How to show that $L^\infty(0,T;L^\infty(\Omega))$ is complete?

Question

How to show that $L^\infty(0,T;L^\infty(\Omega))$ is complete? I did the usual: let $u_n$ be a Cauchy sequence, then we get $$\text{esssup}_t \;\text{esssup}_x |u_n(x,t)-u_m(x,t)| \leq \epsilon$$ Now I don't know what to do.

Answer 1

Consider $\{f_n\}_{n\geq 0} \subset L^{\infty}(0,T;L^\infty(\Omega))$ be a Cauchy sequence.

Now that $\implies$ given $\epsilon > 0$ there exists a $n_0$ s.t. $$ \|f_n - f_m\|_{\infty} \leq \epsilon $$ for all $n,m \geq n_0$.

We have now : in particular $\|f_n\|_\infty \leq \|f_{n_0}\|_\infty + \epsilon $ for all $n \geq n_0$.

This shows that all $f_n$'s are bounded by a finite quantity. So we can safely take pointwise limit of $f_n$, i.e. $$ f(t,g) = \lim_{n \rightarrow \infty} f_n(t, g). $$

The last thing you have to prove is $\| f_n - f \|_\infty \rightarrow 0$ as $n \rightarrow \infty$ which follows from the definition of $f$.