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

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

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Cauchy-Schwarz. – Yimin Mar 1 '13 at 21:25
I tried, but it doesn't work. – Guillermo Mar 1 '13 at 22:08
So $f$ is differentiable...and $f'^2$ is integrable? – 1015 Mar 1 '13 at 22:51
Why LHS can be around $0$? I think that $f'$ blow up like delta function around $1/2$ – Guillermo Mar 1 '13 at 23:17
@Ryuichi I have seen my mistake, but I have some idea.Take $g(x)$ as any differentiable function, and select $c$ such that $f(x) = g(x) + c$, satisfying $$2\int_{0}^{1/2} f(x)dx = \int_{1/2}^1 f(x)dx$$ we can see $$c = \int_{1/2}^1g(x)dx -2\int_{0}^{1/2}g(x)dx = u-2v$$ if we take $$u = \int_{1/2}^1 g(x)dx$$ $$v = \int_0^{1/2} g(x)dx$$ $$\int_0^1 f(x)dx = \int_0^1 (g(x)+c)dx = u+v+u-2v = 2u-v$$ RHS = $$4(\int_{0}^1f(x)dx)^2 = 4(2u-v)^2$$ LHS = $$3\int_0^1 (f'(x))^2dx = 3\int_0^1 (g'(x))^2dx$$ which only has something to do with $g(x)$, without any restrictions. – Yimin Mar 2 '13 at 6:29

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.
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}