UPDATE: Chapter 4, review will start.

Author: gitcordier

FA_DM.pdf

@@ -24,35 +24,35 @@
 %\end{align}
 We now easily obtain
 \begin{align}\label{4_13_a_3}
-\| T_{j\,} \tilde{x}_{j\,} -T_{\, k} \tilde{x}_{\, k}\| \<
+\| T_{j\,} \tilde{x}_{j\,} -T_{\, k} \tilde{x}_{\, k}\| \leq
 \| T_{j\,} \tilde{x}_{j\,}- y_{j\,}\| +
 \| y_{j\,} -T_{j\,} \tilde{x}_{k\,}\| +
 \| T_{j\,} - T_{ k\,}\| \underset{k> j\to \infty}{\longrightarrow}0\quad .
 \end{align}
 $\{T_{ j\,} \tilde{x}_{\, j} \}$ is then a Cauchy sequence. So is $\{T \,\tilde{x}_{\, j}\}$, since $\| T-T_{j}\,\| \to 0$. On the other hand, $Y$ is complete: (a) is then proved and we now establish the counterpart in a Hilbert space.\\
 \\
 %: b
-Fix $\eps$ as a positive scalar. Since $T$ is compact, $Y$ contains a finite set $C$ such that 
+Fix $\epsilon$ as a positive scalar. Since $T$ is compact, $Y$ contains a finite set $C$ such that 
 \begin{align}\label{4_13_b_1}
-T(U\,)\subseteq \bigcup_{c\in C} B(c,\, \eps)\quad .
+T(U\,)\subseteq \bigcup_{c\in C} B(c,\, \epsilon)\quad .
 \end{align}
 As a Hilbert space, $Y$ contains a \textsl{maximal orthonormal set} (or \textsl{Hilbert basis}) $M$. This implies that $\text{span}(M)$ is dense in $Y$; \cf 4.18 \& 4.22 of \cite{Big_Rudin}. The finiteness of $C$ forces $M$ to enclose a finite set $S$ so that 
 \begin{align}\label{4_13_b_2}
-\forall c\in C, \, \exists s(c\,) \in \text{span} (S\,):\, \|c - s(c\,)\| <\eps\quad .
+\forall c\in C, \, \exists s(c\,) \in \text{span} (S\,):\, \|c - s(c\,)\| <\epsilon\quad .
 \end{align}
 Let $x$ be in $U$. It follows from (\ref{4_13_b_1}) that 
 \begin{align}\label{4_13_b_3}
-\|Tx - c_x \| < \eps
+\|Tx - c_x \| < \epsilon
 \end{align}
 for some $c_x$ of $C$. We now combine (\ref{4_13_b_2}) and (\ref{4_13_b_3}) to obtain
 \begin{align}
-\|Tx - s(c_x) \| \< \| Tx - c_x \| + \| c_x - s(c_x)\| < 2\eps
+\|Tx - s(c_x) \| \leq \| Tx - c_x \| + \| c_x - s(c_x)\| < 2\epsilon
 \end{align}
 As a finite-dimensional subspace, $\text{span}(S\,)$ is closed (see footnote 4, Exercise 1.10). We so obtain
 \begin{align}
 Y=\text{span}(S\,)\oplus \text{span}(S\,)^\bot \quad ,
 \end{align}
-by [12.4]. There so exists a unique projection projection $\pi=\pi(\eps)$ of $Y$ onto itself (see [5.6] for the definition) such that
+by [12.4]. There so exists a unique projection projection $\pi=\pi(\epsilon)$ of $Y$ onto itself (see [5.6] for the definition) such that
 \begin{align}
 \pi (Y) = \text{span}(S\,)\, ,\, \, (I-\pi)(Y\,) = \text{span}(S\,)^\bot\quad .
