# How to prove Integral inequality with Hardy's inequality

Let $f\in C_{0}^{\infty}((-1,1))$. Prove that for any $t\in (-1,1)$ we have $$(f(t))^4\le \left(\int_{-1}^{1}\dfrac{[2(1-|x|)f'(x)-f(x)][2(1-|x|)f'(x)+f(x)]}{4(1-|x|^2)}dx\right)\cdot\left(\int_{-1}^{1}|f(x)|^2dx\right)$$

Thank

• Are there things you've tried to solve this? – Hetebrij Oct 17 '15 at 16:49
• For $f(x)=1$, the integral in the middle is $-\frac{1}{2}\int_{-1}^1\frac{dx}{1-x^2}$, which does not converge. There are many other cases like that. What does your inequality state in these cases? – uniquesolution Oct 23 '15 at 17:17
• @uniquesolution think the function space specified means that the functions must go to zero at $(-1,1)$. Or maybe have compact support in the domain. So I don't think a constant function like that is allowed (unless it's zero). – amcalde Oct 23 '15 at 18:06