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 $(0,1)$.


Does the integral $$\int_0^1 u''(x)v(x) \, dx$$ always converge (as an improper Riemann integral) for all functions $u, v \in \mathcal C_0^2 \big( (0,1) \big)$?


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

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

  • $\begingroup$ 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

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 $1$?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.