# proof of uniform convergence of sequence of functions

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?

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.
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 $$\int_0^1 f_n(x)dx= \frac{n}{2+2n} \to \frac 12 \ne 0.$$