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

Question

Que $f : \mathbb{R} \to \mathbb{R} $ ser una función diferenciable. Supongamos que $2\int{0}^{\frac{1}{2}}f(x)\,\mathrm dx=\int{\frac{1}{2}}^{1}f(x) \,\mathrm dx$

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Answer 1

Argumentamos sobre la función $u(t):=f'(t)$. Entonces uno tiene $$f(x)=c+\int_{1/2}^x u(t)\ dt$ $ $c\in\Bbb R$. Calcular $$ \eqalign {\int_0^ {1/2} f (x) \ dx y = {c\over 2}-\int0^ {1/2} t u (t) \ dt = {c\over 2}-a\cr \int{1/2}^1 f (x) \ dx y = {c\over 2} + \int_ {1/2} ^ 1(1-t) u (t) \ dt = {c\over2} + b\cr} $$ con $$a:=\int0^{1/2} t u(t)\ dt, \qquad b:=\int{1/2}^1(1- t) u(t)\ dt\ .$$ The condition ${c\over 2} + b = 2\bigl ({c\over2}-a\bigr) $ enforces $c = 2b +4a$, por lo que obtenemos $$\int_0^1 f(x)\ dx=\biggl({c\over 2}-a\biggr)+\biggl({c\over2}+b\biggr)=3(a+b)\ ,$ $ o $$\int_0^1 f(x)\ dx= 3\int_0^1 g(t)>u(t)\ dt\ ,\quad{\rm with}\quad g(t):=\cases{t&$ (0\leq t\leq{1\over2})$\cr 1-t\quad&$ ({1\over2} \leq t\leq1)$\cr}\ .$$ por Schwarz' desigualdad $$\int_0^1 u^2(t)\ dt\cdot\int_0^1 g^2(t)\ dt\geq \left(\int_0^1 g(t) u(t)\ dt\right)^2={1\over9}\left(\int_0^1 f(x)\ dx\right)^2\ .$ $ desde $$\int_0^1 g^2(t)\ dt=2 \int_0^{1/2} t^2\ dt={1\over12}$ $ sigue la desigualdad indicada.

Answer 2

Integración por partes da $$ \int_0^{1/2} tf'(t) \,\mathrm {d} t = \frac12f\left (\frac12\right)-\int0^ {1/2} f (t) \,\mathrm {d} t\tag {1} $ y $$ \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} $$ Agregar $(1)$ y $(2)$ y aplicando la hipótesis rendimientos $$\begin{align} \int{1/2}^1(1-t)f'(t)\,\mathrm{d}t+\int0^{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{} $$ $(3)$ y Cauchy-Schwarz dicen que $$\begin{align} \frac13\left|\,\int_0^1f(t)\,\mathrm{d}t\,\right| &\le\left(\int0^{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 {alinee el} $$