# Prove that a uniformly convergent sequence of bounded functions is uniformly bounded

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?

• Because it is a sequence of bounded functions.
– BenL
May 7, 2017 at 14:35
• I mean, both are vacuously true; they differ at most by $\epsilon$. That case, we can say that $M_n < \sup{|f|} + \epsilon/2$ and $\sup|f| < M_n + \epsilon/2$, as the inequalities hold for all $x$. At any rate, I'm not sure how to proceed. Any suggestions? The only sample proofs I can find seem to have a major error: they only consider N very large.
– BenL
May 7, 2017 at 15:31
• @Masacroso I think I fixed it with my edits.
– BenL
May 7, 2017 at 15:48
• @Masacroso There's no problem with that inequality, orther than the $\varepsilon/2$. We have $$|f(x)|-|f_n(x)|\leq\left| |f(x)|-|f_n(x)|\right|<\varepsilon,$$ so $|f(x)|<|f_n(x)|+\varepsilon$. May 7, 2017 at 16:09
• @Aweygan right, after all it is the same :p Silly me... May 7, 2017 at 16:13

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 $M$.