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Let $a,b$ be real numbers.

There is given a sequence of functions $(f_{n})_{n\ge 1}$. Where $f_{n}:[a,b]\rightarrow \mathbb{R}$ and these functions are smooth. The same with $f:[a,b]\rightarrow\mathbb{R}$

Is it true that:

If this sequence converges pointwise to function $f$, then also this sequence converges uniformly to $f$ ?

I think this is true because of the fact that these functions are bounded. Nevertheless i hope for your help.

mkultra
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  • The answer is no. See https://math.stackexchange.com/questions/3306545/pointwise-convergence-of-uniformly-continuous-functions-to-zero-but-not-uniform – Robert Z Jul 28 '19 at 16:01
  • Another related question: https://math.stackexchange.com/questions/232408/dinis-theorem-uniform-convergence-and-bolzano-weierstrass?rq=1 – Robert Z Jul 28 '19 at 16:02

1 Answers1

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No. Take$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&nx^n(1-x).\end{array}$$Each $f_n$ is a smooth function and $(f_n)_{n\in\mathbb N}$ converges pointwise to the null function. But the convergence is not uniform, since$$(\forall n\in\mathbb N):f_n\left(\frac n{n+1}\right)=\left(\frac n{n+1}\right)^{n+1}$$and $\lim_{n\to\infty}\left(\frac n{n+1}\right)^{n+1}=e^{-1}$.

José Carlos Santos
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  • I see that $\frac{n}{n+1} \rightarrow 1$. What if we take instead of $[0,1]$ interval $[0,1-\epsilon]$? – mkultra Jul 28 '19 at 15:11
  • Then I will have to change my example, right?! – José Carlos Santos Jul 28 '19 at 15:19
  • I thought that is it possible to find interval small enough to have a uniform convergence. I would like to have a simple condition so that if it is satisifed then functions on interval convergence uniformly. – mkultra Jul 28 '19 at 15:29
  • Yes, my sequene convers uniformly on any interval of the form $[0,1-\varepsilon]$. That doesn't change the fact that it does not converge uniformly on the interval in which I defined it. Therefore, what I wrote answers your question. – José Carlos Santos Jul 28 '19 at 15:32