Given a finite measure space $(X,\Sigma,\mu)$, for $1<p<\infty$,
if $f \in L^p(X)$, then $f \in L^1(X)$.
Can anyone show me how to start the proof?
Thanks.
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1Holder inequality – RealAnalysis Apr 16 '15 at 21:03
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2$|f|^p > |f|$ when $|f| > 1.$ Or use Holder. – zhw. Apr 16 '15 at 21:04
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Intuitively, we have that if $f \in L^p(X)$ the function does not have a bad singularity. The way $f$ would fail to be in $L^1$ if it does not converge to zero fast enough as $|x| \to \infty$. Notice that on a set of finite measure we will not have such a problem. – user7090 Apr 21 '15 at 05:02