3
$\begingroup$

Let $\{f_n\}_{n\in \mathbb{N}}$ be a sequence of measurable functions on a measure space and $f$ measurable. Assume the measure space $X$ has finite measure. If $f_n$ converges to $f$ in $L^{\infty}$-norm , then $f_n$ converges to $f$ in $L^{1}$-norm.

This is my approach:

We know $||f_n-f||_{\infty} \to 0 $ and by definition $||f_n-f||_{\infty} =\inf\{M\geq 0: |f_n-f|\leq M \}.$ Then \begin{align} ||f_n-f||_1\ &=\int |f_n-f| dm\ &\leq \int|f_n|dm+\int|f|dm\ \end{align}

I don't know how to proceed after that, any help would be appreciated.

$\endgroup$
  • $\begingroup$ You might be also interested in the embeddings of the $L^p$ spaces. $\endgroup$ – Jack Mar 26 '12 at 5:53
2
$\begingroup$

For any function $g$, $||g||_1 = \int_X|g(m)|dm \leq \int_X||g||_\infty dm = \mu(X)*||g||_\infty$ (as $|g(m)| \leq ||g||_\infty$ almost everywhere); $||g||_\infty \geq \frac{||g||_1}{\mu(X)}$, so if $||f_n-f||_\infty$ tends to zero, then $||f_n-f||_1$ tends to zero as well.

$\endgroup$
1
$\begingroup$

Your mistake is using the triangle inequality when you don't have to. Here, $$ \|f_n-f\|_1=\int|f_n-f|dm\leq\|f_n-f\|_\infty\,\int1dm=\|f_n-f\|_\infty\,m(X)\to0. $$

$\endgroup$
  • $\begingroup$ The same also holds for $L^p, p \in (0,\infty)$ instead of $L^1$, right? $\endgroup$ – user563311 Sep 5 '18 at 9:38
  • $\begingroup$ Yes, indeed. $\ $ $\endgroup$ – Martin Argerami Sep 5 '18 at 11:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.