Equivalence of weak and strong form of ODE According to Wikipedia https://en.wikipedia.org/wiki/Finite_element_method there is a simple proof of the following (apparently using MVT for integrals).
We have an ODE:$$u''(x) = f(x)$$ with boundary conditions $u(0)=u(1)=0$. Then if the following relation holds
$$\int_{0}^{1}v(x)u''(x)dx=\int_{0}^{1}v(x)f(x)dx $$
for all smooth $v$, then $u$ solves the ODE, i.e. $u''(x)=f(x)$ on $(0,1).$
I am interested in the proof in the simplest case, i.e. the assumptions on the functions that make it easiest to prove. This is self study.
 A: If $u''(x) = f(x)$ pointwise, then
$$\int_{0}^{1}v(x)u''(x)dx=\int_{0}^{1}v(x)f(x)dx $$
follows immediately.
Conversely, if you know a function $u''$ which satisfies the integral constraint, which can be rewritten as
$$\int_{0}^{1}v(x)\left(u''(x)-f(x)\right)dx=0,$$
what can we say about it? Well, this condition implies that the function $u''(x)-f(x)$ is zero almost everywhere (meaning, it is zero on every set of positive measure). This statement is weaker than $u''(x) = f(x)$ pointwise (hence, the name, weak form).
However, if you add the requirement that $u''(x)$ and $f(x)$ be continuous, then $u''(x)-f(x)=0$ almost everywhere is equivalent to $u''(x)-f(x)=0$ pointwise.
A: You could use a proof by contradiction to show that $u''-f=0$ almost everywhere (with the requirement that $u''$ and $f$ are continuous).  Assume $u''-f$ is non-zero on a sub-interval $[c,d]$ of $[0,1]$.  The continuity of $u''$ and $f$ then implies that $u''-f$ has a single sign on $[c,d]$, with a minimum absolute value of $m_1\gt 0$ on $[c,d]$.  Let $v(x)$ be a function that is non-zero and has the same sign as $u''-f$ on $(c,d)$ but is zero everywhere else.  Also require that $|v(x)|\gt m_2$ for some positive number $m_2$ for $x\in[c+(d-c)/4, d-(d-c)/4]$. Then the definite integral of $v(u''-f)$ over $[0,1]$ must be greater than $m_1m_2(d-c)/2$ (since $v(u''-f) \gt m_1m_2$ on an interval with length $(d-c)/2$ and is non-negative everywhere else).  This contradicts the requirement that the integral is 0, so $u''-f$ must be 0 everywhere except possibly at isolated points.
