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Let $(f_n)$ be a sequence of continuous functions on $[0,1]$ converges pointwise everywhere. Let $f$ be the limit function on $[0,1]$. Show that $\{t\in [0,1]:f\mbox{ is continuous at }t\}\neq\emptyset$.

I tried to solve by the following method.

Let $F_n=\{x\in [0,1]:O(f,x)\geq 1/n\}$. It is clear that $F_n$ is closed. If we can show that it has non empty interior then by Baire Category Theorem the desired set is dense in $[0,1]$. To prove that $F_n$ has empty interior I am facing problem.

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  • $\begingroup$ How do you define $O(f,x)$? $\endgroup$ – Robert Israel Dec 3 '17 at 8:16
  • $\begingroup$ $O(f,x)=\inf_{\delta \to 0}\sup\{|f(y)-f(z)|:y,z\in (x-\delta,x+\delta)\}$. $\endgroup$ – harisankar pal Dec 3 '17 at 8:44
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Let $A(\epsilon, N) = \{x \in [0,1]:\; \forall n,m\ge N, \; |f_n(x) - f_m(x)| \le \epsilon \}$. This is closed, and $\bigcup_N A(\epsilon, N) = [0,1]$. By the Baire Category Theorem, some $A(\epsilon, N)$ has nonempty interior. Moreover, by repeating this argument with $[0,1]$ replaced by a closed interval in the interior of $A(\epsilon, N)$ and $\epsilon$ replaced by $\epsilon/2$, some $A(\epsilon/2, N')$ has nonempty interior in $A(\epsilon, N)$. In this way we get a nested sequence of closed intervals $[a_k, b_k]$ and such that $[a_k, b_k]$ is contained in the interior of $A(2^{-k}, N_k)$ for some $N_k$.

Take $t$ in the intersection of these intervals. I claim $f$ is continuous at $t$.

For any $\epsilon > 0$ we can take $k$ so $\epsilon/3 > 2^{-k}$, and $\delta > 0$ so $(t-\delta, t+\delta)$ is contained in $A(2^{-k}, N_k)$, and also so that $|x - t| < \delta$ implies $|f_{N_k}(x) - f_{N_k}(t)| < 2^{-k}$. For $n \ge N_k$ and $x \in (t-\delta, t+\delta)$ and we have $|f_n(x) - f_{N_k}(x)| < 2^{-k}$, so that $|f(x) - f_{N_k}(x)| \le 2^{-k}$. Thus if $|x - t| < \delta$, $$ |f(x) - f(t)| \le |f(x) - f_{N_k}(x)| + |f_{N_k}(x) - f_{N_k}(t)| + |f_{N_k}(t) - f(t)| < 3 \cdot 2^{-k} < \epsilon$$

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