2

Prove $\lim_{n \to \infty} \int_{-\pi}^{\pi} \sin(nx) f(x) \, dx = 0$, with $f: [-\pi, \pi] \to \mathbb{R}$ differentiable.

Note: There are proofs of stronger statements on this site and elsewhere. My goal is to verify my (simple) proof, and to get answers to the two questions in discussion below.

Proof: Let $v(x) = \frac{-\cos nx}{n}, v'(x) = \sin nx$. Through integration by parts, we have $$ \int_{-\pi}^{\pi} \sin(nx) f(x) \, dx = f(\pi)v(\pi) - f(-\pi)v(-\pi) + \int_{-\pi}^{\pi} \frac{\cos nx}{n} f'(x) \, dx. $$

Since $f(\pi), f(-\pi), f'(\pi), f'(-\pi)$ are each constant, and $\cos$ is bounded, all the terms go to zero as $n \to \infty$.

Discussion:

  1. The original question specified $f$ as continuously differentiable. However, unless I'm in error, that is not required.
  2. Intuitively, this reminds of Fourier analysis (which I've yet to study), and says If you beat any nice periodic function with a high enough frequency wave, you'll get a constant zero. Alternatively: As frequency rises, the corresponding component of a nice periodic function will eventually go to zero. Is my intuition correct? Can you clarify?
SRobertJames
  • 4,278
  • 1
  • 11
  • 27
  • For $2)$, yes. The Riemann-Lebesgue lemma is saying that the amplitudes of the $n$th frequency tone vanishes as $n\to\infty$ – FShrike Oct 23 '22 at 20:29
  • 1
    For 1): the Riemann–Lebesgue lemma even holds if this $f$ is only integrable on $[-\pi,\pi].$ But for your integration by parts, I don't think $f$ differentiable is sufficient: what would your last integral mean? For 2): another interpretation is: the Fourier coefficients of a periodic (integrable) function form a null sequence. – Anne Bauval Oct 23 '22 at 20:34
  • @AnneBauval My understanding is that any derivative of a function has at most countable discontinuities, and that we can take the integral of any function with countable discontinuities. – SRobertJames Oct 23 '22 at 20:40
  • @AnneBauval Why can't we take integral of function with countable discontinuities? – SRobertJames Oct 23 '22 at 21:41
  • We can, but $\cos(nx)f'(x)$ is not such a function. It can be very discontinuous: see How discontinuous can a derivative be?. I wonder to which function you are thinking when talking about "countable discontinuities", since $f$ has none. – Anne Bauval Oct 23 '22 at 22:06
  • The proof holds only under the assumption that $f'$ is integrable (so not too unbounded if you want) and that is generally not true as $f'$ can be non integrable (see $x^2\sin 1/x^2$ on $[-\pi, \pi]$) the usual approximation proof is needed – Conrad Oct 23 '22 at 23:34
  • 1
    A derivative is not necessarily integrable (Riemann sense). For example it may be unbounded. But there are bounded derivatives too which are not integrable. See https://en.wikipedia.org/wiki/Volterra%27s_function – Paramanand Singh Oct 24 '22 at 02:15

0 Answers0