frac{1}{p}-\frac{m}{d}$. + \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: + \begin{multline*} + \abs{u(x_1,x_2)}\leq \int_{-\infty}^{x_1}\abs{\partial_{x_1}u(s,x_2)}\dd{s}\leq\\\leq\int_{\RR}\abs{\partial_{x_1}u(s,x_2)}\dd{s}=:v_1(x_2) + \end{multline*} + Similarly, $\abs{u(x_1,x_2)}\leq v_2(x_1)$. So: + \begin{multline*} + {\norm{u}_2}^2\leq \int_{\RR^2} \abs{v_1(x_2)}\abs{v_2(x_1)}\dd{x_1}\dd{x_2}=\norm{v_1}_1 \norm{v_2}_1=\\=\norm{\partial_{x_1}u}_1\norm{\partial_{x_2}u}_1\leq {\norm{\grad u}_1}^2 + \end{multline*} + For the case $1\leq p<2$, we apply the result to the function $u_t:=\abs{u}^{t-1}u$. This function satisfies $\abs{\grad u_t}=t\abs{u}^{t-1}\abs{\grad u}$ and so: + \begin{multline*} + {\norm{u}_{2t}}^t=\norm{u_t}_2\leq \norm{u_t}_1=t\norm{\abs{u}^{t-1}\grad u}\leq \\\leq t\norm{\abs{u}^{t-1}}_{p'}\norm{\grad u}_p=t{\norm{u}_{(t-1)p'}}^{t-1} \norm{\grad u} + \end{multline*} + where $p'$ is the Hölder conjugate of $p$. Now, we choose $t$ so that $2t=(t-1)p'$, that is, $t=\frac{p}{2-p}$ and so: + $$ + \norm{u}_{\frac{2p}{2-p}}\leq \frac{p}{2-p}\norm{\grad u}_p + $$ + \end{proof} + \begin{definition}[Hölder continuity] + Let $k\in\NN$ and $0\leq\theta\leq 1$. We say that a map $u:\Omega\to\CC$ is \emph{$\mathcal{C}^{k,\theta}$-Hölder continuous} if there is $C\geq 0$ such that $\forall\abs{\alpha}=k$ and all $x,y\in\Omega$ we have: + $$ + \abs{\partial^\alpha u(x)-\partial^\alpha u(y)}\leq C\abs{x-y}^\theta + $$ + The set of such functions is denoted by $\mathcal{C}^{k,\theta}(\Omega)$. + \end{definition} + \begin{remark} + Note that $\mathcal{C}^{k,0}(\Omega)=\mathcal{C}^k(\Omega)$ and that $\mathcal{C}^{0,1}(\Omega)$ is the set of Lipschitz continuous functions. + \end{remark} + \begin{remark} + Note that $\mathcal{C}^{k,\theta}(\Omega)$ together with the norm + $$ + \norm{u}_{\mathcal{C}^{k,\theta}(\Omega)}:=\sup_{x\ne y}\sup_{\abs{\alpha}=k}\frac{\abs{\partial^\alpha u(x)-\partial^\alpha u(y)}}{\abs{x-y}^\theta} + $$ + is a Banach space. + \end{remark} + \begin{theorem}[Morrey's embedding]\label{ATFAPDE:morrey_embedding} + Let $m\geq 1$ and $p>\frac{d}{m}$. Then, $W^{m,p}(\RR^d)\subset L^\infty(\RR^d)$. In addition, let: + $$ + k:=\left\lfloor m-\frac{d}{p}\right\rfloor\qquad\theta:=m-\frac{d}{p}-k\in [0,1) + $$ + If $\theta\ne 0$, then $W^{m,p}(\RR^d)\subset \mathcal{C}^{k,\theta}(\RR^d)$. + \end{theorem} + \begin{theorem} + For all $1\leq p\leq \infty$, $W^{1,p}(\RR)\hookrightarrow L^\infty(\RR)\cap \mathcal{C}^{0}(\RR)$. + \end{theorem} + \begin{proof} + The proof for $p>1$ comes from \mnameref{ATFAPDE:morrey_embedding}. For $p=1$ we have that $\forall u\in \mathcal{C}_0^\infty(\RR)$ we have: + $\abs{u(x)}\leq \int_{-\infty}^x{\abs{u'}}\leq \norm{u'}_1$. So $\norm{u}_\infty\leq \norm{u'}_1$. By density, this proves that $\forall u\in W^{1,1}(\RR)$ we have $u\in L^\infty(\RR)$. Now, let $(u_n)\in \mathcal{C}_0^\infty(\RR)$ be such that $u_n\to u$ in $W^{1,1}(\RR)$. Then, $u_n\to u$ in $L^\infty(\RR)$ and so $u$ is continuous because it is the uniform limit of continuous functions. + \end{proof} + \subsubsection{Extension operators} + \begin{definition} + Let $\Omega\subseteq \RR^d$ be an open set. An \emph{extension} of $u\in W\in W^{m,p}(\Omega)$ is a function $\tilde{u}\in