updated complex analysis

Mathematics/4th/Harmonic_analysis/Harmonic_analysis.tex

@@ -75,10 +75,10 @@
     Let $f,g\in L^1(\RR)$ and $\alpha,\beta\in\RR$. Then:
     \begin{enumerate}
       \item $\widehat{(\alpha f+\beta g)}(\xi)=\alpha\widehat{f}(\xi)+\beta \widehat{g}(\xi)$
-      \item\label{HA:FTprop2} Let $h\in\RR$. We define $T_hf(x)=f(x+h)$. Then: $$\widehat{T_hf}(\xi)=\exp{2\pi\ii \xi h}\widehat{f}(\xi)$$
-      \item\label{HA:FTprop3} If $g(x)=\exp{2\pi\ii x h}f(x)$, then: $$\widehat{g}(\xi)=\widehat{f}(\xi-h)$$
-      \item\label{HA:FTprop4} If $\lambda\in\RR^*$, then: $$\frac{1}{\lambda}\widehat{f\left(\frac{x}{\lambda}\right)}(\xi)=\widehat{f}(\lambda\xi)$$
-      \item\label{HA:FTprop5} If $g(x)=\overline{f(x)}$, then: $$\widehat{g}(\xi)=\overline{\widehat{f}(-\xi)}$$
+            \item\label{HA:FTprop2} Let $h\in\RR$. We define $T_hf(x)=f(x+h)$. Then: $$\widehat{T_hf}(\xi)=\exp{2\pi\ii \xi h}\widehat{f}(\xi)$$
+            \item\label{HA:FTprop3} If $g(x)=\exp{2\pi\ii x h}f(x)$, then: $$\widehat{g}(\xi)=\widehat{f}(\xi-h)$$
+            \item\label{HA:FTprop4} If $\lambda\in\RR^*$, then: $$\frac{1}{\lambda}\widehat{f\left(\frac{x}{\lambda}\right)}(\xi)=\widehat{f}(\lambda\xi)$$
+            \item\label{HA:FTprop5} If $g(x)=\overline{f(x)}$, then: $$\widehat{g}(\xi)=\overline{\widehat{f}(-\xi)}$$
     \end{enumerate}
   \end{proposition}
   \begin{sproof}
@@ -139,12 +139,12 @@
     $$\lim_{n\to\infty}f^{(k-1)}(a_n)=\lim_{n\to\infty}f^{(k-1)}(b_n)=0$$
     Hence using integration by parts:
     \begin{align*}
-      \widehat{f^{(k)}}(\xi) & =\lim_{n\to\infty}\int_{a_n}^{b_n} f^{(k)}(x)\exp{-2\pi\ii \xi x}\dd{x}   \\
+      \widehat{f^{(k)}}(\xi) & =\lim_{n\to\infty}\int_{a_n}^{b_n} f^{(k)}(x)\exp{-2\pi\ii \xi x}\dd{x}  \\
       \begin{split}
-         & =\lim_{n\to\infty}  f^{(k-1)}(x)\exp{-2\pi\ii \xi x}\Big|_{a_n}^{b_n}\dd{x} +                   \\
-         & \hspace{2cm}+2\pi\ii \xi\lim_{n\to\infty}\int_{a_n}^{b_n}f^{(k-1)}(x)\exp{-2\pi\ii \xi x}\dd{x}
+        & =\lim_{n\to\infty}  f^{(k-1)}(x)\exp{-2\pi\ii \xi x}\Big|_{a_n}^{b_n}\dd{x} +                   \\
+        & \hspace{2cm}+2\pi\ii \xi\lim_{n\to\infty}\int_{a_n}^{b_n}f^{(k-1)}(x)\exp{-2\pi\ii \xi x}\dd{x}
       \end{split} \\
-                             & =\left(2\pi\ii \xi\right)\widehat{f^{(k-1)}}(n)                           \\
+                             & =\left(2\pi\ii \xi\right)\widehat{f^{(k-1)}}(n)                          \\
                              & ={\left(2\pi\ii \xi\right)}^k\widehat{f}(\xi)
     \end{align*}
   \end{proof}
@@ -310,7 +310,7 @@
             \begin{multline*}
               \lim_{R\to \infty}\sup_{\abs{x}\geq \delta}F_R(x)=\lim_{t\to 0}\sup_{\abs{x}\geq \delta}P_t(x)=\\=\lim_{t\to 0}\sup_{\abs{x}\geq \delta}W_t(x)=0
             \end{multline*}
-      \item\label{HA:propsKernelsitem4} For all $\delta>0$, we have:
+            \item\label{HA:propsKernelsitem4} For all $\delta>0$, we have:
             \begin{multline*}
               \lim_{R\to \infty}\int_{\abs{x}\geq \delta}F_R(x)\dd{x}=\lim_{t\to 0}\int_{\abs{x}\geq \delta}P_t(x)\dd{x}=\\=\lim_{t\to 0}\int_{\abs{x}\geq \delta}W_t(x)\dd{x}=0
             \end{multline*}
@@ -371,8 +371,8 @@
