Why the convergence is uniform here?

Question

I was reading a proof that creates a function to prove the completion of $C[a,b]$ with the maximum norm and the author wrote:

Let $(f_{n})$ be a Cauchy sequence in $C[a,b].$ Fix $x \in [a,b],$ then by the definition of a Cauchy sequence we have $$\forall \epsilon >0, \exists N \in \mathbb{N}: \|f_{n} - f_{m}\| _{max} < \epsilon \quad \forall m,n \geq N. \quad (*) $$ But by definition of the maximum norm $|f_{n}(x) - f_{m}(x)| \leq \|f_{n} - f_{m}\| _{max}.$ Therefore , there exists $N$ such that $$ |f_{n}(x) - f_{m}(x)| < \epsilon \quad \forall m,n \geq N. \quad (**)$$ Hence, $(f_{n}(x))$ is a Cauchy sequence in $\mathbb{R}.$

Then the author said by $(**),$ there exists $N$ such that $$ |f_{n}(x) - f_{m}(x)| < \epsilon \quad \forall x \in [a,b] \quad \forall m,n \geq N.$$

And hence the convergence is uniform.

My question is :

How $(**)$ give us this uniform convergence? could anyone explain this for me, please?

Answer 1

The property $(\ast)$ gives you an $N$ dependent only on $\varepsilon$ for the Cauchyness of the sequence $(f_n)$ in the $\|\cdot\|_{\text{max}}$ metric.

So that $N$ also works for all $x$ simultaneously for that $\varepsilon$ because for any $x$ we have $|f_n(x)-f_m(x)| \le \|f_n -f_m\|_{\text{max}}$, so the sequence $(f_n(x))_n$ is Cauchy in $\Bbb R$ for any fixed $x$, and thus converges to some point which we define to be $f(x) \in \Bbb R$.

Answer 2

Take $\varepsilon>0$. You know that there is a $N\in\mathbb N$ such that$$(\forall x\in[a,b])(\forall m,n\in\mathbb N):m,n\geqslant N\implies\bigl\lvert f_n(x)-f_m(x)\bigr\rvert<\frac\varepsilon2.$$But then, if $x\in[a,b]$ and $n\geqslant N$,\begin{align}\bigl\lvert f_n(x)-f(x)\bigr\rvert&=\left\lvert f_n(x)-\lim_{m\to\infty}f_m(x)\right\rvert\\&=\lim_{m\to\infty}\bigl\lvert f_n(x)-f_m(x)\bigr\rvert\\&\leqslant\frac\varepsilon2\\&<\varepsilon.\end{align}