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What would be an example of a sequence of uniformly continuous functions on a compact domain which converges pointwise to $0$, but not uniformly?

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  • $\begingroup$ The functions $f_n(x)=x^n$ defined on $[0,1)$ form an example. $\endgroup$ – Suzet Jul 28 '19 at 14:27
  • $\begingroup$ @Suzet I forgot, I want the domain to be compact. $\endgroup$ – Jannik Pitt Jul 28 '19 at 14:30
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Consider $$f_n(x)=1-\min(1,n|x-1/n|)=\begin{cases} nx&\text{ if }x<\frac{1}{n}\\2-nx&\text{ if }\frac1{n}\leq x\leq\frac{2}{n}\\0&\text{ otherwise.}\end{cases}$$ Each $f_n$ is continuous on the compact set $[0,1]$ and therefore it is also uniformly continuous. Moreover, $f_n(x)\to 0$ for any $x\in [0,1]$, but the convergence is not uniform on $[0,1]$ because $\max_{x\in[0,1]}|f_n(x)|=f(1/n))=1$.

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The another standard one is the growing steeple on $[0,1]$:

$$f_n(x)=\begin{cases}n^2 x &\text{if}\;0 \leq x \leq \frac{1}{n}\\ 2n-n^2 x &\text{if}\;\frac{1}{n} \leq x \leq \frac{2}{n}\\ 0 &\text{if}\;\frac{2}{n} \leq x \leq 1 \end{cases}$$

Then each $f_n$ is uniformly continuous and also the limit is zero, but the convergence is not uniform.

Added: It is easy to visualize the graph of $f_n$. Actually each $f_n$ is a triangle with height $n$ attained at $1/n$

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Take$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&nx^n(1-x).\end{array}$$The sequence $(f_n)_{n\in\mathbb N}$ converges pointwise to the null function, but not uniformly, since$$(\forall n\in\mathbb N):f_n\left(\frac n{n+1}\right)=\left(\frac n{n+1}\right)^{n+1}$$and $\lim_{n\to\infty}\left(\frac n{n+1}\right)^{n+1}=e^{-1}$.

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