Context
I am reading a book that does not discuss the convergence of integrals for a certain class of functions, and I am not convinced that these integrals always converge, even if the authors proceed as though they do.
Define $\mathcal C_0^2\big( (0,1) \big)$ to be the space of continuous functions $u \colon [0,1] \to \mathbf R$ such that $u(0) = u(1) = 0$ and $u''$ exists and is continuous on $(0,1)$.
Question
Does the integral $$\int_0^1 u''(x)v(x) \, dx$$ always converge (as an improper Riemann integral) for all functions $u, v \in \mathcal C_0^2 \big( (0,1) \big)$?
Progress
I have proved the following:
If $a<b$ are real numbers, the function $f \colon (a,b] \to \mathbf R$ is Riemann integrable on $[c,b]$ for all $a < c < b$, the function $g \colon [a,b] \to \mathbf R$ is Riemann integrable, and if $\int_a^b f(x) \, dx$ converges absolutely, then $\int_a^b f(x)g(x) \, dx$ converges absolutely.
(And a similar result for the upper endpoint.)
In particular:
If $u \in \mathcal C_0^2 \big( (0,1) \big)$ and $\int_0^1 u''(x) \, dx$ converges absolutely, then $\int_0^1 u''(x)v(x) \, dx$ converges absolutely for every $v \in \mathcal C_0^2 \big( (0,1) \big)$,
since continuous functions on compact intervals are Riemann integrable.
Therefore, a necessary condition for $\int_0^1 u''(x)v(x) \, dx$ to diverge is that $\int_0^1 u''(x) \, dx$ diverges or converges conditionally. Does such a function exist?
Not a Counterexample
I managed to find a function $u \in \mathcal C_0^2 \big( (0,1) \big)$ with unbounded second derivative, but according to WolframAlpha, the integral of $u''$ converges absolutely. The function is defined by
$$u(x) = \frac{\pi}2x(x-1) - (x-1) \int_0^x \arcsin \big( (2t-1)^2 \big) \, dt, \quad x \in [0,1],$$
and thus
$$u''(x) = \pi - 2 \arcsin \big( (2x-1)^2 \big) - \frac{4(x-1)(2x-1)}{\sqrt{1 - (2x-1)^4}}, \quad x \in (0,1).$$
Try the function $u(x) = v(x) = \sqrt{x - x^2}$. Are you sure there wasn't an assumption about limits of $u'$ at $0$ and $1$?