 \end{align}
@@ -66,13 +66,13 @@
 \end{align}
 Then,
 \begin{align}
-\|( I- \pi )(T x) \| \< \| I- \pi \| \, \| Tx -s(c_x)\| < 4 \eps \quad (x\in U)\quad 
+\|( I- \pi )(T x) \| \leq \| I- \pi \| \, \| Tx -s(c_x)\| < 4 \epsilon \quad (x\in U)\quad 
 \end{align}
 (the fact that $\pi$ has norm $1$ is hidden in the right side inequality). We have just so proved that 
 \begin{align}
-\| T-\pi\circ T\, \| \in \underset{\eps \sim 0}{O}(\eps) \quad . 
+\| T-\pi\circ T\, \| \in \underset{\epsilon \sim 0}{O}(\epsilon) \quad . 
 \end{align}
-That is particularly true whether $\eps=\eps_0,\, \eps_1,\, \eps_2,\, \dots ,\, \eps_n \underset{n\to \infty}{\longrightarrow} 0\,$. Let so $T_n$ be $ \pi(\eps_n) \circ T\,$ and conclude that these (compact) operators approximate $T$ in the desired fashion, \ie
+That is particularly true whether $\epsilon=\epsilon_0,\, \epsilon_1,\, \epsilon_2,\, \dots ,\, \epsilon_n \underset{n\to \infty}{\longrightarrow} 0\,$. Let so $T_n$ be $ \pi(\epsilon_n) \circ T\,$ and conclude that these (compact) operators approximate $T$ in the desired fashion, \ie
 \begin{align}
 \| T-T_n \| \underset{n \to \infty}{\longrightarrow} 0\quad .
 \end{align}

chapter_4/4_13.tex

@@ -7,9 +7,9 @@
 \renewcommand{\labelenumi}{(\alph{enumi})}
 \item Prove that $T\in \mathscr{B}(L^2(\mu))$ and that 
 \begin{align*}
-\| T\, \|^2 \< \int_\Omega\int_\Omega \lvert K (s,t\,)\rvert ^2 d\mu (s\,) d\mu(t\,).
+\| T\, \|^2 \leq \int_\Omega\int_\Omega \lvert K (s,t\,)\rvert ^2 d\mu (s\,) d\mu(t\,).
 \end{align*}
-\item Suppose $a_i$, $b_i$ are members of $L^2(\mu)$, for $1\< i\< n$, put $K_1=\sum a_i(s\,)b_i(t\,)$ and define $T_1$ in terms of $K_1$ a $T$ was defined in terms of $K$. Prove that $\dim \mathscr{R}(T_1)\< n$.
+\item Suppose $a_i$, $b_i$ are members of $L^2(\mu)$, for $1\leq i\leq n$, put $K_1=\sum a_i(s\,)b_i(t\,)$ and define $T_1$ in terms of $K_1$ a $T$ was defined in terms of $K$. Prove that $\dim \mathscr{R}(T_1)\leq n$.
 \item Deduce that $T$ is a compact operator in $L^2(\mu)$. Hint: Use exercise 13.
 \item Suppose $\lambda\in \C,\, \lambda\neq 0$. Prove: Either the equation 
 \begin{align*}Tf-\lambda f=g\end{align*}
@@ -20,28 +20,28 @@
 %: a 
 \paragraph{PROOF.} Let $X$ (respectively $P\,$) be the Banach space $L^2(\mu)$ (respectively $L^2(\mu\times \mu)\,$). A consequence of the Radon-Nikodym theorem (\cf 6.16 of \cite{Big_Rudin}\,) is that there exists a group isomorphism $\rho:\, X\to \, X^{\,\ast},\, f\mapsto f^{\,\,\ast}$ such that
 \begin{align}
-\langle u ,\, f^{\,\,\ast} \rangle =\int_\Omega u\mdot f \, \d\mu \quad (u\in X,\, f\in X)\quad .
+\langle u ,\, f^{\,\,\ast} \rangle =\int_\Omega u\cdot f \, \d\mu \quad (u\in X,\, f\in X)\quad .
