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.

  • 2
    $\begingroup$ This is just the continuity of the norm $\|\cdot\|_p$. It can be proven using the reverse triangle inequality: $$\left|\|f\|_p - \|f_k\|_p\right| \le \|f - f_k\|_p \xrightarrow{k \to\infty} 0$$ $\endgroup$ – mechanodroid Dec 22 '17 at 10:45
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    $\begingroup$ The difference of the norms is always smaller than the norm of the difference. $\endgroup$ – Gribouillis Dec 22 '17 at 10:46
  • $\begingroup$ hi @mechanodroid: Does this inequality hold when $p\in (0,1)$ ? since $L^p$ is not a norm for p between 0 and 1. $\endgroup$ – user1992 Dec 22 '17 at 11:19
  • $\begingroup$ @user1992 I missed that detail. If you define $\|\cdot\|_p = \int |f|^p$, then it still holds because $d_p(f, g) = \|f - g\|_p$ is a metric for $p \in (0,1)$. If you define $$\|\cdot\|_p = \left(\int |f|^p\right)^\frac1p$$ then I don't think it holds. $\endgroup$ – mechanodroid Dec 22 '17 at 11:24
  • $\begingroup$ @mechanodroid:okay,I understand,thanks $\endgroup$ – user1992 Dec 22 '17 at 11:36

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