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I don't think such a sequence exists.

I have a proof (given in class) that if the sequence is equicontinuous and converges point-wise (non-uniformly) for every rational on $[a,b]$ then it converges uniformly on all of $[a,b]$.

I also have a proof (written by me) that if a sequence of continuous functions defined on $[a,b]$ converges point-wise for every rational on $[a,b]$, then it converges.

EDIT: "uniformly" change to "pointwise"

on all of $[a,b]$.

But I was told that equicontinuity is essential. I'm having hard time seeing why. Can someone give me a counterexample to demonstrate the need for equicontinuity?

My understanding is at the level of Rudin's principles [...] chapters 1-6 and most of chapter 7.

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$f_n(x) = (-1)^n(x/\sqrt 2)^n$ converges pointwise to $0$ everywhere on $[0,\sqrt 2),$ hence at each rational in $[0,\sqrt 2],$ but does not converge at $\sqrt 2.$

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  • $\begingroup$ Upvote! This example covers the problem and then some, as convergence is at every point, not just rationals. $\endgroup$ – coffeemath Mar 5 '16 at 18:54
  • $\begingroup$ sorry, i mis typed that. please see the edit. $\endgroup$ – user282002 Mar 6 '16 at 1:37
  • $\begingroup$ I changed my proof to answer the restated question. $\endgroup$ – zhw. Mar 6 '16 at 17:40
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The following example is about why for an arbitrary sequence of functions (without equicontinuity assumed) one cannot conclude uniform continuity from pointwise convergence at every rational.

On $[0,2]$ let $f(x)=0$ if $x\le \sqrt{2}$ and $1$ otherwise. Take each $f_n$ to be this $f.$ Note this $f$ is not continuous at $\sqrt{2}.$

I'll add to this if I can think up such a sequence of functions each of whih is continuous.

NOTE I just saw user zhw.'s answer above which is a sequence of continuous functions on $[0,1]$ which converges at every point (rational or not) and yet the convergence is not uniform (and limit function discontinuous). So I won't look further.

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  • $\begingroup$ Can we relax the assumption to continuity, rather than equicontinuity, then? $\endgroup$ – Vik78 Mar 5 '16 at 18:29
  • $\begingroup$ @Vik78 I'll work on that, don't see it right away... $\endgroup$ – coffeemath Mar 5 '16 at 18:36
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Define the sequence of functions $(f_n)_n$ as follows: for any $n\geq 1$,

  • $f_n(x) = 0$ for $x\in[0,\frac{1}{\sqrt{2}}-\frac{1}{10n}]\cup [\frac{1}{\sqrt{2}}+\frac{1}{10n},1]$;

  • $f_n(\frac{1}{\sqrt{2}}) = n$;

  • $f_n$ is piecewise affine and continuous;

so that $f_n$ is a continuous function looking like a very narrow triangle centered at $\frac{1}{\sqrt{2}}$.

Then $f_n(x) \xrightarrow[n\to\infty]{}0$ for any $x\in[0,1]\setminus\{\frac{1}{\sqrt{2}}\}$, and a fortiori at all rationals. But there is not pointwise convergence on $[0,1]$, since $f_n(\frac{1}{\sqrt{2}}) \xrightarrow[n\to\infty]{}\infty$.

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