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Let $f$ be a twice continuously differentiable function from $[0,1]$ into $R$,Give that $$f(0)+2f(\frac{1}{2})+f(1)=0$$ show that $$\int_{0}^{1}(f''(x))^2dx\ge 1920\left(\int_{0}^{1}f(x)dx\right)^2$$

I tried some methods ,such Cauchy-Schwarz inequality $$\int_{0}^{1}(f''(x))^2dx\cdot\int_{0}^{1}g^2(x)dx\ge\left(\int_{0}^{1}f''(x)g(x)dx\right)^2$$ where $g(x)$ is polynomial function.and $\max{\deg{(g(x))}}\le 2$

and By parts intgeral we have $$\int_{0}^{1}f''(x)g(x)dx=f'(1)g(1)-f'(0)g(0)-f(1)g'(1)+f(0)g'(0)+\int_{0}^{1}f(x)g''(x)dx= f'(1)g(1)-f'(0)g(0)-f(1)g'(1)+f(0)g'(0)+C\int_{0}^{1}f(x)dx$$

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  • $\begingroup$ Use the technique described here twice $\endgroup$ – RE60K Mar 7 '15 at 9:07
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1) Let $g_1(x)=x(x-1/2)$, $g_2(x)=(x-1)(x-1/2)$. By two integration by parts, we have

$$\int_0^{1/2}f^{\prime\prime}(x)g_1(x)dx=-\frac{f(1/2)+f(0)}{2}+2\int_0^{1/2}f(x)dx$$ and $$\int_{1/2}^{1}f^{\prime\prime}(x)g_2(x)dx=-\frac{f(1/2)+f(1)}{2}+2\int_0^{1/2}f(x)dx$$ Hence $$\int_0^{1/2}f^{\prime\prime}(x)g_1(x)dx+\int_{1/2}^{1}f^{\prime\prime}(x)g_2(x)dx=2\int_0^1f(x)dx$$

2) By Cauchy-Schwarz:

$$(\int_{0}^{1/2}f^{\prime\prime}(x)g_1(x)dx)^2\leq (\int_{0}^{1/2}f^{\prime\prime}(x)^2dx)\frac{1}{15.2^6}$$

$$(\int_{1/2}^{1}f^{\prime\prime}(x)g_2(x)dx)^2\leq (\int_{1/2}^{1}f^{\prime\prime}(x)^2dx)\frac{1}{15.2^6}$$

3) Use now $\sqrt{U}+\sqrt{V}\leq \sqrt{2}\sqrt{U+V}$ with $\displaystyle U=\int_{0}^{1/2}f^{\prime\prime}(x)^2dx$ and $\displaystyle V=\int_{1/2}^{1}f^{\prime\prime}(x)^2dx$ to finish the proof.

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  • $\begingroup$ (+1)'ed nice !! $g(x) = \begin{cases} g_1(x) &, x \in [0,1/2] \\ g_2(x) &, x \in [1/2,1] \end{cases}$ is continuous and $f''$ is continuous as well, you could apply CS ineq directly on $g(x)f''(x)$ :) $\endgroup$ – r9m Apr 3 '15 at 17:21

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