${f_n}$ is a sequence of continuous functions that converges uniformly on [0,1]. Show that there is an M such that $|f_n(x)|\leq M$ for all n and x.
My thoughts: since the functions are continuous on a closed bounded interval, they must be bounded. So, we can take the maximum bound and it will satisfy the condition. But this seems too simple. Is this right, or am I missing something here?
Uniform convergence says:
$\forall \epsilon > 0$, $\forall n \geq N$, such that $|f_n(x) - f(x)| < \epsilon, \forall n \geq N$ and $\forall x \in [0,1].$
Then: $$|f_n(x)|=|f_n(x) - f(x)+f(x)| \leq \epsilon + f(x), \forall n \geq N \forall x \in [0,1]$$
Finally a continous function over a interval is bounded pick for $f$ that bound $B$ and you have:
$$|f_n(x)|=|f_n(x) - f(x)+f(x)| \leq \epsilon + f(x)\leq B + \epsilon, \forall n \geq N \forall x \in [0,1]$$
Since then take $M=max\{sup(f_1(x)),sup(f_2(x)),...,sup(f_{N-1}(x)),B+\epsilon\}$. (That because for $\forall n \geq N$ the bound is $B+\epsilon$ and for the first $N-1$ is easy take the maximum of their supremes because are finite)
After answering that I noted that the question have already an answer: HERE
