# showing an inequality using Holder's inequality

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.

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