 \end{align}
 Define a.e $K_s,\, K_t:\, \Omega\to \C$ by setting
 \begin{align}
 K_s(t\,)\Def K_t(s\,)\Def K(s,\, t\,)\, \, \text{ a.e}\quad \left((s,t\,)\in \Omega\right)\quad .
 \end{align}
 $T$ is clearly linear. Moreover, 
 \begin{align}
-\lvert (T f\,)(s\,) \rvert = \lvert \langle K_s,\, f^{\,\,\ast} \rangle \rvert \< \| K_s\|_X\quad(\|\,f\,\,\|_X < 1) \quad
+\lvert (T f\,)(s\,) \rvert = \lvert \langle K_s,\, f^{\,\,\ast} \rangle \rvert \leq \| K_s\|_X\quad(\|\,f\,\,\|_X < 1) \quad
 \end{align}
 (the latter inequality is a Cauchy-Schwarz one). Now apply the Fubini's theorem with $\lvert K\,\rvert^2$ to obtain
 \begin{align}
-\|Tf\,\,\|^2_X \< \int_\Omega \| K_s\|^2_X \, \,\d \mu (s\,) =\| K\,\|_P^2 \,< \infty \quad(\|\,f\,\,\|_X < 1)\quad .
+\|Tf\,\,\|^2_X \leq \int_\Omega \| K_s\|^2_X \, \,\d \mu (s\,) =\| K\,\|_P^2 \,< \infty \quad(\|\,f\,\,\|_X < 1)\quad .
 \end{align}
 (a) is then proved. \\
 \\
 %: b
 To show (b), remark that
 \begin{align}
-\int_\Omega a_i (s\,) \mdot b_i \mdot f \,\,\d\mu \,\,\in \C\mdot a_i(s\,)\,\,\,\text{a.e}\quad \quad (f\in X, \, s\in \Omega) \quad.
+\int_\Omega a_i (s\,) \cdot b_i \cdot f \,\,\d\mu \,\,\in \C\cdot a_i(s\,)\,\,\,\text{a.e}\quad \quad (f\in X, \, s\in \Omega) \quad.
 \end{align}
-It is now clear that $T$ maps any $f$ of $X$ into $\C \mdot a_1+\dotsb+ \C\mdot a_n$. We so conclude that $\dim R(T_1)\< n $. \\
+It is now clear that $T$ maps any $f$ of $X$ into $\C \cdot a_1+\dotsb+ \C\cdot a_n$. We so conclude that $\dim R(T_1)\leq n $. \\
 \\
 %: c
 We now aim at (c). The current part refers to Exercise 4.13. $X$ is also a Hilbert space and so contains a Hilbert basis $M$. Define a.e
@@ -51,25 +51,25 @@
 \end{align}
 whenever $b$ ranges $M$. Hence,
 \begin{align}
-K_s= \sum_{b\in M} a_b(s\,) \mdot b \, \text{ a.e} \quad (s\in \Omega) \quad.
+K_s= \sum_{b\in M} a_b(s\,) \cdot b \, \text{ a.e} \quad (s\in \Omega) \quad.
 \end{align}
-Provided any positive scalar $\eps$, there so exists a finite subset $S=S(\eps)$ of $M$ such that
+Provided any positive scalar $\epsilon$, there so exists a finite subset $S=S(\epsilon)$ of $M$ such that
 \begin{align}
-\| K_s - \sum_{b\in S} a_b(s\,) \mdot b\,\|_X < \eps \quad (s\in \Omega)\quad .
+\| K_s - \sum_{b\in S} a_b(s\,) \cdot b\,\|_X < \epsilon \quad (s\in \Omega)\quad .
 \end{align}
-Remark that $\underset{b\in S}{\sum} a_b \mdot b$ matches the definition of $K_1$; \cf(b): from now on, 
+Remark that $\underset{b\in S}{\sum} a_b \cdot b$ matches the definition of $K_1$; \cf(b): from now on, 
 \begin{align}
-K_1\Def \sum_{b\in S} a_b \mdot b\quad .
