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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

$$u''(x) = \pi - 2 \arcsin \big( (2x-1)^2 \big) - \frac{4(x-1)(2x-1)}{\sqrt{1 - (2x-1)^4}}, \quad x \in (0,1).$$

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  • $\begingroup$ Is this a calculus of variations lemma? I recall seeing something like this in Gelfand and Fomin's book. I don't remember the proof though. $\endgroup$ – lordoftheshadows Feb 13 '17 at 2:09
  • $\begingroup$ It appears implicitly as part of a lemma in "Introduction to Partial Differential Equations" by A. Tveito and R. Winther. In connection with the one-dimensional Poisson equation, they set out to prove that the Laplace operator (in one dimension) satisfies a certain symmetry property involving integrals of the kind described above, but their convergence is not addressed. $\endgroup$ – Qeeko Feb 13 '17 at 2:35
  • $\begingroup$ That makes sense. The lemma I was thinking of had some stronger boundary conditions and a slightly different formulation. $\endgroup$ – lordoftheshadows Feb 13 '17 at 3:31
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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 $1$?

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  • $\begingroup$ A simple and effective counterexample - thanks! The lemma is formulated under no additional hypotheses on the functions, and is therefore wrong. However, in all problems of interest in the corresponding chapter (the one-dimensional Poisson equation), the function $u''$ has a continuous extension to the boundary, in which case the question of convergence is trivial. $\endgroup$ – Qeeko Feb 13 '17 at 3:25
  • $\begingroup$ $\sqrt x -x$ also works. $\endgroup$ – zhw. Feb 13 '17 at 3:37

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