# Integral inequality :$\int_0^1(f'(x))^2dx\geq 32\int_0^1(f(x))^2dx + 16\left(\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right)^2$

Assume $f:[0,1]\to \mathbb{R}$ is differentiable and $f'$ is integrable. Given $f\left(\frac{1}{4}\right)=f\left(\frac{3}{4}\right)=f(1)-f(0)=0$, then prove that $$\int_0^1(f'(x))^2dx\geq 32\int_0^1(f(x))^2dx + 16\left(\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right)^2$$ My attempt: I tried to make the most out of the given information. Consider the integrals $$I_1=\int_0^{\frac{1}{4}}\left(x-\frac{1}{8}\right)f'(x)dx=\frac{f(0)}{8} - \int_0^{\frac{1}{4}}f(x)dx$$ $$I_2=\int_{\frac{1}{4}}^{\frac{1}{2}}\left(x-\frac{3}{8}\right)f'(x)dx=\frac{f\left(\frac{1}{2}\right)}{8} - \int_{\frac{1}{4}}^{\frac{1}{2}}f(x)dx$$ $$I_3=\int_{\frac{1}{2}}^{\frac{3}{4}}\left(x-\frac{5}{8}\right)f'(x)dx=\frac{f\left(\frac{1}{2}\right)}{8} - \int_{\frac{1}{2}}^{\frac{3}{4}}f(x)dx$$ $$I_4=\int_{\frac{3}{4}}^{1}\left(x-\frac{7}{8}\right)f'(x)dx=\frac{f\left(1\right)}{8} - \int_{\frac{3}{4}}^{1}f(x)dx$$ Notice that $$I_1+I_2-I_3-I_4=-\left(\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right)$$ Hence $$\left|\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right|^2\leq \left(I_1+I_2-I_3-I_4\right)^2\leq 4\left(|I_1|^2+|I_2|^2+|I_3|^2+|I_4|^2\right)$$ Using C-S inequality, we get $$\left|I_k\right|^2\leq \frac{1}{2^8\times 3}\int_{\frac{k}{4}}^{\frac{k+1}{4}}(f'(x))^2dx ,\quad k\in \{0,1,2,3\}$$ Hence $$\left(\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right)^2\leq \frac{4}{2^8\times 3}\int_0^1(f'(x))^2dx$$ Upto this seems fine , but i am having trouble estimating the second term i.e $\int_0^1(f(x))^2dx$. I tried considering integrals like $\int_0^{\frac{1}{4}}xf'(x)f(x)dx,\quad \int_{\frac{1}{4}}^{\frac{1}{2}}\left(x-\frac{1}{2}\right)f'(x)f(x)dx$ etc, but they are not giving satisfactory results. Please see if i can improve this or is there some other approach.

## 2 Answers

My favourite technique for such kind of problems is to project them over $\ell^2$.

Let us see how that works in this case.

Let we define four ausiliary functions over $[0,1]$: $$\left\{\begin{array}{rcl} g_1(x) &=& f\left(\frac{x}{4}\right) \\ g_2(x) &=& f\left(\frac{x+1}{4}\right) \\ g_3(x) &=& f\left(\frac{3-x}{4}\right) \\ g_4(x) &=& f\left(\frac{4-x}{4}\right) \end{array}\right.\tag{1}$$ together with their Fourier series: $$g_k(x) = a_0^{(k)}+\sum_{n\geq 1} c_n^{(k)}\cos(2\pi n x)+\sum_{n\geq 1} s_n^{(k)}\sin(2\pi n x). \tag{2}$$ The initial constraints become $g_1(0)=g_4(0)$ and $g_2(0)=g_3(0)=0$; moreover: $$\int_{0}^{1/2}f(x)\,dx -\int_{1/2}^{1}f(x)\,dx = \frac{1}{4}\sum_{k=1}^{4}a_0^{(k)}, \tag{3}$$ while by Parseval's identity: $$\int_{0}^{1}f(x)^2\,dx = \sum_{k=1}^{4}\left[\frac{{a_0^{(k)}}^2}{4}+\frac{1}{8}\sum_{n\geq 1}\left({c_n^{(k)}}^2+{s_n^{(k)}}^2\right)\right]\tag{4}$$ as well as: $$\int_{0}^{1}f'(x)^2\,dx = \frac{\pi^2}{2}\sum_{k=1}^{4}\sum_{n\geq 1}n^2\left({c_n^{(k)}}^2+{s_n^{(k)}}^2\right).\tag{5}$$ If now we translate the initial constraints as (the pointwise convergence of the Fourier series of $g_k$ is granted by $f'\in L^2$): $$a_0^{(k)}+\sum_{n\geq 1}c_n^{(k)}=0\qquad (k=2,3)$$ $$a_0^{(1)}+\sum_{n\geq 1}c_n^{(1)}=a_0^{(4)}+\sum_{n\geq 1}c_n^{(4)}\tag{6}$$ and write down the continuity conditions $g_1(1)=g_2(0),g_2(1)=g_3(1),g_3(0)=g_4(1)$ the initial problem is finally projected over $\ell^2$ and we just need to apply the Cauchy-Schwarz inequality to prove the given integral inequality. As already remarked by Marco Cantarini, this is just a particular version of Wirtinger's inequality and the given constants are not optimal, but the outlined approach gives the chance to find both the optimal constants and the functions fulfilling the equality in the optimal inequality.

Thanks for mentioning Wirtinger's inequality, i was not aware of it( have not studied fourier analysis yet). I will use two of its applications : $1)$ If $f(a)=f(b)=0$ then $$\frac{(b-a)^2}{\pi^2}\int_a^b(f'(x))^2dx\geq\int_0^a(f(x))^2dx$$ $2)$ If either one of $f(a)$ or $f(b)$ is zero then $$\frac{4(b-a)^2}{\pi^2}\int_a^b(f'(x))^2dx\geq\int_a^b(f(x))^2dx$$ Now using $(1)$ for the interval $\left[\frac{1}{4},\frac{3}{4}\right]$ we get $$\frac{1}{4\pi^2}\int_{\frac{1}{4}}^{\frac{3}{4}}(f'(x))^2dx\geq\int_{\frac{1}{4}}^{\frac{3}{4}}(f(x))^2dx$$ and using $(2)$ for intervals $\left[0,\frac{1}{4}\right]$ and $\left[\frac{3}{4},1\right]$ $$\frac{1}{4\pi^2}\int_{0}^{\frac{1}{4}}(f'(x))^2dx\geq\int_{0}^{\frac{1}{4}}(f(x))^2dx$$ $$\frac{1}{4\pi^2}\int_{\frac{3}{4}}^{1}(f'(x))^2dx\geq\int_{\frac{3}{4}}^{1}(f(x))^2dx$$ adding these three $$\frac{1}{4\pi^2}\int_{0}^{1}(f'(x))^2dx\geq\int_{0}^{1}(f(x))^2dx$$ and i have already established that $$\frac{4}{2^8\times 3}\int_0^1(f'(x))^2dx\geq\left(\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right)^2$$ Hence $$32\int_0^1(f(x))^2dx + 16\left(\int_0^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^1f(x)dx\right)^2\leq\left(\frac{1}{12}+\frac{8}{\pi^2}\right)\int_0^1(f'(x))^2dx$$ and $\left(\frac{1}{12}+\frac{8}{\pi^2}\right)<1$