LEC ANA2 10

Discontinuities of pointwise limit of continuous functions

We have seen that the pointwise limit of continuous maps is not continuous; we will now characterize the set of discontinuities of such maps.

Lemma 314.1(Lemma).

Let $X$ be a complete metric space. $f_n:X\to \mathbb{C}$ are continuous functions, and $f_n\to f$ pointwise. Given any ball $B\subseteq X$ and $ϵ>0$, there is another ball $B_0\subseteq B$ and $m\in \mathbb{N}$ such that $|f_m(x)-f(x)|<ϵ$ for all $x\in B_0$.

Proof.

$\overline{B}$ is complete. Define

$$F_n:=\{x\in \overline{B}:|f_k(x)-f_l(x)|\le ϵ\forall k,l\ge n\}.$$

$F_n$ is closed since¹

$$F_n=\bigcap _{k,l\ge n}\{x\in \overline{B}:|f_k(x)-f_l(x)|\le ϵ\}.$$

Since $f_n\to f$ pointwise, we have

$$\bigcup _{n=1}^{\infty }{F}_n=\overline{B}.$$

By Corollary 168.9, there exists $m$ such that $F_m$ is not nowhere dense, that is, $F_m$ has nonempty interior. Thus, there exists an open ball $B_0\subseteq F_m$ such that for $x\in B_0$, $|f_m(x)-f_l(x)|\le ϵ$ for all $l\ge n_0$. As $l\to \infty$, $|f_m(x)-f(x)|\le ϵ$, for all $x\in B_0$.□

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Theorem 314.2(Theorem).

Let $X$ be a complete metric space and $f_n:X\to \mathbb{C}$ be continuous for all $n\in \mathbb{N}$ and $f_n\to f$ pointwise. Then, $D=\{x\in X|f\text{ dicontinuous at }x\}$ is meagre.

Proof.

Recall Definition 195.7. Define

$$D_n:=\{x\in X:o(f,x)\ge 1/n\}.$$

By Theorem 195.9, $D_n$ is closed. By Theorem 195.8, $D={\bigcup }_{n=1}^{\infty }{D}_n$. It suffices to show that each $D_n$ is nowhere dense.

Fix $D_{n_0}$. FTSOC, suppose there exists an open ball $B\subseteq D_{n_0}$. Then, by Lemma 1 there exists $B_0\subseteq B$ and $m\in \mathbb{N}$ such that

$$\begin{array}{rlr}|f_m(x)-f(x)|<1/3n_0& & (\forall x\in B_0).\end{array}$$

Since $f_m$ is continuous, there exists another ball $B_0^{\prime }\subseteq B_0$ such that

$$\begin{array}{rlr}|f_m(y)-f_m(z)|<1/3n_0& & (\forall y,z\in B_0).\end{array}$$

For $y,z\in B_0^{\prime }$, we have

$$\begin{array}{rl}|f(y)-f(z)|& \le |f(y)-f_m(y)|+|f_m(y)-f_m(z)|+|f_m(z)-f(z)|\\ & <1/n_0,\end{array}$$

a contradiction.□

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Uniform boundedness theorem

Theorem 314.3(Theorem).

Let $X$ be a Banach space, $Y$ a normed vector space and $B(X,Y)$ the space of all continuous² linear operators from $X$ to $Y$ equipped with the operator norm. Suppose that $\{T_i{\}}_{i\in I}$ is a collection of continuous linear operators form $X$ to $Y$. If, for every $x\in X$,

$$\underset{i\in I}{\sup }|T_i(x)|<\infty ,$$

then

$$\underset{i\in I}{\sup }\Vert T_i\Vert <\infty .$$

Proof.

Let $F_n=\{x\in X:|T_{i}x|\le n\forall i\in I\}$. As usual, each $F_n$ is closed, since³

$$F_n=\bigcap _{i\in I}\{x\in X:|T_{i}x|\le n\},$$

and we have ${\bigcup }_{n=1}^{\infty }{F}_n=X$. By Corollary 168.9, there exists $F_m$ with non-empty interior. Let $B\subseteq F_m$ be a closed ball with center $v_0$ and radius $r$. If $|x|=1$,

$$\begin{array}{rlr}|T_{i}x|=\left|T_i\left(\frac{(v_0+rx)-v_0}{r}\right)\right|=\left|\frac{T_i(v_0+rx)-T_i(v_0)}{r}\right|\le \frac{2m}{r}& & (\forall T_i)\end{array}$$

Thus, $\Vert T_i\Vert \le 2m/r$ for all $i$, and ${\sup }_{i\in I}\Vert T_i\Vert \le 2m/r$.□

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Footnotes

1. Note that $|f_k-f_l|$ is a continuous function, and that $[-ϵ,ϵ]$ is closed. ↩

2. Continuity and boundedness are equivalent here; see Proposition 172.6. ↩

3. Again, the map $x\mapsto |T_{i}x|$ is continuous by Proposition 172.6 and Lemma 172.9, and $[-n,n]$ is closed. ↩