Let $f,f_{k}\in L^p,$ where $,0<p\le\infty.$
Show that $\lVert f_{k}\rVert_{p}\longrightarrow\lVert f\rVert_{p}$ whenever $\lVert f-f_{k}\rVert_{p}\longrightarrow 0$ as $k\longrightarrow\infty$
Here's my attempt :
For each $p>0$( possibly $p$ is infinity )$~$,we have $f_{k}$ converges to $f$ in measure provided that $\lVert f-f_{k}\rVert_{p}\longrightarrow 0$ as $k\longrightarrow\infty~$.Hence $f_{k_{n}}$ converges to $f$ in measure for arbitrary subsequence $f_{k_{n}}$ of $f_{k}.$ Thus there exists an subsequence $f_{k_{n_{\,i}}}$ of $f_{k_{n}}$ such that $\lim_{{i}\rightarrow\infty}f_{k_{n_{\,i}}}=f\,$ a.e.
Henceforth, $~~\lim_{{i}\rightarrow\infty}|f_{k_{n_{\,i}}}|^p=|\lim_{{i}\rightarrow\infty}f_{k_{n_{\,i}}}|^p=|f|^p\,$ a.e. ,where the second equality holds by the function$~|\cdot|^p$ is continuous.
Therefore, one has
$$\limsup_{i\longrightarrow\infty}\int |f_{k_{n_{\,i}}}|^{p}\le \int|f|^{p}\le \liminf_{i\longrightarrow\infty}\int|f_{k_{n_{\,i}}}|^{p}$$
Now , for each subsequence $\bigg(\int|f_{k_{n}}|^{p}\bigg)_{n=1}^{\infty}$ of $~~\bigg(\int|f_{k}|^{p}\bigg)_{k=1}^{\infty}$, we have a convergent
sub-subsequence, that is,
$$\lim_{i\longrightarrow\infty}\int|f_{k_{n_{\,i}}}|^{p}=\int|f|^{p}$$ Whence,we must have $$\lim_{k\longrightarrow\infty}\lVert f_{k}\rVert_{p}^{p}=\lim_{k\longrightarrow\infty}\int|f_{k}|^{p}=\int|f|^{p}=\lVert f\rVert_{p}^{p}$$
So,one has $$\lim_{k\longrightarrow\infty}\lVert f_{k}\rVert_{p}=\bigg(\lim_{k\longrightarrow\infty}\lVert f_{k}\rVert_{p}^{p}\bigg)^{1/p}=\bigg(\lVert f\rVert_{p}^{p}\bigg)^{1/p}=\lVert f\rVert_{p}$$
If you have the time , please check this for validity . Any suggestion or advice will be appreciated . Thanks for considering my request.