                                     & \leq\int_{-\infty}^{\infty}\!\!{\left[\int_{-\infty}^{\infty}\!{\phi_\varepsilon(\vf{y})}^p\abs{f(\vf{x}-\vf{y})-f(\vf{x})}^p\dd{\vf{x}}\right]}^{\!\frac{1}{p}}\!\!\dd{\vf{y}} \\
                                     & = \int_{-\infty}^{\infty}\phi_\varepsilon(\vf{y})\norm{f-T_{-\vf{y}}f}_p\dd{\vf{y}}                                                                                             \\
       \begin{split}
-         & \leq \int_{\abs{\vf{y}}< \delta}\phi_\varepsilon(\vf{y})\norm{f-T_{-\vf{y}}f}_p\dd{\vf{y}}+ \\
-         & \hspace{3cm}+2\norm{f}_p\int_{\abs{\vf{y}}\geq\delta}\phi_\varepsilon(\vf{y})\dd{\vf{y}}
+        & \leq \int_{\abs{\vf{y}}< \delta}\phi_\varepsilon(\vf{y})\norm{f-T_{-\vf{y}}f}_p\dd{\vf{y}}+ \\
+        & \hspace{3cm}+2\norm{f}_p\int_{\abs{\vf{y}}\geq\delta}\phi_\varepsilon(\vf{y})\dd{\vf{y}}
       \end{split}
     \end{align*}
     Given $\varepsilon>0$, by \mcref{HA:translated} $\exists\delta>0$ such that the first integral is bounded by $\varepsilon$. Now use this $\delta$ and \mcref{HA:propsKernelsitem4} to conclude that the second integral goes to 0 as $R\to\infty$.
@@ -619,7 +619,7 @@
   \begin{definition}
     Let $f\in L^2(\RR)$. We say that $f$ is \emph{bandlimited} if $\exists B\in\RR$ such that $\supp\widehat{f}\subseteq[-B,B]$.
   \end{definition}
-  \begin{theorem}[Nyquist-Shannon sampling theorem]
+  \begin{theorem}[Nyquist-Shannon sampling theorem]\label{HA:nyquistShannon}
     Let $f\in L^2(\RR)$ be bandlimited with constant $B$. Then:
     $$f(x)\overset{L^2}{=}\sum_{k\in\ZZ}f\left(\frac{k}{2B}\right)\frac{\sin(\pi(2Bx-k))}{\pi(2Bx-k)}$$
     Moreover: $${\norm{f}_2}^2=\frac{1}{2B}\sum_{k\in\ZZ}\abs{f\left(\frac{k}{2B}\right)}^2$$
@@ -636,6 +636,9 @@
     $${\norm{f}_2}^2={\norm{\widehat{f}}_2}^2=\frac{1}{2B}\sum_{k\in\ZZ}\abs{f\left(\frac{k}{2B}\right)}^2$$
     because by a similar argument as before, the Fourier coefficients of $\widehat{f}(k)$ (thought as periodically extended) are $\frac{1}{2B}f\left(\frac{-k}{2B}\right)$.
   \end{proof}
+  \begin{remark}
+    In the context of signal processing, \mnameref{HA:nyquistShannon} tells us that if a function $f$ contains no frequencies higher than $B$ hertz, then it can be completely determined from its ordinates at a sequence of points spaced less than $\frac{1}{2B}$ seconds apart.
+  \end{remark}
   \subsubsection{Discrete Fourier transform}
   \begin{definition}
     Consider a function $f$ with support $\{0,\ldots,N-1\}$. We can think $f$ as:
@@ -839,8 +842,8 @@
     \begin{align*}
       \abs{\phi_\varepsilon(\varphi)-\delta_{\vf{0}}(\varphi)} & \leq \int_\Omega \abs{\phi_\varepsilon(x)}\abs{\varphi(\vf{x})-\varphi(\vf{0})}\dd{\vf{x}} \\
       \begin{split}
-         & =\int_{\norm{\vf{x}}<\delta} \abs{\phi_\varepsilon(x)}\abs{\varphi(\vf{x})-\varphi(\vf{0})}\dd{\vf{x}}+               \\
-         & \hspace{1cm}+\int_{\norm{\vf{x}}\geq\delta} \abs{\phi_\varepsilon(x)}\abs{\varphi(\vf{x})-\varphi(\vf{0})}\dd{\vf{x}}
+        & =\int_{\norm{\vf{x}}<\delta} \abs{\phi_\varepsilon(x)}\abs{\varphi(\vf{x})-\varphi(\vf{0})}\dd{\vf{x}}+               \\
+        & \hspace{1cm}+\int_{\norm{\vf{x}}\geq\delta} \abs{\phi_\varepsilon(x)}\abs{\varphi(\vf{x})-\varphi(\vf{0})}\dd{\vf{x}}
       \end{split}
     \end{align*}
     Now use the properties of approximation of identity to see that each interval goes to zero as $\varepsilon\to 0$.
