Question: Let $f_n$ be a sequence of bounded functions on a set $S\subset\mathbb R$. Suppose $f_n\to f$ uniformly on $S$. Prove $f_n$ is uniformly Cauchy on $S$.

Attempt: I proved this easily by ignoring the first sentence (i.e. $f_n$ be a sequence of bounded functions). Can someone please tell me what is the importance that first sentence and how would it help me in constructing my proof?

My version: Let $\varepsilon>0$ be arbitrary. Choose $N$ s.t. $\frac1N < \varepsilon$. Let $n,m > N$. Then $$ \begin{align*} |f_n(x) - f_m(x)| &< |f_n(x) - f(x) + f(x) - f_m(x)|\\ &< |f_n(x) - f(x)| + |f_m(x) - f(x)|\\ &< \frac\varepsilon2 + \frac\varepsilon2 = \varepsilon.\end{align*}$$ QED

Thanks for all the advice!

  • $\begingroup$ Why would either $ | f_n(x) - f(x) | < \frac{\epsilon}{2}$ or $| f_m(x) - f(x) | < \frac{\epsilon}{2}$? $\endgroup$ – Ilham Apr 24 '15 at 1:10

Your proof is invalid, and I questioned the step in your comments. (Also you should be careful about other inequalities).

How do you prove a general sequence in a metric space is Cauchy if it is convergent?

Let $\epsilon > 0$.

Then also $\dfrac \epsilon 2 > 0$.

Because $f_n$ converges uniformly to $f$, $\exists N$ s.t.$ \forall n > N$, $\forall x \in S$, $|f_n(x) - f(x)| < \dfrac \epsilon 2$ and $\forall m > N, \forall x \in S, |f_m(x) - f(x)| < \dfrac \epsilon 2$

So if $m > N$ and $n > N$, then $\forall x \in S$: $$|f_n (x) -f_m(x)| \leq |f_n(x) - f(x)| + |f_m(x) - f(x)| < \dfrac \epsilon 2 + \dfrac \epsilon 2 = \epsilon$$

So $f_n$ converges uniformly to $f$.

  • $\begingroup$ Umm, I haven't learned metric space yet. Moreover, I would set it to |fn(x)−f(x)|<ϵ/2 or |fm(x)−f(x)|<ϵ/2 because the question is already telling me that the sequence of functions is converging uniformly to f and so I just need to choose some epsilon such that it would lead me to the definition of cauchy. $\endgroup$ – Jenkin Tsui Apr 24 '15 at 1:20
  • $\begingroup$ You have got the right idea, but you're not choosing the right $N$. $\endgroup$ – Ilham Apr 24 '15 at 1:21
  • $\begingroup$ Ah, I see what N you chose there! So do we really need to apply any knowledge about "Let fn be a sequence of bounded functions on a set S⊂R. " which is stated in the question? $\endgroup$ – Jenkin Tsui Apr 24 '15 at 1:27
  • $\begingroup$ Not that I can see. Fun exercise: can you prove the converse? $\endgroup$ – Ilham Apr 24 '15 at 1:27
  • 1
    $\begingroup$ Cool! Yep. Write f(x) = lim fk(x) and then go by definition. $\endgroup$ – Jenkin Tsui Apr 24 '15 at 1:34

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