Lim Supremum fn(x) = sup f(x)

Question

We have a sequence of function converging uniformly to f(x) which is bound to [0;1] I want to proove that the lim of the sup fn(x) = sup f(x) .

Can someone help me ?

Answer 1

Let $\epsilon > 0$. There exists some $N \geq 1$ such that for all $n \geq N$, $\|f_n-f\|_{\infty} \leq \epsilon$.

Thus, for each $n \geq N$, and each $0 \leq x \leq 1$, $\sup\,f_n - f(x) \geq f_n(x)-f(x) \geq -\epsilon$. In particular, $(\sup\,f_n) - (\sup\,f) \geq -\epsilon$.

You can get the other inequality in a similar way.