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I know that there are functions in $L^2$ that are integrable but not continuous. Is there any function in $L^2$ that is not even integrable?

3x89g2
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2 Answers2

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Yes, for instance if you define $f : \Bbb R \to \Bbb R$ by $f(x)=0$ if $x<1$ and $f(x)=1/x$ if $x≥1$, then $f \in L^2(\Bbb R)$ but $f \not \in L^1(\Bbb R)$.

Watson
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$$f(x)=\frac{1}{x}$$ on $[1,\infty [$. Notice that on $(a,b)$, $$L^2(a,b)\subset L^1(a,b).$$ Therefore, such a result would be impossible on bounded set.

Surb
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