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Consider a sequence of functions $\{f_n\}$ on some closed interval $I \subset \mathbb{R}$. Let's assume that $f_n$ is bounded on $I$ by $M \in \mathbb{R}$ for each $n \in \mathbb{N}$. If $\{f_n\}$ converges pointwise to some function $f$ on $I$ does it necessarily follow that $\{f_n\}$ converges uniformly to $f$ on $I$?

Intuitively I think this is the case, which I believe is supported by the following argument. As $\{f_n\}$ is pointwise convergent on $I$ we can construct a set for each $\epsilon>0$: $$K_\epsilon =\{K(x) \in \mathbb{N}\ : x \in I\ \text{and}\ \forall n>K(x)\ ,|f_n(x)-f(x)|<\epsilon\}$$ Due to the fact that $\{f_n\}$ is bounded on a closed interval, and is pointwise convergent to $f$, I believe that $max (K_\epsilon)$ exists for each $\epsilon >0$. Then for $\epsilon > 0$ we have $n>max(K_\epsilon )$ implies $$|f_n(x)-f(x)|< \epsilon \ \ \forall x \in I$$ and so by definition $\{f_n\}$ is uniformly convergent on $I$. My only issue is that I'm not able to prove that the maximum of the set exists, I just intuitively feel that it does under the imposed conditions. If anyone could confirm or explain where I've gone wrong with my reasoning I'd be much obliged.

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The answer is no. Take $f_n(x)=x^n$ on $[0,1]$. $|f_n(x)|\le1$ for all $x\in[0,1]$ and all $n\in\mathbb{N}$. $f_n$ converges pointwise to the function $f(x)=0$ if $0\le x<1$, $f(1)=1$, but the convergence is not uniform. Even more, no subsequence converges uniformly.

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  • $\begingroup$ Thank you. Obviously in this case the $Kmax$ I mentioned doesn't exist, because you can get arbitrarily close to $1$ and so the $K$ value can become arbitrarily large to ensure the difference term is lower than $\epsilon$ correct? $\endgroup$ – K.Power Jan 29 '16 at 17:32
  • $\begingroup$ The definition of $K(\epsilon)$ is not clear. Does it depend on $x$, as the notation $K(x)$ inside the brackets suggests? $\endgroup$ – Julián Aguirre Jan 31 '16 at 20:10
  • $\begingroup$ Each $K_\epsilon$ is the set of all $K(x)$ values such that for each $x \in I$ if $n>K(x)$ then $|f(x)-f_n(x)|< \epsilon$ for that particular $\epsilon$ value. I'm struggling to get across my idea. Basically in each distinct set there is a $K$ value for each $x$ in the interval called $K(x)$, so that the difference between $fn$ and $f$ evaluated at $x$ is less than the $\epsilon$ value of that set when $n>K(x)$. $\endgroup$ – K.Power Feb 1 '16 at 20:17
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Let $\mathbb Q = \{q_1,q_2, \dots\}.$ Set $f_n=\chi_{\{q_1,\dots, q_n\}}.$ Then $|f_n|\le 1$ on $\mathbb R$ for all $n.$ We have $f_n\to \chi_{\mathbb Q}$ pointwise on $\mathbb R,$ but the convergence fails to be uniform on any interval of positive length.

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