Show that $f=0$

Question

>Let $f$ be continuous on $[a,b]$.

Assume that for every function $h\in C^2[a,b]$ with $h(a)=h(b)=0$ we have $\int _a^b h(x)f(x)dx=0$

Show that $f=0$

Assume that $f\neq 0\implies \exists c\in [a,b] $ such that $f(c)\neq 0$. Without loss of generality let $f(c)>0$. Since $f$ is continuous so $\exists \delta >0$ such that $f(x)>0\forall x\in (c-\delta,c+\delta$.

But I cant proceed after that. Can someone help.

Answer 1

You can say more than $f(x)>0$ on $(c-\delta,c+\delta)$, you can say there exists $\epsilon>0$ and $\delta>0$ such that $f(x) \geq \epsilon$ on $(c-\delta,c+\delta)$ (do you see why?).

Now find $h\in C^2[a,b]$ such that:

• $h=0$ outside of $(c-\delta/2,c+\delta/2)$
• $h \geq d $ for some constant $d$ on $(c-\delta/2,c+\delta/2)$
• $\int_a^b h(x) \ dx >0$.

Show that for such an $h$, $\int_a^b f(x)h(x) \ dx >0$.