let $\delta\in(0,1)$, and $f\in C_{0}^{1}(R_{+})$,show that $$\int_{0}^{\infty}(f(t))^2t^{-\delta}dt\le\dfrac{4}{(1-\delta)^2}\int_{0}^{\infty}(f'(t))^2t^{2-\delta}dt$$

My idea: I think we must use Cauchy-Schwarz inequality

$$\int_{0}^{\infty}(f'(t))^2t^{2-\delta}dt\int_{0}^{\infty}t^{-2-\delta}dt \ge\left(\int_{0}^{\infty}f'(t)t^{-\delta}dt\right)^2$$

But I can't know this coefficient $\dfrac{4}{(1-\delta)^2}$

How do have it? Thank you

  • $\begingroup$ $C_0^1$ here is $C^1$ functions with compact support? $\endgroup$ – Simon S Nov 5 '14 at 3:55

In $C_0^1(R_+)$, $\lim_{t\to\infty} f(t)=0$, so you can use fundamental theorem of calculus to write $f^2(t)=\int_t^\infty 2f(s)f'(s)ds$. The left hand side then equals to $$\int_0^\infty \int_t^\infty 2f(s) f'(s) t^{-\delta}dsdt $$ Then interchange the order of integration $$\int_0^\infty\int_0^s 2t^{-\delta}dt f(s) f'(s)ds=\frac{2}{1-\delta} \int_0^\infty f(s)f'(s) s^{1-\delta} ds$$ By Cauchy-Schwarz inequality $$ \int_0^\infty f(s)f'(s) s^{1-\delta} ds\le\left(\int_0^\infty f(s)^2s^{-\delta}ds \right)^{1/2}\left(\int_0^\infty (f'(s))^2s^{2-\delta}ds \right)^{1/2} $$ Combine everything altogether $$\int_0^\infty (f(t))^2t^{-\delta} dt\le\frac{2}{1-\delta} \left(\int_0^\infty f(s)^2s^{-\delta}ds \right)^{1/2}\left(\int_0^\infty (f'(s))^2s^{2-\delta}ds \right)^{1/2}$$ This yields the inequality.

  • $\begingroup$ very nice !! (+1) $\endgroup$ – r9m Nov 6 '14 at 0:54

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