+K_1\Def \sum_{b\in S} a_b \cdot b\quad .
 \end{align}
 It follows from (b) that
 \begin{align}
 \dim R(K_1) < \infty \quad.
 \end{align}
 Now turn back to (a), with $K-K_1$ playing the role of $K$, and so obtain
 \begin{align}
-\|T-T_1\| < \eps \mu(\Omega)\< \infty \quad .
+\|T-T_1\| < \epsilon \mu(\Omega)\leq \infty \quad .
 \end{align}
-For if $\mu$ is finite, use (a) of Exercise 4.13 to conclude that $T$ is compact. Assume henceforth that $\mu$ is not (necessarily) finite and pick $\delta$ in $\R_+$. The simple functions (with finite measure support\,) form a dense family of an $L^p$ space ($1\< p<\infty$); \cf 3.13 of \cite{Big_Rudin}. It then exists a simple function $K_\delta$ of $L^2(\mu\times \mu)$ such that 
+For if $\mu$ is finite, use (a) of Exercise 4.13 to conclude that $T$ is compact. Assume henceforth that $\mu$ is not (necessarily) finite and pick $\delta$ in $\R_+$. The simple functions (with finite measure support\,) form a dense family of an $L^p$ space ($1\leq p<\infty$); \cf 3.13 of \cite{Big_Rudin}. It then exists a simple function $K_\delta$ of $L^2(\mu\times \mu)$ such that 
 \begin{align}
 (\mu\times\mu)\left(\{K_\delta\neq 0\}\right) <\infty \,, \,\, \| K-K_\delta \|_P <\delta\quad .
 \end{align}
@@ -79,13 +79,13 @@
 \end{align}
 The key ingredient is that $K_\delta$ can be identified with an element of the finite measure space $L^2(\{K_\delta\neq 0\},\mu\times\mu)\,$. What we have attempted to approximate $T$ by $T_1$ can therefore be reiterated (with $K_\delta$ playing the role of $K$) to achieve an approximation $T_{\delta,1}$ of $T_\delta$ so that
 \begin{align}\label{4_15_14}
-\|T_{\delta}-T_{\delta,1}\| < \eps\quad .
+\|T_{\delta}-T_{\delta,1}\| < \epsilon\quad .
 \end{align}
 It now follows from (\ref{4_15_13}) and (\ref{4_15_14}) that
 \begin{align}
-\|T-T_{\delta,1}\|\< \|T-T_{\delta}\| +\|T_{\delta}-T_{\delta,1}\| < \eps+\delta\quad .
+\|T-T_{\delta,1}\|\leq \|T-T_{\delta}\| +\|T_{\delta}-T_{\delta,1}\| < \epsilon+\delta\quad .
