# Convergence in $L^{\infty}$ norm implies convergence in $L^1$ norm

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.

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

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.
Your mistake is using the triangle inequality when you don't have to. Here, $$\|f_n-f\|_1=\int|f_n-f|dm\leq\|f_n-f\|_\infty\,\int1dm=\|f_n-f\|_\infty\,m(X)\to0.$$
• The same also holds for $L^p, p \in (0,\infty)$ instead of $L^1$, right? – user563311 Sep 5 '18 at 9:38
• Yes, indeed. $\$ – Martin Argerami Sep 5 '18 at 11:08