proof of uniform convergence of sequence of functions

Question

Let be $f_n$ a sequence of functions with $f_n:\mathbb{R}\supset[\alpha, \beta] \to \mathbb{R}$ and $f_n$ continuous and bounded for all $n\in \mathbb{N}$. Further, $f_n$ converges point-wisely: $\lim\limits_{n\to\infty}f_n(x) = f(x)$, where $f(x)$ is also continuous and bounded for all $x\in[\alpha, \beta]$.

Can I conclude that $f_n$ converges uniformly?

I tried to construct a finite covering of $[\alpha,\beta]$ which should have delivered a $n_0$ such that for all $n>n_0$ the condition of uniform convergence holds. However, it didn't work...

May be it is not possible with further assumptions. In that case I would be interested in which further assumptions I need to prove uniform convergence?

Answer 1

No, that is not true. Take, for instance, $f_n\colon[0,1]\longrightarrow\Bbb R$ defined by $f_n(x)=nx(1-x)^n$. Each $f_n$ is continuous and bounded and $(f_n)_{n\in\Bbb N}$ converges uniflmly to the null function. But the convergence is not uniform, since$$(\forall n\in\Bbb N):f\left(\frac1{n+1}\right)=\left(\frac n{n+1}\right)^{n+1}\text{ and }\lim_{n\to\infty}\left(\frac n{n+1}\right)^{n+1}=\frac1e\ne0.$$However, if, for each $x\in[\alpha,\beta]$, the sequence $(f_n(x))_{n\in\Bbb N}$ is increasing (or if, for each $x\in[\alpha,\beta]$, the sequence $(f_n(x))_{n\in\Bbb N}$ is decreasing), then, yes, the convergence is uniform; that's Dini's theorem.

Answer 2

The statement is not true. For instance, $f_n(x)=n x (1-x^2)^n$ is pointwize convergent to $f = 0$, but the corresponding integrals satisfy $$ \int_0^1 f_n(x)dx= \frac{n}{2+2n} \to \frac 12 \ne 0. $$