 \end{align}
-Since $\eps$ and $\delta$ were arbitrary, the $\sigma$-finite case is proved. We now establish (d).\\
+Since $\epsilon$ and $\delta$ were arbitrary, the $\sigma$-finite case is proved. We now establish (d).\\
 \\
 %: d
 Provided $g$ of $X$, let $E_g$ be the following equation on $X$
@@ -103,14 +103,14 @@
 \\
 Our last step is the description of $T^{\,\ast}$. Let $S:\, X\to X$ be such that
 \begin{align}
-(Sf\,)(t\,)\Def \int_\Omega K_t \mdot f \,\,\, \text{a.e}\quad \quad (\,f\in X,\, t\in \Omega)
+(Sf\,)(t\,)\Def \int_\Omega K_t \cdot f \,\,\, \text{a.e}\quad \quad (\,f\in X,\, t\in \Omega)
 \end{align}
 Proceed as in (a), with $S$ instead of $T$: $S$ lies in $\mathscr{B}(X)$. Next, we claim that 
 \begin{align}
 \langle u,\, T^{\,\ast} f^{\,\,\ast} \rangle = &\, \langle Tu,\,f^{\,\,\ast} \rangle\\
-=& \int_{\Omega} (T u ) \mdot f \,\, \,\d\mu \\
-\label{int_Fubini}=& \int_{\Omega^2} K\mdot f \mdot u \,\,\, \d(\mu\times\mu) \\
-=& \int_{\Omega} (S f\, ) \mdot u \,\,\, \d\mu\, \\
+=& \int_{\Omega} (T u ) \cdot f \,\, \,\d\mu \\
+\label{int_Fubini}=& \int_{\Omega^2} K\cdot f \cdot u \,\,\, \d(\mu\times\mu) \\
+=& \int_{\Omega} (S f\, ) \cdot u \,\,\, \d\mu\, \\
 =&\, \langle u,\, (S f\, )^\ast \rangle \quad ,
 \end{align}
 whenever $u$ and $f$ run through the closed unit ball of $X$. Since $\|T\, \|$, $\| T^{\,\ast} \|$ are equal and finite, only exactness of (\ref{int_Fubini}) is possibly in doubt; the below justification dissipates it. In conclusion,
@@ -124,18 +124,18 @@
 \\
 \underline{Justification of (\ref{int_Fubini})}. The current proof shall be complete once we have justified (\ref{int_Fubini}). To do so, keep $u$ and $f$ as above. Let us introduce
 \begin{align}
-A(s\,)\Def \int_\Omega \lvert K_s(t\,) \mdot u (t\,)\rvert \,\, \d\mu(t\,) \, \,\,\text{a.e} \quad (s\in \Omega)\quad ,
+A(s\,)\Def \int_\Omega \lvert K_s(t\,) \cdot u (t\,)\rvert \,\, \d\mu(t\,) \, \,\,\text{a.e} \quad (s\in \Omega)\quad ,
 \end{align}
 to make hold the following Cauchy-Schwarz inequality
 \begin{align}
-A(s\,)\< \| K_s\|_X \quad (s\in \Omega)\quad .
+A(s\,)\leq \| K_s\|_X \quad (s\in \Omega)\quad .
 \end{align}
 Thus,
 \begin{align}
  \int_{\Omega^2} \lvert K(s,\,t\,) \, u(t\,)\, f\,(s\,) \rvert \,\d\mu(s\,)\d\mu(t\,) 
  = & \int_\Omega \lvert \,f\,(s\,) \rvert \,A(s\,) \, \d\mu(s\,) \\
- \< & \int_\Omega \lvert \,f\,(s\,) \rvert \, \| K_s\|_X\,\d\mu(s\,) \\
- \label{4_15_complement_2} \< & \left[ \int_\Omega \| K_s\|_X^2\,\,\d\mu(s\,) \right]^{\frac{1}{2}} 
+ \leq & \int_\Omega \lvert \,f\,(s\,) \rvert \, \| K_s\|_X\,\d\mu(s\,) \\
+ \label{4_15_complement_2} \leq & \left[ \int_\Omega \| K_s\|_X^2\,\,\d\mu(s\,) \right]^{\frac{1}{2}} 
  = \| K\,\|_P < \infty \quad .
  \end{align}
 The inequality in (\ref{4_15_complement_2}) is a Cauchy-Schwarz one, the following equality follows from the Fubini's theorem. This achieves the proof.\QED

chapter_4/4_15.tex

@@ -8,11 +8,11 @@ \section{Exercise 1. Basic results}
 %
 
 \section{Exercise 13. Operator compactness in a Hilbert space}
-%\input{\ROOT/chapter_4/4_13.tex}
+\input{\ROOT/chapter_4/4_13.tex}
 \newpage
 %
 \setcounter{section}{14}
 %
 \section{Exercise 15. Hilbert-Schmidt operators}
-%\input{\ROOT/chapter_4/4_15.tex}
+\input{\ROOT/chapter_4/4_15.tex}
 \newpage