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

Question

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 $f\in \mathcal C([0,1])$.

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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 ?

Answer 1

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.