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Let $f_n$ be a sequence of functions, $f_n : S\rightarrow T$, not necessarily continuous and suppose that $f_n \rightarrow f$ as $n \rightarrow \infty$. Let $f$ be uniformly continuous. I.e. for all $\epsilon \gt 0, \exists \delta \gt 0$ such that $\forall x\in S, |f(x) - f(y)| \lt \epsilon$ whenever $|x - y| \lt \delta, y\in S$.

We want to show that for all $\epsilon \gt 0, \ \exists N$ such that for all $x \in S, n\gt N$, we have $|f_n(x) - f(x)|\lt \epsilon$.

If we need it, $S$ is compact and $T = \mathbb{C}$.

Since $S$ is compact, by the extreme value theorem there exists $c,d \in S$ such that $f(c) \leq f(x) \leq f(d), \ \forall x \in S$.

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This is false. For instance, consider the sequence of functions (which are even continuous)

$$f_n(x) = \begin{cases}{nx \quad(0 \leq x \leq 1/2n) \\ 1/2-nx \quad (1/2n \leq x \leq 1/n) \\ 0 \quad ( 1/n \leq x \leq 1) }\end{cases}$$

it converges to $0$ on the compact space $[0,1]$, but not uniformly.

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  • $\begingroup$ What are sufficient conditions for the convergence to be uniform? $\endgroup$ Commented Oct 4, 2013 at 3:44
  • $\begingroup$ @EnjoysMath There aren't really any. Uniform convergence is a very strong property! $\endgroup$ Commented Oct 4, 2013 at 3:47
  • $\begingroup$ My Fourier analysis book is not complete then. It says "If $f$ is continuous everywhere, then it is uniformly continuous, and $\delta$ can be chosen independent of $x$. This provides the desired conclusion that $f * K_n \rightarrow f$ uniformly." without proof. I will just write that the limit is uniformly continuous in my notes so that an error won't crop up later. $\endgroup$ Commented Oct 4, 2013 at 3:52
  • $\begingroup$ @EnjoysMath The convergence properties of $fK_n \to f$ depend on those of $K_n$, not of $f$. Does $K_n$ converge uniformly? $\endgroup$ Commented Oct 4, 2013 at 3:54
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    $\begingroup$ $K_n$ is a "Good kernel" and $f*K_n$ is convolution of two functions on $\mathbb{R}$. I'll just move on. Not all books are the best. $\endgroup$ Commented Oct 4, 2013 at 3:56
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No, e.g. Let $S=[0,1]$, and $f_n$ be the characteristic function of $\{1/n\}$, which converges to the zero function, but with $|f_n(1/n)-f(1/n)|=1$ for all $n$ showing that the convergence is not uniform. You could modify this to make each $f_n$ continuous (differentiable, etc.) if desired.

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People make examples pointlessly complicated. How about this one: $$ f_n(x) = \arctan(x-n). $$ Then $f_n(x) \to-\pi/2$ as $n\to\infty$, and the function constantly equal to $-\pi/2$ is uniformly continuous, but the sequence does not converge uniformly.

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Consider $f_n(x) = x^n$ on $(0,1)$. Then $f_n(x) \to 0$, and the function $x \mapsto 0$ is definitely uniformly continuous, but clearly the convergence is not uniform (since $f_n (\frac{1}{\sqrt[n]{2}}) = \frac{1}{2}$).

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This is not going to work out well for you; sorry. For example, let $f_n\colon [0,1]\to [0,1]$ be given by $$f_n(x)=\begin{cases} 0 & 0<x<\frac 1 n \\ 1 & (x=0) \lor \left(\frac1n\le x\right). \end {cases} $$ Then $(f_n)$ converges pointwise to $x\mapsto 1$, but it does not converge uniformly.

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