I have read the theorem which says:
If continuous sequence $(f_n(x))$ converges uniformly to function $f(x)$ in some interval of real numbers, than $f(x)$ must be also continuous.
I was just wondering if the statement written in the question is true or not?
because if the proposed statement is not true ,then i can always say that if $f(x)$ comes out to be discontinuous then,$f_m(x)$ does not converge uniformly.
Best Answer
The statement is true and your conclusion is correct. Take, for instance, for each natural $n$,$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&x^n.\end{array}$$Then, if you define$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x<1\\1&\text{ otherwise,}\end{cases}\end{array}$$you have$$\bigl(\forall x\in[0,1]\bigr):\lim_{n\to\infty}f_n(x)=f(x).$$So, since each $f_n$ is continuous and $f$ is discontinuous, the convergence cannot possibly be uniform.