# 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.

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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.

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Thanks, that was easy. –  Lyapunov Mar 26 '12 at 6:27

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

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Thank you, I feel silly! –  Lyapunov Mar 26 '12 at 6:27

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. $f_n\to f$ in the $L^1$-norm , if and only if $f_n\to f$ in measure? Is this true? Thanks.

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This is a question, not an answer. To ask a question, use the ask question button, right under the MATHEMATICS logo. –  tomasz Apr 1 '13 at 12:02
When you post this as a separate question, please use the MathJax that I have added. –  robjohn Apr 1 '13 at 15:20