Greetings Mathematics Community. I believe that I am thinking too hard about the following problem and would like some guidance in solving it.
Let $X$ have finite measure and let $f_n:X \to \mathbb{C}$ and $f:X \to \mathbb{C}$ be measurable functions. Show that if $f_n$ converges to $f$ in $L^{\infty}$norm, then $f_n$ also converges to $f$ in $L^1$norm.
The definitions that I am working with are as follows:
We say that $f_n$ converges to $f$ uniformly almost everywhere, essentially uniformly, or in $L^{\infty}$norm if, for every $\epsilon>0$, there exists $N$ such that for every $n \geq N, |f_n(x)-f(x)| \leq \epsilon$ for $\mu-$almost every $x\in X$. The $L^{\infty}$norm $||f||_{L^{\infty}(\mu)}$ of a measurable function $f:X\to \mathbb{C}$ is defined to be the infimum of all the quantities $M\in [0,+\infty]$ that are essential upper bounds for $f$ in the sense that $|f(x)|\leq M$ for almost every $x$. Then $f_n$ converges to $f$ in $L^{\infty}$ norm if and only if $||f_n-f||_{L^{\infty}(\mu)} \to 0$ as $n \to \infty$.
We say that $f_n$ converges to $f$ in $L^1$norm if the quantity $||f_n-f||_{L^1(\mu)} = \int_{X} |f_n(x)-f(x)| d\mu$ converges to 0 as $n \to \infty$.
Here is my attempt at solving this problem:
Since $f_n$ converges to $f$ in $L^{\infty}$ norm, then we have
$||f_n(x)-f(x)||_{\infty} \to 0$ as $n \to \infty$ and by definition, $||f_n(x)-f(x)||_{\infty} = \inf\{M \geq 0 : |f_n(x)-f(x)| \leq M\}$. Then $||f_n(x)-f(x)||_1 = \int |f_n(x)-f(x)|dx$.
Here is where I am stuck: Could I create another function $g$ such that $||g(x)||_1 \leq ||g(x)||_{\infty}$ and somehow include Markov's Inequality to show that the $L^1$ norm converges?
Alternatively, could I simply say that $\int|f_n(x)-f(x)|dx \leq ||f_n(x)-f(x)||_{\infty}$ and since we are given that $f_n$ converges to $f$ in $L^{\infty}$norm, then $\int|f_n(x)-f(x)|dx \leq 0$?
I would be grateful for any advice or guidance. Thanks in advance.
Answer
$$\|f_n-f\|_1=\int_X|f_n-f|d\mu\le\|f_n-f\|_{\infty}\mu(X).$$
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