@@ -1063,15 +1066,15 @@
   \begin{proof}
     Clearly $T*\psi$ is linear. Let $\varphi_n\overset{\mathcal{S}}{\longrightarrow}0$. Then, it suffices to see that $\tilde\psi*\varphi_n\overset{\mathcal{S}}{\longrightarrow}0$. For the sake of simplicity we only do the case $d=1$. For all $\alpha,\beta\in{\NN\cup\{0\}}$ we have:
     \begin{align*}
-      \abs{\vf{x}^\alpha \partial^\beta(\tilde\psi*\varphi_n)(\vf{x})} & =\abs{\vf{x}^\alpha (\partial^\beta\tilde\psi*\varphi_n)(\vf{x})}                                                                  \\
-                                                                       & \leq \int_{\RR^d} \abs{\vf{x}^\alpha \partial^\beta\psi(\vf{y})\varphi_n(\vf{x}-\vf{y})}\dd{\vf{y}}                                \\
+      \abs{\vf{x}^\alpha \partial^\beta(\tilde\psi*\varphi_n)(\vf{x})} & =\abs{\vf{x}^\alpha (\partial^\beta\tilde\psi*\varphi_n)(\vf{x})}                                                                 \\
+                                                                       & \leq \int_{\RR^d} \abs{\vf{x}^\alpha \partial^\beta\psi(\vf{y})\varphi_n(\vf{x}-\vf{y})}\dd{\vf{y}}                               \\
       \begin{split}
-         & \leq 2^m\!\!\int_{\RR^d}\!\abs{\vf{x}-\vf{y}}^\alpha \abs{\partial^\beta\psi(\vf{y})}\abs{\varphi_n(\vf{x}-\vf{y})}\!\dd{\vf{y}} \\
-         & \quad\;+2^m\int_{\RR^d} \abs{\vf{y}}^\alpha\abs{\partial^\beta\psi(\vf{y})} \abs{\varphi_n(\vf{x}-\vf{y})}\dd{\vf{y}}            \\
+        & \leq 2^m\!\!\int_{\RR^d}\!\abs{\vf{x}-\vf{y}}^\alpha \abs{\partial^\beta\psi(\vf{y})}\abs{\varphi_n(\vf{x}-\vf{y})}\!\dd{\vf{y}} \\
+        & \quad\;+2^m\int_{\RR^d} \abs{\vf{y}}^\alpha\abs{\partial^\beta\psi(\vf{y})} \abs{\varphi_n(\vf{x}-\vf{y})}\dd{\vf{y}}            \\
       \end{split} \\
       \begin{split}
-         & \leq 2^m\sup_{\vf{x}\in\RR^d}\abs{\vf{x}}^\alpha\abs{\varphi_n(\vf{y})}\int_{\RR^d}  \abs{\partial^\beta\psi(\vf{y})}\dd{\vf{y}}  \\
-         & \;\;\;+2^m\sup_{\vf{x}\in\RR^d}\abs{\varphi_n(\vf{y})}\int_{\RR^d} \abs{\vf{y}}^\alpha\abs{\partial^\beta\psi(\vf{y})}\dd{\vf{y}} \\
+        & \leq 2^m\sup_{\vf{x}\in\RR^d}\abs{\vf{x}}^\alpha\abs{\varphi_n(\vf{y})}\int_{\RR^d}  \abs{\partial^\beta\psi(\vf{y})}\dd{\vf{y}}  \\
+        & \;\;\;+2^m\sup_{\vf{x}\in\RR^d}\abs{\varphi_n(\vf{y})}\int_{\RR^d} \abs{\vf{y}}^\alpha\abs{\partial^\beta\psi(\vf{y})}\dd{\vf{y}} \\
       \end{split}
     \end{align*}
     where in the second inequality we have used \mcref{HA:lemma_aMbM} with $m=\abs{\alpha}+1$. Note that this latter terms tend to zero as $n\to\infty$ because of the properties of the Schwartz space.
@@ -1435,9 +1438,9 @@
     \begin{align*}
       \begin{split}
         \widehat{f^2}+2\F{(\H(f\H f))} & =\int_{\RR} \widehat{f}(\eta)\widehat{f}(\xi-\eta)[1+m(\xi)\cdot \\
-                                       & \hspace{2cm}\cdot(m(\xi-\eta)+m(\eta))]\dd{\eta}
+          & \hspace{2cm}\cdot(m(\xi-\eta)+m(\eta))]\dd{\eta}
       \end{split} \\
-       & =\int_{\RR} \widehat{f}(\eta)\widehat{f}(\xi-\eta)m(\xi-\eta)m(\eta)\dd{\eta}                              \\
+       & =\int_{\RR} \widehat{f}(\eta)\widehat{f}(\xi-\eta)m(\xi-\eta)m(\eta)\dd{\eta}                  \\
        & =\widehat{\H f}*\widehat{\H f}=\widehat{(\H f)^2}
     \end{align*}
     where the second equality follows for all $\xi,\eta\in\RR^2\setminus\{(0,0)\}$