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I'm reading Real analysis(Stein) and confused by a statement: If $f\in L^2$, then $f\chi_R$ is in both $L^1$ and $L^2$, where $\chi_R$ is the characteristic function of $[-R, R]$.

But $f$ in $L^2$ does not mean that $f\in L^1$. Such an example given by Almost everywhere continuous functions. How could this function on $[-R, R]$ be in $L^1$?

John Griffin
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Russell Hiwayor
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1 Answers1

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Due to the Cauchy-Schwarz inequality, we have $$ \|f\chi_R\|_1 \leq \|f\|_2\|\chi_R\|_2 = (2R)^{1/2}\|f\|_2 < \infty. $$

John Griffin
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  • Oh thank u!! I must be confused by that example. – Russell Hiwayor Aug 16 '17 at 04:10
  • @RussellHiwayor The example shows only that there are nonmeasurable functions whose square is measurable. Such a function will not be in $L^2$ because $L^2$ consists of all measurable functions that are square integrable. However, this isn't the main point. There are elements in $L^2$ that are not in $L^1$, but on a finite measure space (which is what we are really working with by multiplying by $\chi_R$), then $L^p \subseteq L^q$ for all $p>q$. To prove this statement, just use Holder's inequality as I have used the Cauchy-Schwarz inequality here. – John Griffin Aug 16 '17 at 15:29