Suppose that $\int _0^1 f(x)v(x)=0$ for every $v \in C^{\infty}([0,1])$ for which $v'(0)=v(1/2)=0$. Show that $f(x)=0$ for all $x\in [0,1]$.

Question

Suppose that $\int _0^1 f(x)v(x)=0$ for every $v \in C^{\infty}([0,1])$ for which $v'(0)=v(1/2)=0$. Show that $f(x)=0$ for all $x\in [0,1]$.Suggestion: take u to be the suitable cut off version of $\bar{f}$, i.e. $\bar{f}$ multiplied by an appropriate cutoff function. I really don't know how to do.

Answer 1

Given $f$ let $v=x^2(2x-1)^2f(x)$. Then $$ \begin{align} 0 &=\int_0^1f(x)v(x)\,\mathrm{d}x\\ &=\int_0^1x^2(2x-1)^2f(x)^2\,\mathrm{d}x\tag{1} \end{align} $$ Suppose that $f(x_0)=a\ne0$. Because $f$ is continuous, there is a $\delta$ so that for $|x-x_0|\le\delta$ $|f(x)|\ge\frac{a}{2}$.

For $\delta\le1/3$, outside of $S_\delta=[0,\delta/2]\cup[\frac12-\delta/4,\frac12+\delta/4]$, we have $x^2(2x-1)^2\ge\delta^2/25$. There must be pieces of $|x-x_0|\le\delta$ outside of $S_\delta$ of length at least $\delta/2$. Thus, $$ \int_0^1x^2(2x-1)^2f(x)^2\,\mathrm{d}x\ge\frac{\delta^2}{25}\frac{a^2}{4}\frac{\delta}2\tag{2} $$ which contradicts $(1)$.