Finite products of Banach spaces are Banach

If $x^{(m)}$ is Cauchy for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\max }$, then each coordinate sequence $(x_k^{(m)}{)}_m$ is Cauchy in $E_k$, because $\mathrm{\parallel }{x}_k^{(m)}-x_k^{(\mathrm{\ell })}\mathrm{\parallel }\le \mathrm{\parallel }{x}^{(m)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_{\max }$ for every $k<n$.

Since each $E_k$ is Banach, [L1] gives a point $x_k\in E_k$ with $x_k^{(m)}\to x_k$. Put $x:=(x_0,\dots ,x_{n-1})$.

Given $\epsilon >0$, choose $M_0$ so that $\mathrm{\parallel }{x}^{(m)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_{\max }<\epsilon \mathrm{/}2$ for $m,\mathrm{\ell }\ge M_0$, and for each $k<n$ choose $M_k$ so that $\mathrm{\parallel }{x}_k^{(m)}-x_k\mathrm{\parallel }<\epsilon \mathrm{/}2$ for $m\ge M_k$. For $M:=\max \{M_0,M_1,\dots ,M_{n-1}\}$ and $m\ge M$, taking $\mathrm{\ell }\to \mathrm{\infty }$ in a fixed coordinate gives $\mathrm{\parallel }{x}_k^{(m)}-x_k\mathrm{\parallel }\le \epsilon \mathrm{/}2$ for every $k<n$, hence $\mathrm{\parallel }{x}^{(m)}-x{\mathrm{\parallel }}_{\max }\le \epsilon \mathrm{/}2<\epsilon$.

Thus the product is complete for the maximum norm, hence Banach for that norm by [L1].

The inequalities in [L3] show that a sequence is Cauchy or convergent for one standard product norm exactly when it is so for the others. Therefore completeness for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_{\max }$ is equivalent to completeness for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_2$ and for $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$, so the product is Banach for all three norms.