無聊之下嘅產物

喺 Royden (4th edition) p.452 有一 Theorem 將 locally compact Hausdorff space 有嘅 property 集埋一齊：

其中證明 (iii) 嘅主要工具係 Urysohn's lemma，但由 Urysohn's lemma 又可推出 Tietze Extension Theorem， 從 (iii) 我地應該可以做得更多。

Modification of (iii). Let $X$ be locally compact Hausdorff. If $\mathcal O$ is a neighborhood of a compact subset $K$ of $X$, then the continuous function $f:K\to \mathbb R$ may be extended to a function $F\in C_c(X)$ for which $F$ vanishes outside $\mathcal O$.
If $f$ is bounded, say $|f|\leq M$ for some $M>0$, then the extension above can be chosen so that $|F|\leq M$ on $X$.

Proof. As $K$ is compact and $\mathcal O\supseteq K$, by (ii) of theorem 7 above there is an open $V$ such that $K\subseteq V\subseteq \overline{V}\subseteq \mathcal O$ with $\overline{V}$ compact. Once again by (ii) of  theorem 7 above there is an open $U$ such that
$K\subseteq U\subseteq \overline{U}\subseteq V\subseteq \overline{V}\subseteq \mathcal O$.

Now we do our extension, as $\overline{V}$ is a compact Hausdorff space, $f$ is continuous (w.r.t. subspace topology) on $K$, by the Tietze extension theorem there is a $\mathcal F\in C (\overline{V})$ such that $\mathcal F|_{K}= f|_K$, we extend $\mathcal{F}$ on $X\setminus \overline{V}$ by defining $\mathcal F|_{X\setminus \overline{V}}\equiv 0$. The change of $\mathcal F$ between $\overline{V}$ and $X\setminus \overline{V}$ may not be continuous, we will try to ``smooth" this transition. As $K$ is compact and $U\supseteq K$, by (iii) of theorem 7 there is a $\psi \in C_c(X)$ such that $0\leq \psi \leq 1$, $\psi=1$ on $K$ and $\psi=0$ on $X\setminus U$. Now we claim that the product of continuous functions $F:= \mathcal F \cdot \psi$ will do.

Clearly $\mathop{\mathrm{supp}} F \subseteq \mathop{\mathrm{supp}} \psi$, hence $F$ has compact support. To show $F$ is continuous on $X$ we use the following fact:

Fact. Let $X=\cup_\alpha X_\alpha$ be a union of open subsets. Then $f:X\to Y$ is continuous if and only if the restrictions $f|_{X_\alpha}:X_\alpha \to Y$ are continuous, where $X_i$ has the subspace topology.

Proof. It follows from the observation that: For any subset $A$ of $Y$, $f^{-1}(A)= \cup_{\alpha} (f|_{X_\alpha})^{-1}(A)$.$\qed$

Observe that both $X_1:=V$ and $X_2:=X\setminus \overline{U}$ are open, $X=X_1\cup X_2$. It is enough to argue $F|_{X_i}$'s are continuous. On $X_1$, since $\mathcal F$ is a continuous function on $\overline{V}$, $\mathcal F|_V$ is therefore a continuous function on $V$. And as $\psi$ is continuous on $X$, so $F|_{X_1}$ is continuous. On $X_2$, since $\psi|_{X\setminus U}\equiv 0$ $\implies$ $\psi|_{X\setminus \overline{U}} \equiv 0$, and thus $F|_{X\setminus \overline{U}} \equiv 0$, hence $F|_{X_2}$ is continuous on $X_2$. We also note that $F|_{X\setminus \mathcal O}\equiv 0$.

When $|f|\leq M$ on $K$, we repeat the proof above but that time $\mathcal F$ can be chosen such that $|\mathcal F|\leq M$ by the following version of Tietze extension theorem.$\qed$

建立呢種 extension 嘅原因係為左證明 Lusin's theorem on $(X,\mathcal B(X),\mu)$，其中 $X$ 為 locally compact Hausdorff，$B(X)$ 為 Borel $\sigma$-algebra on $X$ 及 $\mu$ 為 Radon measure (Royden's definition: A Borel measure such that Borel set is outer regular and open set is inner regular)，我嘅 approach (某習題) 係先證明 Lusin's theorem 對 simple function 成立，從而利用 simple functions $\{\phi_n\}$, $\phi_n\to f$ pointwise 及 Egoroff's theorem 及再利用上述 extension 完成證明 (已證明若 $E\in \mathcal B(X),\mu(E)<\infty$，那麼 $E$ 是 inner regular)。

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Some problem for entertainment:
Problem. Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function so that $\displaystyle\left|f(x)-\sin(x^2)\right|\le\frac{1}{4}$ for any $x\in\mathbb{R}$. Prove that there exists a sequence of real numbers $\{x_n\}_{n=1}^\infty$ for which $\lim_{n\to\infty} f'(x_n)=+\infty$ .