# Boundedness leading to pointwise convergence implying uniform convergence?

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.

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.
• 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? – K.Power Jan 29 '16 at 17:32
• The definition of $K(\epsilon)$ is not clear. Does it depend on $x$, as the notation $K(x)$ inside the brackets suggests? – Julián Aguirre Jan 31 '16 at 20:10
• 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 $n>K(x)$. – K.Power Feb 1 '16 at 20:17
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.