Definite integral implies integrand is the same

Question

Let $H_0^1$ be the space of all functions $g:[0,1]\to \mathbb R$ such that $g'$ is in $L^2[0,1]$ and $g(0)=g(1)=0$, given that $H_0^1$ is a Hilbert space. Let $f\in C[0,1]$. Given that $u\in C^{(2)}[0,1] \cap H_0^1,$ and satisfies for all $v\in H^1_0$ that $$ \int_0^1 u'v'dx = \int_0^1 fvdx. $$

Need to show $-u'' = f.$

My try: Using integration by parts and since $v\in H_0^1$, we get $$ \int_0^1 v(f+u'') dx = 0.$$ Taking $v=f+u'',$ we can conclude $-u''=f.$ So, all we need to show is $f+u'' \in H_0^1.$ I am not able to show this with the given information. Any hints or help is appreciated!

Answer 1

Take $\{v_n(x)=1-e^{2\pi inx}:n\in \mathbb Z\}$. Then $v_n\in H^1_0,$ and $\int^1_0(f+u'')dx=\int^1_0e^{2\pi inx}(f+u'')dx=c_n$, the $n^{\text{th}}$ Fourier coefficient of $f+u''.$ But $c_n\to 0$ by the Riemann-Lebesgue Lemma so in fact $c_n=\int^1_0(f+u'')dx=0$ for all $n\in \mathbb Z$ so $f+u''=0$ almost everywhere. But as $f+u''$ is continuous, it is actually zero everywhere.

Answer 2

Rather than showing $f + u'' \in H^1_0$, let's just note that $f + u''$ is certainly in $L^2$. And since $H^1_0$ is dense in $L^2$, having $\int v (f+u'') = 0$ for all $v \in H^1_0$ implies $f+u'' = 0$ almost everywhere. Finally, as Matematleta noted, since $f, u''$ are continuous, $f+u''$ is actually zero everywhere.