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.


Thanks.



Fourier transform either changes sign infinitely often far out or is continuous at $x=0$
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 $0$. 


