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There is a known fact says: if $f^2$ is integrable then $f$ not necessary integrable ($f:[a,b]\to \mathbb{R}$).

But since $f^2=|f|^2$, then one may expect that also $|f|$ not necessary integrable. Is this true, or this previous fact not correct for $|f|$?

Thanks.

user31009
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2 Answers2

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Yes it is, since you are specifying the domain to be the bounded interval $[a,b]$.

Since $f^2$ is integrable it is measurable, which means that $|f| = \sqrt{f^2}$ is also measurable. (The same cannot be said about $f$!) The conclusion is just the Cauchy-Schwarz inequality: $$\int_a^b |f| \, dx \le (b-a)^{1/2} \left( \int_a^b f^2 \, dx \right)^{1/2}.$$

Umberto P.
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  • Don't you need to know that $|f|$ (or $f$) is integrate to begin with ? The space of which this inequality is deduced is the space of integrable functions $f$ – Belgi Apr 23 '15 at 16:43
  • No, as long as $f^2$ has finite integral, so must $|f|$. – Umberto P. Apr 23 '15 at 16:46
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$$|f(x)|\leq1+f(x)^2$$ so: $$\int_a^b|f(x)|dx\leq \int_a^b1+f(x)^2dx$$

drhab
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