que $f\in C^1[0,2]$ y el tal $\int_{0}^{2}f(x)dx=0,f(0)=f(2)$,
Mostrar que %#% $ #%
Creo que debemos utilizar la desigualdad de $$\int{0}^{2}f^2(x)dx\le\int{0}^{2}f'^2(x)dx$
mi idea: he veo este % que $Cauchy$y $f(x)\in C^1([a,b],R)$, muestran que: $f(a)=f(b)=0$ $ pf: $$\displaystyle\int{a}^{b}f^2(x)dx\le\dfrac{(b-a)^2}{8}\displaystyle\int{a}^{b}[f'(x)]^2dx$ $ después $ de $$|f(x)|=|f(x)-f(a)|\le\sqrt{x-a}\left(\displaystyle\int{a}^{x}[f'(t)]^2dt\right)^{\frac{1}{2}}$ para que podamos $$f^2(x)\le(x-a)\displaystyle\int{a}^{x}[f'(t)]^2dt\le(x-a)\displaystyle\int{a}^{b}[f'(t)]^2dt$ $a$ que tenemos; $b$ $ entonces utilizamos $$\displaystyle\int{a}^{b}f^2(x)dx\le\displaystyle\int{a}^{b}\left[(x-a)\displaystyle\int{a}^{b}[f'(t)]^2dt\right]dx=\dfrac{(b-a)^2}{2}\displaystyle\int{a}^{b}[f'(x)]^2dx$ $\dfrac{a+b}{2}$, entonces %#% $ de #% otra parte, para cualquier $b$, que $$\displaystyle\int{a}^{\frac{a+b}{2}}f^2(x)dx\le\dfrac{(b-a)^2}{8}\displaystyle\int{a}^{\frac{a+b}{2}}[f'(x)]^2dx$ $ podemos $x\in[\frac{a+b}{2},b],f(x)=-\displaystyle\int{x}^{b}f'(t)dt$ $$f^2(x)=\left(\displaystyle\int{x}^{b}f'(x)dx\right)^2\le(b-x)\displaystyle\int{x}^{b}[f'(t)]^2dt$ tiene:\begin{align} &\displaystyle\int{\frac{a+b}{2}}^{b}f^2(x)dx\le\displaystyle\int{\frac{a+b}{2}}^{b}(b-x)\left(\displaystyle\int{a}^{b}[f'(t)]^2dt\right)dx\le \displaystyle\int{\frac{a+b}{2}}^{b}(b-x)dx\left(\displaystyle\int{\frac{a+b}{2}}^{b}[f'(t)]^2dt\right)dx\ &=\left(\displaystyle\int{\frac{a+b}{2}}^{b}(b-x)dx\right)\left(\displaystyle\int{\frac{a+b}{2}}^{b}[f'(x)]^2dx\right)\ &=\dfrac{(b-a)^2}{8}\displaystyle\int{\frac{a+b}{2}}^{b}[f'(x)]^2dx \end {Alinee el} entonces $\dfrac{a+b}{2}$ $
que $b$, entonces tenemos %#% $ #%