Let $f$ be a function continuous on the real line such that $f(x) = 0$ for all $|x|\geq T$ (T being some positive number). I want to show the following inequality: $\int_R |f(x)|dx \leq [\int_R(1+|x|)^2|f(x)|^2dx]^{1/2} [\int_R(1+|x|)^{-2}dx]^{1/2}$.

I know that I have to use the Holder's inequality, but I don't know how to deal with the second $(1+|x|)^{-2}$. If it was just +2 as it's power I could have taken the (1+|x|)^2 out and just used the Holder's inequality on f(x) and 1.

Any help is appreciated.

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Holder inequality tells you that $$ \int |g(x)|\cdot |h(x)|\ dx \le (\int |g(x)|^2)^{1/2}(\int |h(x)|^2)^{1/2}. $$

Write $$ |f(x)|=\frac{1}{1+|x|}\cdot (1+|x|)|f(x)|. $$ Now use Holder to $g(x)=\frac{1}{1+|x|}$ and $h(x)=(1+|x|)|f(x)|$.

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  • $\begingroup$ I got it. I'm just curious what is the use for $f(x)=0$ for all $|x| \geq T$. $\endgroup$ – Jack Jan 30 '19 at 21:40
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    $\begingroup$ With that assumption, you have $f\in L^p$ for all $p\geq 1$. $\endgroup$ – user587192 Jan 30 '19 at 21:45

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