Prove the following integral inequality: $\int_{0}^{1}(f''(x))^2dx\ge 1920\left(\int_{0}^{1}f(x)dx\right)^2$

Question

Let $f$ be a twice continuously differentiable function from $[0,1]$ into $\mathbb R$. Given 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 as 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 integral we have $$\int_{0}^{1}f''(x)g(x)dx=f'(1)g(1)-f'(0)g(0)-f(1)g'(1)+f(0)g'(0)+\int_{0}^{1}f(x)g''(x)dx= f'(1)g(1)-f'(0)g(0)-f(1)g'(1)+f(0)g'(0)+C\int_{0}^{1}f(x)dx.$$

Answer 1

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:

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

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

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.

Answer 2

Let $g(x)$ be the piecewise differentiable function defined as, $$g(x)=\begin{cases} g_1(x),\quad x\in\left[0,\frac{1}{2}\right)\\ g_2(x),\quad x\in\left[\frac{1}{2},1\right] \end{cases}$$ By integration by parts, we find \begin{align*} \int_0^{\frac{1}{2}}g_1(x)f''(x)\mathrm{d}x &=g_1(x)f'(x)\bigg|_0^{\frac{1}{2}}-\int_0^{\frac{1}{2}}g_1'(x)f'(x)\mathrm{d}x\\ &=g_1(x)f'(x)\bigg|_0^{\frac{1}{2}}-g_1'(x)f(x)\bigg|_0^{\frac{1}{2}}+\int_0^{\frac{1}{2}}g_1''(x)f(x)\mathrm{d}x \end{align*} and \begin{align*} \int_{\frac{1}{2}}^1g_2(x)f''(x)\mathrm{d}x &=g_2(x)f'(x)\bigg|_{\frac{1}{2}}^1-\int_{\frac{1}{2}}^1g_2'(x)f'(x)\mathrm{d}x\\ &=g_2(x)f'(x)\bigg|_{\frac{1}{2}}^1-g_2'(x)f(x)\bigg|_{\frac{1}{2}}^1+\int_{\frac{1}{2}}^1g_2''(x)f(x)\mathrm{d}x \end{align*} First, we need $g_1''(x)=g_2''(x)=\text{constant}$, so, $g_1(x)$ and $g_2(x)$ are quadratic polynomials $$\begin{cases} g_1(x)=cx^2+a_1x+a_0,\quad x\in[0,\frac{1}{2})\\ g_2(x)=cx^2+b_1x+b_0,\quad x\in[\frac{1}{2},1] \end{cases}$$ Next, Let $$g_1(x)f'(x)\bigg|_0^{\frac{1}{2}}+g_2(x)f'(x)\bigg|_{\frac{1}{2}}^1=0$$ we get, $$g(1)=g(0)=0\Rightarrow a_0=0,c+b_1+b_0=0 \tag{1}$$ and $$g_1\left(\frac{1}{2}\right)=g_2\left(\frac{1}{2}\right)\Rightarrow \frac{1}{2}a_1=\frac{1}{2}b_1+b_0\tag{2}$$ Next, Let $$g_2'(x)f(x)\bigg|_{\frac{1}{2}}^1+g_1'(x)f(x)\bigg|_0^{\frac{1}{2}}=g'(1)f(1)+\left(g_1'\left(\frac{1}{2}\right)-g_2'\left(\frac{1}{2}\right)\right)f\left(\frac{1}{2}\right)-g'(0)f(0)=0$$ Since, $f(0)+2f(\frac{1}{2})+f(1)=0$, we get $$g_1'\left(\frac{1}{2}\right)-g_2'\left(\frac{1}{2}\right)=-2g'(0)=2g'(1)$$ so $$a_1-b_1=-2a_1=4c+2b_1 \tag{3}$$ from $(1),(2),(3)$, we get $$a_0=0,\quad a_1=-\frac{1}{2}c,\quad b_1=-\frac{3}{2}c,\quad b_0=\frac{1}{2}c$$ Therefore, $$\int_0^1g(x)f''(x)\mathrm{d}x =\int_0^1g''(x)f(x)\mathrm{d}x=2c\int_0^1f(x)\mathrm{d}x$$ by Cauchy-Schwarz inequality, we get $$\left(2c\int_0^1f(x)\mathrm{d}x\right)^2\leqslant \int_0^1(g(x))^2\mathrm{d}x\int_0^1(f''(x))^2\mathrm{d}x$$ Among, $$\int_0^1(g(x))^2\mathrm{d}x=\int_0^{\frac{1}{2}}(g(x))^2\mathrm{d}x+\int_{\frac{1}{2}}^1(g(x))^2\mathrm{d}x=\frac{c^2}{480}$$ Finally, we get $$\int_{0}^{1}(f''(x))^2\mathrm{d}x\geqslant 1920\left(\int_{0}^{1}f(x)\mathrm{d}x\right)^2. \tag*{$\Box$}$$