$3\int_{0}^{1}(f'(x))^2dx \geq (2\int_{0}^{1}f(x)dx)^2 \impliedby 2\int_{0}^{\frac{1}{2}}f(x)\,\mathrm dx=\int_{\frac{1}{2}}^{1}f(x) \,\mathrm dx$

Question

Let $f : \mathbb{R} \to \mathbb{R} $ be a differentiable function. Suppose that $2\int_{0}^{\frac{1}{2}}f(x)\,\mathrm dx=\int_{\frac{1}{2}}^{1}f(x) \,\mathrm dx$

Show that $$3\int_{0}^{1}(f'(x))^2 \,\mathrm dx \geq (2\int_{0}^{1}f(x)\,\mathrm dx)^2$$

Answer 1

We argue about the function $u(t):=f'(t)$. One then has $$f(x)=c+\int_{1/2}^x u(t)\ dt$$ for some $c\in\Bbb R$. Compute $$\eqalign{\int_0^{1/2}f(x)\ dx&={c\over 2}-\int_0^{1/2} t u(t)\ dt={c\over 2}-a\cr \int_{1/2}^1 f(x)\ dx&={c\over 2}+\int_{1/2}^1(1- t) u(t)\ dt={c\over2}+b\cr}$$ with $$a:=\int_0^{1/2} t u(t)\ dt, \qquad b:=\int_{1/2}^1(1- t) u(t)\ dt\ .$$ The condition ${c\over 2}+b=2\bigl({c\over2}-a\bigr)$ enforces $c=2b+4a$, so that we obtain $$\int_0^1 f(x)\ dx=\biggl({c\over 2}-a\biggr)+\biggl({c\over2}+b\biggr)=3(a+b)\ ,$$ or $$\int_0^1 f(x)\ dx= 3\int_0^1 g(t)\>u(t)\ dt\ ,\quad{\rm with}\quad g(t):=\cases{t&$(0\leq t\leq{1\over2})$\cr 1-t\quad&$({1\over2}\leq t\leq1)$\cr}\ .$$ By Schwarz' inequality $$\int_0^1 u^2(t)\ dt\cdot\int_0^1 g^2(t)\ dt\geq \left(\int_0^1 g(t) u(t)\ dt\right)^2={1\over9}\left(\int_0^1 f(x)\ dx\right)^2\ .$$ Since $$\int_0^1 g^2(t)\ dt=2 \int_0^{1/2} t^2\ dt={1\over12}$$ the stated inequality follows.

Answer 2

Integration by parts gives $$ \int_0^{1/2}tf'(t)\,\mathrm{d}t=\frac12f\left(\frac12\right)-\int_0^{1/2}f(t)\,\mathrm{d}t\tag{1} $$ and $$ \int_{1/2}^1(1-t)f'(t)\,\mathrm{d}t=-\frac12f\left(\frac12\right)+\int_{1/2}^1f(t)\,\mathrm{d}t\tag{2} $$ Adding $(1)$ and $(2)$ and applying the hypothesis yields $$ \begin{align} \int_{1/2}^1(1-t)f'(t)\,\mathrm{d}t+\int_0^{1/2}tf'(t)\,\mathrm{d}t &=\int_{1/2}^1f(t)\,\mathrm{d}t-\int_0^{1/2}f(t)\,\mathrm{d}t\\ &=\frac23\int_0^1f(t)\,\mathrm{d}t-\frac13\int_0^1f(t)\,\mathrm{d}t\\ &=\frac13\int_0^1f(t)\,\mathrm{d}t\tag{3} \end{align} $$ $(3)$ and Cauchy-Schwarz say that $$ \begin{align} \frac13\left|\,\int_0^1f(t)\,\mathrm{d}t\,\right| &\le\left(\int_0^{1/2}t^2\,\mathrm{d}t+\int_{1/2}^1(1-t)^2\,\mathrm{d}t\right)^{1/2}\left(\int_0^1f'(t)^2\,\mathrm{d}t\right)^{1/2}\\ &=\frac1{\sqrt{12}}\left(\int_0^1f'(t)^2\,\mathrm{d}t\right)^{1/2}\\ 4\left(\int_0^1f(t)\,\mathrm{d}t\right)^2&\le3\int_0^1f'(t)^2\,\mathrm{d}t\tag{4} \end{align} $$

Answer 3

by integrating by parts, $$\int_0^{\frac{1}{2}}g_1(x)f'(x)\,\mathrm dx=g_1(x)f(x)\bigg|_0^{\frac{1}{2}}-\int_0^{\frac{1}{2}}g_1'(x)f(x)\,\mathrm dx \tag{1}$$ $$\int_{\frac{1}{2}}^1g_2(x)f'(x)\,\mathrm dx=g_2(x)f(x)\bigg|_{\frac{1}{2}}^1-\int_{\frac{1}{2}}^1g_2'(x)f(x)\,\mathrm dx \tag{2}$$ Since $2\int_{0}^{\frac{1}{2}}f(x)\,\mathrm dx=\int_{\frac{1}{2}}^{1}f(x) \,\mathrm dx$, we let $g_1'(x)=a_1\neq0$ and $g_2'(x)=b_1\neq0$. $$g(x)=\begin{cases} g_1(x)=a_1x+a_0,x\in[0,\tfrac{1}{2})\\ g_2(x)=b_1x+b_0,x\in(\tfrac{1}{2},1] \end{cases}$$ Let $g_1(x)f(x)\bigg|_0^{\frac{1}{2}}+g_2(x)f(x)\bigg|_{\frac{1}{2}}^1=0$, we get $$\begin{cases} g_1(0)=g_2(1)=0\\ g_1(\tfrac{1}{2})=g_2(\tfrac{1}{2}) \end{cases} \Rightarrow\quad \begin{cases} a_0=b_1+b_0=0\\ \tfrac{1}{2}a_1+a_0=\tfrac{1}{2}b_1+b_0 \end{cases}$$ Hence, we get $a_0=0,a_1=a_1,b_0=-a_1,b_1=a_1$, $$g(x)=\begin{cases} g_1(x)=a_1x,x\in[0,\tfrac{1}{2})\\ g_2(x)=-a_1x+a_1,x\in(\tfrac{1}{2},1] \end{cases}$$ from $(1)+(2)$, we have \begin{align*} \int_0^1g(x)f'(x)\,\mathrm dx&=-\int_0^{\frac{1}{2}}g_1'(x)f(x)\,\mathrm dx-\int_{\frac{1}{2}}^1g_2'(x)f(x)\,\mathrm dx\\ &=-\frac{a}{3}\int_0^1f(x)\,\mathrm dx+\frac{2a}{3}\int_0^1f(x)\,\mathrm dx=\frac{a}{3}\int_0^1f(x)\,\mathrm dx \end{align*} by Cauchy-Schwarz inequality, we get $$\left(\frac{a}{3}\int_0^1f(x)\,\mathrm dx\right)^2\leqslant \int_0^1g^2(x)\,\mathrm dx\int_0^1(f'(x))^2\,\mathrm dx$$ and $$\int_0^1g^2(x)\,\mathrm dx=\frac{a_1^2}{24}+\frac{a_1^2}{24}=\frac{a_1^2}{12}$$ Therefore, $$\left(\int_0^1f(x)\,\mathrm dx\right)^2\leqslant \frac{3}{4}\int_0^1(f'(x))^2\,\mathrm dx$$