Problem: $$X=L_2((1,\infty))\cap L_\infty((1,\infty))$$ $$\|x\|=\max\{\|x\|_2,\|x\|_\infty\}$$ Show $(X,\|\cdot\|)$ is Banach space.
Attempt:
In Banach space every cauchy sequence converges: $$\|f_n-f_m\|=\max\{\|f_n-f_m\|_2,\|f_n-f_m\|_\infty\}<\varepsilon$$ $$ \Rightarrow\|f_n-f_m\|_2<\varepsilon\ \mathrm{and}\ \|f_n-f_m\|_\infty<\varepsilon$$ This means that $(f_n)$ converges in $(L_2,\|\cdot\|_2)$ and $(L_\infty,\|\cdot\|_\infty)$, because they are both Banach spaces.
The convergence in L spaces implies: $$\exists (f_{n_k})\subset (f_n), f_1\in L_2:\ |f_{n_k}-f_1|<\varepsilon$$ $$\exists (f_{n_k})\subset (f_n), f_2\in L_\infty:\ |f_{n_j}-f_2|<\varepsilon$$
I know, that I can have $|f_{n_k}-f_{n_j}|<\varepsilon$, so using the triangle inequality: $$|f_1-f_2|=|(f_1-f_{n_k})-(f_2-f_{n_j})+(f_{n_k}-f_{n_j})|<|f_{n_k}-f_1|+|f_{n_j}-f_2|+|f_{n_k}-f_{n_j}|<3\varepsilon$$ so $|f_1(x)-f_2(x)|<\varepsilon\ \forall x\in(1,\infty) \mathrm{(almost)}$.
Is my proof correct? Is it missing something? I think the final result should be $f_1=f_2$, not just that they are near.