# proof verification: Proving an upper bound for a uniformly convergent sequence of functions on $\mathbb{R}$

Assume $$f_n: \mathbb{R}\to\mathbb{R}$$ is a sequence of functions that converges uniformly to $$f$$. Assume that there exists $$M>0$$ such that for all $$n\in \mathbb{N}$$ and $$x\in \mathbb{R}$$ one has $$|f_n(x)|\leq M.$$ Prove that for all $$x\in \mathbb{R}, |f(x)|\leq M.$$

My proof: By hypothesis, there exists $$\epsilon>0, \epsilon=M$$ such that there is an $$N$$ $$n>N \implies |f_n(x)-f(x)|<\epsilon=M$$ for all $$x\in \mathbb{R}.$$ Hence, $$|f(x)|\leq |f_N(x)|\leq M$$ Since $$f_n(x)\to f(x)=f_N(x)$$ as $$n\to \infty,$$ one has $$|f(x)|\leq M$$

Is this proof correct? I'm a bit worried about my claim that "Since $$f_n(x)\to f(x)=f_N(x)$$ as $$n\to \infty,$$ one has $$|f(x)|\leq M$$".

## 1 Answer

You have no reason to suppose that you will have $$\bigl\lvert f(x)\bigr\rvert\leqslant\bigl\lvert f_N(x)\bigr\rvert$$. If, for instance, $$f_n(x)=1-\frac1n$$ and $$f(x)=1$$, then you don't have that.

You can prove it this way: if $$(f_n)_{n\in\mathbb N}$$ converge uniformly to $$f$$, then it converges pointwise to $$f$$. So, for each $$x\in\mathbb R$$, $$f(x)=\lim_{n\to\infty}f_n(x)$$. But this implies that , for each $$x\in\mathbb R$$, $$\bigl\lvert f(x)\bigr\rvert=\lim_{n\to\infty}\bigl\lvert f_n(x)\bigr\rvert\leqslant M$$.