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

Question

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.

Screenshot of the exercise

Answer 1

As Daniel Fischer noted, in the present form the exercise is trivial: if $f$ is even and one-sided limits at $0$ exist, they are automatically equal. I think what the authors really mean is this:

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

Having a continuous representative means there is a continuous function $g$ such that $f=g$ almost everywhere.

The proof then goes as hinted in the book. Since $f$ is even and real-valued, $\hat f$ is also even and real-valued. We suppose $\hat f$ has only finitely many changes of sign; so the goal is to prove that $f$ has a continuous representative.

Since $\hat f$ is continuous, the question of its integrability is one of behavior at infinity. At infinity, $\hat f$ is of constant sign (its evenness ensures it's the same sign at $-\infty$ as $+\infty$). The existence of the limit $$\lim_{t \to 0} \int_{-\infty}^{\infty} \exp(-2 \pi^2 \gamma^2 t) \hat f(\gamma) \, \mathrm{d}\gamma$$ then implies $\hat f\in L^1(\mathbb R)$. By the $L^1$ Fourier inversion, the inverse Fourier transform of $\hat f$ agrees with $f$ almost everywhere. Since this transform is continuous, the conclusion is proved.