If $f_n\to f$ pointwise, $f$ is continuous and $f$ is continuous, then $f_n \to f$ uniformly.

Question

Let $(f_n)$ a sequence of continuous function on $[a,b]$ that converges pointwise to $f$. We suppose that $f$ is continuous on $[a,b]$. Prove that the convergence is uniform.

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I'm stuck at some point :

Since $|f_n(x)-f(x)|$ is continuous on $[a,b]$, for all $n$, there is $x_n\in [a,b]$ s.t. $$\sup_{[a,b]}|f_n(x)-f(x)|=|f_n(x_n)-f(x_n)|.$$ Since $(x_n)$ is bounded, there is a subsequence that convergence (let denote $x\in [a,b]$ the limit, and still denote the subsequence $(x_n)$). Now $$|f_n(x_n)-f(x_n)|\leq |f_n(x_n)-f_n(x)|+|f_n(x)-f(x_n)|+|f(x_n)-f(x)|.$$ The fact that $|f_n(x)-f(x_n)|\to 0$ and $|f(x_n)-f(x)|\to 0$ is clear. But I can't manage to prove that $|f_n(x_n)-f_n(x)|\to 0$. Any idea ?

Answer 1

The statement cannot be proved, since it is false. Suppose, for instance, that you define, for each $n\in\mathbb N$,$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&x^n-x^{2n}.\end{array}$$Then each $f_n$ is continuous and $(f_n)_{n\in\mathbb N}$ converges pointwise to the null function (which is continuous too), but the convergence is not uniform:$$(\forall n\in\mathbb N):f_n\left(\sqrt[n]{\dfrac12}\right)=\frac14.$$

Answer 2

Let $f=0$ and $f_n$ be the function whose graph is given by the line joining the points $(0,0),({1 \over 2n}, 0)$, $({1 \over n}, 1), ({3 \over 2n}, 0), (1,0)$. The $f,f_n$ are continuous, and $f_n(x) \to f(x)$ everywhere, but the convergence is not uniform since $f_n({1 \over n})-f({1 \over n}) = 1$ for all $n$.