1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Because it is a sequence of bounded functions. $\endgroup$ – BenL May 7 '17 at 14:35
  • $\begingroup$ 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. $\endgroup$ – BenL May 7 '17 at 15:31
  • $\begingroup$ @Masacroso I think I fixed it with my edits. $\endgroup$ – BenL May 7 '17 at 15:48
  • 1
    $\begingroup$ @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 $|f(x)|<|f_n(x)|+\varepsilon$. $\endgroup$ – Aweygan May 7 '17 at 16:09
  • $\begingroup$ @Aweygan right, after all it is the same :p Silly me... $\endgroup$ – Masacroso May 7 '17 at 16:13
1
$\begingroup$

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 $M$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.