W^{m,p}(\RR^d)$ so that $\tilde{u}\almoste{=}u$ in $\Omega$. An \emph{extension operator} is a bounded linear operator $E:W^{m,p}(\Omega)\to W^{m,p}(\RR^d)$ so that $Eu$ is an extension of $u$ $\forall u\in W^{m,p}(\Omega)$. + \end{definition} + \begin{remark} + From now on, we will denote $\RR_{\pm}^d:=\RR^{d-1}\times\RR_{\pm}$ and $\RR_0^d:=\RR^{d-1}\times\{0\}$. + \end{remark} + \begin{theorem} + For all $m\in\NN$ and all $1\leq p<\infty$, $\mathcal{C}^\infty(\overline{\RR_+^d})$ is dense in $W^{m,p}(\RR_+^d)$. + \end{theorem} + \begin{proof} + Let $$ + \tau_h(u)(x_1,\ldots, x_d):=u(x_1,\ldots, x_{d-1},x_d+h) + $$ + be the translation operator and set $u_\varepsilon:=\tau_{\varepsilon}(u)*\phi_\varepsilon$, where $\varepsilon>0$ and $\phi_\varepsilon$ is an approximation of identity. Then, $u_\varepsilon\in \mathcal{C}^\infty(\overline{\RR_+^d})$ by the properties of the convolution. Moreover: + \begin{multline*} + \norm{\partial^\alpha u_\varepsilon-\partial^\alpha u}_p \leq \norm{\partial^\alpha u_\varepsilon-\partial^\alpha (\tau_{\varepsilon}u)}_p+\norm{\partial^\alpha (\tau_{\varepsilon}u)-\partial^\alpha u}_p \\ + \leq \norm{(\partial^\alpha \tau_\varepsilon u)*\phi_\varepsilon-\partial^\alpha (\tau_{\varepsilon}u)}_p+\norm{\tau_{\varepsilon}(\partial^\alpha u)-\partial^\alpha u}_p + \end{multline*} + The first term goes to zero by the properties of smoothing sequences, and the second goes to zero since translations are continuous in $L^p$ (check \mcref{HA:translated}). + \end{proof} + \begin{remark} + The same proof shows that $\mathcal{C}^\infty(\overline{\Omega})$ is dense in $W^{m,p}(\Omega)$, if $\Omega$ is bounded with $\Fr{\Omega}$ of class $\mathcal{C}^1$. This time, one needs to locally translate u along the normal direction. + \end{remark} + \begin{theorem} + For all $m\in\NN$ and all $1\leq p<\infty$, there is an extension operator $E:W^{m,p}(\RR_+^d)\to W^{m,p}(\RR^d)$. + \end{theorem} + \begin{proof} + We only do the proof for $d=1$ and $m=1$ to highlight the main ideas. Let $u\in W^{1,p}(\RR_+)$. We define the \emph{first order reflection}: + $$ + \bar{u}:=\begin{cases} + u(x) & \text{if }x\geq 0 \\ + -3u(-x)+4u(-x/2) & \text{if }x<0 + \end{cases} + $$ + By density, it is enough to prove the result for $u\in \mathcal{C}^1(\RR_+)$. An easy check shows that $\bar{u}\in \mathcal{C}^1(\RR)$. Moreover, we have: + \begin{align*} + {\norm{\bar{u}}_{W^{1,p}(\RR)}}^p & =\int_{\RR}{\abs{\bar{u}}^p}+{\abs{\bar{u}'}^p} \\ + \begin{split} + &=\!\int_{\RR_+}\!{\abs{u}^p}\!+\!\!{\abs{u'}^p}\!+\!\int_{\RR_-}\![{\abs{-3u(-x)+4u(-x/2)}^p}+\\ + &\hspace{2.75cm}+{\abs{3u'(-x)-2u'(-x/2)}^p}] + \end{split} \\ + & \leq C{\norm{u}_{W^{1,p}(\RR_+)}}^p + \end{align*} + for some constant $C>0$. Thus, $E$ is a bounded extension operator. + \end{proof} + \begin{remark} + The reader can check that the same construction works on $\RR^d$, by setting + $$ + \bar{u}(x_1,\ldots,x_d):= + \begin{cases} + u(x_1,\ldots,x_d) & \text{if }x_d\geq 0 \\ + \begin{split} + -3u(x_1,\ldots,x_{d-1}-x_d)+\\ + +4u(x_1,\ldots,x_{d-1}-x_d/2) + \end{split} & \text{if }x_d<0 + \end{cases} + $$ + The proof for higher derivatives $m \geq 1$ needs to add more terms in order to make the junction smooth