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Suppose that a sequence of continuous function $(f_n)$ converges point wise to $f$ on $[0,1]$.

Prove that if there exists a sequence $(x_n)$ in $[0,1]$ converging to $x^*$ in $[0,1]$ such that $(f_n(x_n))$ does not converge to $f(x^*)$, then $(f_n)$ is not uniformly convergent.

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    $\begingroup$ This is exactly the statement that the uniform limit of continuous functions is continuous. This is a standard theorem which is probably proven in your text. $\endgroup$ – Omnomnomnom Oct 3 '15 at 12:55
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Hint: go back to the definition. Uniform convergence of $(f_n)_n$ to $f$ on $[0,1]$ would by definition mean that $$\sup_{x\in[0,1]} \lvert f_n(x)-f(x)\rvert \xrightarrow[n\to\infty]{} 0$$ Suppose there exists $(x_n)_n$ as stated. Can you use this to give a lower bound on the $\sup$ (for infinitely many $n$) that would be a positive constant? (That would imply the $\sup$ cannot converge to $0$).

(See below for details)


  • Suppose by contradiction $(f_n)_n$ does converge uniformly to $f$. Since each $f_n$ is continuous, this implies $f$ is.

  • By assumption, there exists some $\varepsilon > 0$ and a subsequence $(\varphi(n))_n$ such that for all $n \geq 0$, $\lvert f_{\phi(n)}(x_{\phi(n)}) - f(x^\ast) \rvert \geq \varepsilon$.

  • Also, by the triangle inequality $$\lvert f_n(x_n) - f(x_n)\rvert \geq \lvert f_n(x_n) - f(x^\ast)\rvert - \lvert f(x^\ast) - f(x_n)\rvert $$ for any $n$. For the second term, as $f$ is continuous and $x_n\to x^\ast$, there exists $N \geq 0$ such that for any $n \geq N$, $\lvert f(x^\ast) - f(x_n)\rvert \leq \frac{\varepsilon}{2}$.

  • Combine the two: \begin{align} \lvert f_{\phi(n)}(x_{\phi(n)}) - f(x_{\phi(n)})\rvert &\geq \lvert f_{\phi(n)}(x_{\phi(n)}) - f(x^\ast)\rvert - \lvert f(x_{\phi(n)}) - f(x^\ast)\rvert \\ &\geq \lvert f_{\phi(n)}(x_{\phi(n)}) - f(x^\ast)\rvert - \frac{\varepsilon}{2} \\ &\geq \varepsilon - \frac{\varepsilon}{2} = \frac{\varepsilon}{2} \end{align} for any $n\geq N$ (which implies $\phi(n)\geq N$).

  • For $n\geq N$, this implies that $$ \sup_{x\in[0,1]} \lvert f_n(x)-f(x)\rvert \geq \lvert f_{\phi(n)}(x_{\phi(n)}) - f(x_{\phi(n)})\rvert \geq \frac{\varepsilon}{2} $$ and therefore the $\sup$ cannot converge to $0$.

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