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How to prove that, given a function $f\in L^p$ there exists a function $f_{n}$ compactly supported, in $L^{\infty}$ and such that $f_n \rightarrow f$ in $L^{p}$ ? I think of using the function $f_{n}(x)=\vert f(x) \vert 1_{\{x:\vert f(x\vert)<n\}}$ which is in $L^{\infty}$ since $\vert f_{n} \vert \leq n$. Could I say that $f_n$ is compactly supported in $B(0,n)$? how could I prove that $f_n \rightarrow f$ in $L^{p}$ ?

TZakrevskiy
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user125967
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1 Answers1

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Try $$f_n(x) = 1_{|x|<n} 1_{|f(x)|<n} f(x).$$. It's compactly supported and bounded. Dominated convergence theorem will help you to prove that $f_n\to f$.

TZakrevskiy
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  • I agree with you that it is compactly supported in $B(0,n)$. How could you prove the a.e convergence to $f$? the domination is clear since $\Vert f_n \Vert_{p} < \Vert f \Vert_{p} \in L^{1}$ – user125967 Feb 04 '14 at 16:56
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    @user125967 $f$ is finite a.e.. – David Mitra Feb 04 '14 at 17:05