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For a given continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$, define a sequence of function $\{f_n\}_{n\in \mathbb{N}}$ by $$f_n(x):=f\Big(x+\frac{1}{n}\Big).$$ Now if $f$ is uniformly continuous, then I can show that {$f_n$} converges to $f$ uniformly. At this point, I was thinking about the converse, i,e; if {$f_n$} defined above converges to $f$ uniformly, does that imply the uniform continuity of $f$.
I think the answer is negative, but I am not getting any counter example for that. Does anyone have any counter example?

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  • $\begingroup$ a general result is that uniform convergence of a cont. sequence implies that the limit function is continuous as well if that helps you $\endgroup$ – Simonsays Mar 17 '18 at 16:13
  • $\begingroup$ That sequence of functions already appeared here, but in a more complicated question of pointwise convergence. $\endgroup$ – Giuseppe Negro Mar 17 '18 at 16:41
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Try $f:x\mapsto\sqrt{|x|}$. It is not uniformly continuous, having an unbounded derivative at $0$. But the $f_n$ do converge uniformly to $f$.

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Take $$f_n (x)=x^2+\frac {1}{n+1} $$

$(f_n)$ converge uniformly to $$f:x\mapsto x^2$$ at $\Bbb R $ but $f $ is not uniformly continuous at $\Bbb R $ since

$$\lim_{n\to+\infty}f (n+\frac{1}{n})-f (n)\ne 0 $$

If $f_n $ are continuous and if the convergence is uniform at a compact $[a,b] $, then the limit function is Uniformly continuous at $[a,b] $.

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    $\begingroup$ You cannot choose $f_n$ arbitrary... the questions asks for an example of a continuous function $f$ which is not uniformly continuous but such that $f_n$ which are given by $f_n(x) = f \left( x + \frac{1}{n} \right)$ converge uniformly to $f$. In your case, if $f(x) = x^2$ then $f_n(x) = x^2 + \frac{2x}{n} + \frac{1}{n^2}$ and they don't converge uniformly to $f$. $\endgroup$ – levap Mar 17 '18 at 16:51

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