enough. + \end{remark} + \begin{definition} + We say that a domain $\Omega\subseteq \RR^d$ has boundary of class $\mathcal{C}^k$ if $\forall x\in \Fr{\Omega}$ there is a neighborhood $\varepsilon,\delta>0$ and a $\mathcal{C}^k$-diffeomorphism $\phi:B(x,\varepsilon)\to B(0,\delta)$ so that $\phi(x)=0$ and $\phi(B(x,\varepsilon)\cap \Omega)=B(0,\delta)\cap \RR_+^d$. Note that in particular this implies that $\phi(\Fr{\Omega}\cap B(x,\varepsilon))=B(0,\delta)\cap \RR_0^d$. + \end{definition} + \begin{theorem} + Let $\Omega\subseteq \RR^d$ be a bounded domain with $\mathcal{C}^k$ boundary. Then, $\forall m\leq k$ and all $1\leq p<\infty$, there is an extension operator $E:W^{m,p}(\Omega)\to W^{m,p}(\RR^d)$. + \end{theorem} + \begin{theorem} + Let $\Omega\subseteq \RR^d$ be a bounded domain with $\mathcal{C}^k$ boundary. Then, $\forall m\leq k$, if $1\leq p<\frac{d}{m}$, + % now add the conclusions of the gagliardo theorem + there is an embedding $W^{m,p}(\Omega)\hookrightarrow L^q(\Omega)$, where $\displaystyle\frac{1}{q}=\frac{1}{p}-\frac{m}{d}$. If $p>\frac{d}{m}$, then $W^{m,p}(\Omega)\hookrightarrow \mathcal{C}^{k-m,\theta}(\Omega)$, where $\theta=m-\frac{d}{p}-\ell$ and $\ell:=\left\lfloor m-\frac{d}{p}\right\rfloor$. + \end{theorem} + \begin{theorem}[Reillich-Kondrachov's compactness theorem] + Let $\Omega\subseteq \RR^d$ be a bounded domain with $\mathcal{C}^k$ boundary. Then, $\forall m\leq k$ we have: + \begin{itemize} + \item If $1\leq p<\frac{d}{m}$, $\forall r\in [p,q)$, where $\displaystyle\frac{1}{q}=\frac{1}{p}-\frac{m}{d}$, the embedding $W^{m,p}(\Omega)\hookrightarrow L^r(\Omega)$ is compact. + \item If $p\geq \frac{d}{m}$, then $\forall r\in [p,\infty)$, the embedding $W^{m,p}(\Omega)\hookrightarrow L^r(\Omega)$ is compact. + \item If $p>\frac{d}{m}$, then the embedding $W^{m,p}(\Omega)\hookrightarrow \mathcal{C}^{0}(\overline{\Omega})$ is compact. + \end{itemize} + \end{theorem} + \subsubsection{Trace operators} + \begin{theorem} + Let $1\leq p<\infty$ and $u\in W^{1,p}(\RR_+^d)$. Then, the function $u|_{\RR_0^d}:\RR^{d-1}\to\CC$ belongs to $L^p(\RR^{d-1})$. + \end{theorem} + \begin{definition} + We define the \emph{trace operator} as the map: + \begin{align*} + \function{T}{W^{1,p}(\RR_+^d)}{L^p(\RR^{d-1})}{u}{u|_{\RR_0^d}} + \end{align*} + \end{definition} + \begin{theorem} + Let $1\leq p<\infty$ and $u\in W^{1,p}(\RR_+^d)$. Then, $Tu=0$ if and only if $u\in W_0^{1,p}(\RR_+^d)$. + \end{theorem} + \begin{lemma} + Let $\Omega\subseteq \RR^d$ be an open set and $u\in W^{1,p}(\Omega)$ with $1\leq p\leq \infty$. Then, $\norm{\grad \abs{u}}\almoste{\leq}\norm{\grad u}$. + \end{lemma} + \begin{proof} + $$ + \norm{2\abs{u}\grad\abs{u}}=\norm{\grad \abs{u}^2}=2\norm{\Re(\overline{u}\grad u)}\leq 2\norm{\grad u}\abs{u} + $$ + On the set $\{u\ne 0\}$, we can divide by $2\abs{u}$, which gives the result. The proof on the set $\{u=0\}$ is much difficult. + \end{proof} \end{multicols} \end{document} \ No newline at end of file diff --git a/main_math.tex b/main_math.tex index 8618a04..8c22fdf 100644 --- a/main_math.tex +++ b/main_math.tex @@ -98,7 +98,7 @@ \chapter{Fourth year} \cleardoublepage -\chapter{Fourth year} +\chapter{Fifth year} \newpage \subfile{Mathematics/5th/Advanced_probability/Advanced_probability.tex}