# If $(f_n)$ is a cauchy sequence for the sup-norm, then $(f_n)$ converge in $\mathcal C^0([0,1])$

Let $$(f_n)$$ a cauchy sequence for $$\|f\|_\infty =\sup_{[0,1]}|f|$$ of $$\mathcal C([0,1])$$. Prove that $$(f_n)$$ converges to a function $$f\in \mathcal C([0,1])$$.

I made a different proof than my course. Is it correct ? I just need to prove that there is a function $$f$$ s.t. $$f_n\to f$$ uniformly. So I did as follow :

• Let $$\varepsilon >0$$. There is $$N\in \mathbb N$$ s.t. $$n,m\geq N$$ we have $$\|f_n-f_m\|_\infty <\frac{\varepsilon}{2}$$. In particular, $$(f_n(x))_n$$ is a Cauchy sequence, and thus converges to a $$f(x)$$.

• Let $$n\geq N$$. There is $$x_n\in [0,1]$$ s.t. $$\|f_n-f\|_\infty -\frac{\varepsilon}{2} \leq |f_n(x_n)-f(x_n)|.$$ Let $$m\geq N$$. Then $$|f_n(x_n)-f(x_n)|\leq |f_n(x_n)-f_m(x_n)|+|f_m(x_n)-f(x_n)|\leq \frac{\varepsilon }{2}+\underbrace{|f_m(x_n)-f(x_n)|}_{\to 0, m\to \infty }\to \frac{\varepsilon }{2}.$$

• Therefore $$\|f_n-f\|_\infty -\frac{\varepsilon }{2}\leq \frac{\varepsilon }{2}\implies \|f_n-f\|_\infty \leq \varepsilon ,$$ for all $$n\geq N$$. This prove that $$f_n\to f$$ uniformly.

Does it works ?

• Your proof is incomplete. You did not prove that $f \in C[0,1]$. – Kabo Murphy Aug 9 at 12:21
• @KaviRamaMurthy: I know that if $f_n\to f$ uniformly, then $f$ is continuous. That's why I wrote : I just have to prove that $f_n\to f$ uniformly :) So is the proof of uniform convergence correct ? – John Aug 9 at 12:55
• Yes, your proof seems OK. – Kabo Murphy Aug 9 at 13:05
• @KaviRamaMurthy: For my part, I see one mistake : why $\|f-f_n\|_\infty$ should be finite ? – Surb Aug 9 at 15:30

You also have to prove that $$f$$ is continuous. For this note that $$|f_N(x)-f_n(x)| < \frac {\epsilon} 2$$ for all $$x$$ for all $$n \geq N$$. Let $$n \to \infty$$ to conclude that $$|f_N(x)-f(x)| \leq \frac {\epsilon} 2$$ for all $$x$$. Now $$f_N$$ is uniformly continous, so there exists $$\delta >0$$ such that $$|f_N(x)-f_N(y)| <\frac {\epsilon} 2$$ for $$|x-y| <\delta$$. Use the inequality $$|f(x)-f(y)| \leq |f(x)-f_N(x)|+|f_N(x)-f_N(y)|+|f(y)-f_N(y)|$$ to complete the proof.
• I know that if $f_n\to f$ uniformly, then $f$ is continuous. That's why I wrote : I just have to prove that $f_n\to f$ uniformly :) – John Aug 9 at 12:55