This was asked recently (Show that any uniformly convergent sequence of bounded functions is uniformly bounded), and I suppose due to its poor formatting or something it was ignored. At any rate, my proof is different, and I would like feedback on it.
Proposition
Let $\{f_n(x)\}$ be a uniformly convergent sequence of bounded functions. Then the sequence is uniformly bounded.
Proof
Fix $\epsilon/2 >0$. There exists $N$ such that for all $n > N$,
$|f(x) – f_n(x)| < \epsilon/2$.
Then, by reverse triangle inequality,
$|f(x)| < |f_n(x)| + \epsilon < M_n + \epsilon/2$, where $M_n$ is a number such that $|f_n(x)| < M_n \forall x$.
Because $M_n$ is bounded below by the supremum of $|f(x)|$, ${M_n}$ has an infimum, which we shall denote by $M*$. Thus $|f(x)|< M* + \epsilon/2$
Now, again by uniform convergence, $|f_n(x) – f(x)| < \epsilon/2$ for all but a finite number of functions $\{f_1, f_2, … f_{N-1}\}$, with their respective upper bounds $\{M_1, M_2, … M_{N-1}\}$. Denote the maximum over this finite set of numbers by $M**$.
Then, for $n \geq N$
$|f_n(x)| – |f(x)| \leq |f_n(x) – f(x)| < \epsilon/2$
$|f_n(x)| \leq |f(x)| + \epsilon/2 < M* + \epsilon.$
Thus, because $\epsilon$ was arbitrary, $|f_n(x)| < M*$ for all $n >N$.
Finally, let $M = \max\{M*, M**\}$, and it is clear that $|f_n(x)|$ is uniformly bounded by $M$
Edit: I retrofitted my proof using ideas from the interwebs
Because $f_n(x)$ is uniformly convergent, it is uniformly Cauchy, and thus there exists $N$ such that for all $m, n \geq N$
$|f_m(x) – f_n(x)| < 1$.
so
$|f_m(x)| \leq |f_m(x) – f_N(x)| + |f_N(x)| \leq 1 + M_N(x).$ This applies for all $m > N$. If instead of $M_N$ we take $\max\{M_1, M_2, … M_N\}$ this of course bounds $|f_m|$ for $m<N$ as well. Thus, let $M = 1 + \max\{M_1, M_2, … M_N\}$
Is this legit?
Best Answer
I think everything is alright except how you obtain $M^*$. There is some $N$ such that for all $x\in X$ and all $n\geq N$ we have $$|f(x)|<|f_n(x)|+1\leq M_n+1,$$ thus $f$ is uniformly bounded by some $M^*$. Furthermore, for all $n\geq N$ we have $$|f_n(x)|<|f(x)|+1\leq M^*+1$$ for all $x\in X$. You defined $M^{**}$ fine, so if you set $M=\max\{M^*+1,M^{**}\}$, then we have $\{f_n\}$ is uniformly bounded by $M$.