泛函分析讀書筆記I

Content

• Baire Category Theorem
• Uniform Boundedness Principle
• Sequential Completeness of $\mathscr{D'}(X)$
• ...

Theorem1 (Baire). Let $X$ be a complete metric space. If $\{U_n\}_{n\geq 1}$ be a sequence of open dense subsets in $X$, then their intersection $U = \bigcap_{n = 1}^{\infty} U_n$ is dense.

Proof. Define $U := \bigcap_{n = 1}^{\infty} U_n$. Let $V \subseteq X$ be an open set in $X$. We need to show $U \cap V \neq \emptyset$. We construct a sequence $\{ x_n \}$ as the following:

Step 1. The set $U_1 \cap V$ is nonempty and open. So there exists $x_1 \in U_1\cap V$and $0 < r_1 < 1$ such that $\bar{B}(x_1,r_1)\subseteq{U_n}\cap V$.

Step n. The set $U_n \cap B (x_{n - 1}, r_{n - 1})$ is nonempty and open. Thus we may pick $x_n$ and $0 < r_n < \frac{1}{n}$ such that $\bar{B}(x_n,r_n)\subseteq {U_n}\cap\bar{B}(x_{n-1},r_{n-1})$.

By our construction. $x_n \in \bar{B} (x_m, r_m)$ when $n > m$. Therefore, $d({x_n},{x_m})<{r_m}<\frac{1}{m}$, i.e., $\{ x_n \}$ is Cauchy. Since $X$ is complete, $x_n\rightarrow x\in X$ as $n \rightarrow\infty$. By closedness,
\[ x \in \bar{B} (x_n, r_n) \subset U_n \cap V \]
for all $n$. Thus, $x \in \bigcap^{\infty}_{n = 1} (U_n \cap V) = U \cap V$.

Remark. Let $A\subseteq X$. $A$ is dense in $X$ if and only if for all open set $U\subseteq X$, $A\cap U\neq \emptyset$.



Definition 2. A Banach space is a complete normed vector space.

Notation. Let $E$ and $F$ be two normed vector space. We denote by $\mathcal{L}(E,F)$ the space of continuous linear operators from $E$ into $F$ equipped with the norm 
\[\Vert T\Vert=\sup_{x\in E \\ \Vert x\Vert\leq 1} \Vert Tx\Vert.\]

Theorem 3.  Let $E$ and $F$ be two Banach space and let $\{T_i\}_{i\in I}\subseteq \mathcal{L}(E,F)$ (not necessarily countable). Assume that 
\[sup_{i\in I}\Vert T_i x\Vert<\infty\]
for all $x\in E$. Then 
\[sup_{i\in I}\Vert T_i\Vert<\infty.\]
In other words, there exists a constant $c$ such that
\[\Vert T_i x\Vert\leq c\Vert x\Vert\]
for all $x\in E$, $i\in I$.

Proof. Define an open set
\[U_n:=\{x\in E: \Vert T_i x\Vert> n,i\in I\} \]
for all $n\geq 1$. Note that 
\[\bigcap_n^\infty U_n=\emptyset.\]
By theorem 1, there exists some $U_m$ such that $\bar{U}_m\subsetneq X$. Then there exists $y$ and $r>0$ such that $B(y,r)\subseteq X-U_m$. For all $z\in B(0,1)$, 
\[\Vert T_i(y+rz)\Vert\leq m\]
for all $i\in I$. Since \[T_i z=r^{-1}(T_i(y+rz)-T_i(y)),\]
we get 
\[\Vert T_i\Vert\leq r^{-1}(m+\Vert T_i(y)\Vert)<\infty\]
for all $i\in I$, which completes the proof.