I am reading a book "Fourier Series and Integrals" by Dym & McKean.

There is an exercise (Page 106):

Exercise: Check that if $f$ is a real, even, summable function and if $f(0+)$ and $f(0-)$ exist, then either $f(0-) =f(0+)$ or $\hat f(\gamma)$ changes sign infinitely often as $|\gamma| \to \infty$.

Note that $\hat f(\gamma)$ is a real function, so its "sign" makes sense!

There is a hint for the exercis as follow:

Hint: The function $f$ is summable if it is of one sign far out, as you can see from $$\frac{f(0-) +f(0 + )}{2} = \lim_{t \to 0} \, (P_t * f) (0) = \lim_{t \to 0} \int_{-\infty}^{\infty} \exp(-2 \pi^2 \gamma^2 t) \hat f(\gamma) \, \mathrm{d}\gamma.$$ Here $P_t=P_t(x)=\dfrac{\exp(-\dfrac{x^2}{2})}{\sqrt{2\pi t}}$ is the Gauss Kernel and $P_t * f$ means the convolution of $f$ with Gauss Kernel.

My try:

If f is of one sign far out, then by using $$\frac{f(0-) +f(0 + )}{2} = \lim_{t \to 0} \int_{-\infty}^{\infty} \exp(-2 \pi^2 \gamma^2 t) \hat f(\gamma) \, \mathrm{d}\gamma,$$ and Monotone convergence Theorem we deuce that $\hat f \in L^1(\mathbb{R})$ so $$f(-x)=\hat {\hat f}, $$ is continuous and consequently $f(x)$ will be continuous at $x=0$ and therefore $f(0-) =f(0+)$.

I don't know how to handle the other half.


  • 2
    $\begingroup$ please don't post on different sites at the same time, it's a waste of effort --- math.stackexchange.com/questions/2084541/… $\endgroup$ – Carlo Beenakker Jan 5 '17 at 13:48
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    $\begingroup$ Doesn't f being even already implies your claim? $\endgroup$ – Fan Zheng Jan 5 '17 at 14:45
  • $\begingroup$ The Gauß kernel is missing a $t$, it is $(2\pi t)^{-1/2}\exp(-x^2/2t)$. $\endgroup$ – Jochen Wengenroth Jun 10 at 13:39

The exercise is stated in Dym and McKean with a mistake. The correct statement is

If $f$ is real, even, the finite limits $f(x\pm 0)$ exist for all $x$, and $\hat{f}$ does not change sign for $|x|>A$, then $f$ is continuous at all points.

The statement is in the paper of M. Kac (1938) to which Dym and McKean refer. The proof is simple: inverse Fourier transform of positive function is continuous. If $\hat{f}$ is positive only for $|x|>A$, add to it the function $h(x)=B,\; |x|<A$ and $h(x)=0$ otherwise, to make it positive everywhere. The inverse Fourier transform of $h$ is explicitly computed and it is continuous.

Remark. The assumptions can be relaxed by saying that $f$ is real, even, integrable, and bounded in a neighborhood of $0$.

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