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 $|f(x)|\leq M$".


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 $x\in\mathbb R$, $\bigl\lvert f(x)\bigr\rvert=\lim_{n\to\infty}\bigl\lvert f_n(x)\bigr\rvert\leqslant M$.


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