let $f(n)_{n\geq1}$ be a sequence of functions in $L^P(\mathbb{R})$, $P\in[1,\infty)$ which converge to $f$ in $L^P(\mathbb{R})$.

i have choose a subsequence $(f_{k_{n}})$ such that $$||f_{k_{n}}-f||_P \leq 2^{-2k}$$ for each $k\geq1$. Now we have to show that $$\int \left(\sum _k |f_{k_{n}}(x)-f(x)|\right)^p \ < \infty$$.

Hint is provided

Use Holder's inequality to show $$ \left(\sum _k |f_{k_{n}}(x)-f(x)|\right)^p \leq C \sum_k|f_{k_{n}}(x)-f(x)|^p \ 2^{kp}$$ for some fixed finite $C >0$.Use the definition of $f_{k_{n}}$ to complete the proof.

i am confused how to use the given hint. any help please.

thanks in advanced


\begin{align*} \left(\int\left(\sum|f_{n_{k}}-f|\right)^{p}\right)^{1/p}&=\left(\int\lim_{N\rightarrow\infty}\left(\sum_{k=1}^{N}|f_{n_{k}}-f|\right)^{p}\right)^{1/p}\\ &=\lim_{N\rightarrow\infty}\left(\int\left(\sum_{k=1}^{N}|f_{n_{k}}-f|\right)^{p}\right)^{1/p}\\ &=\lim_{N\rightarrow\infty}\left\|\sum_{k=1}^{N}|f_{n_{k}}-f|\right\|_{L^{p}}\\ &\leq\lim_{N\rightarrow\infty}\sum_{k=1}^{N}\|f_{n_{k}}-f\|_{L^{p}}\\ &=\lim_{N\rightarrow\infty}\sum_{k=1}^{N}\dfrac{1}{2^{2k}}\\ &<\infty, \end{align*} where the second equality comes from Monotone Convergence Theorem and the continuity of $(\cdot)^{1/p}$.

With the hint: \begin{align*} \sum|f_{n_{k}}(x)-f(x)|&=\sum|f_{n_{k}}(x)-f(x)|2^{k}\cdot\dfrac{1}{2^{k}}\\ &\leq\left(\sum|f_{n_{k}}(x)-f(x)|^{p}2^{kp}\right)^{1/p}\left(\sum\dfrac{1}{2^{kq}}\right)^{1/q}\\ &=C\left(\sum|f_{n_{k}}(x)-f(x)|^{p}2^{kp}\right)^{1/p}, \end{align*} of course, here $p>1$. For $p=1$ is trivial.

So we have \begin{align*} \int\left(\sum|f_{n_{k}}-f|\right)^{p}&\leq C\int\sum|f_{n_{k}}-f|^{p}2^{kp}\\ &=\sum\int|f_{n_{k}}-f|^{p}2^{kp}\\ &=\sum\|f_{n_{k}}-f\|_{L^{p}}^{p}2^{kp}\\ &\leq\sum 2^{-2kp}\cdot 2^{kp}\\ &=\sum 2^{-kp}\\ &<\infty. \end{align*}

  • $\begingroup$ from the question, it is expected to follow the hint + may be monotonic convergence $\endgroup$ – dEmigOd Apr 2 '18 at 6:12
  • $\begingroup$ Okay, I will edit that. $\endgroup$ – user284331 Apr 2 '18 at 6:12
  • $\begingroup$ Oh, I overlooked that, I know what you meant. $\endgroup$ – user284331 Apr 2 '18 at 6:22
  • $\begingroup$ @user284331 where does we use holder inequality as hinted $\endgroup$ – paarth Apr 2 '18 at 6:24
  • $\begingroup$ Wait, I will include the hint-oriented solution, I overlooked that. $\endgroup$ – user284331 Apr 2 '18 at 6:24

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