Function that isn't the Fourier transform of periodic integrable function

Question

What's an example of a function $\phi : Z\rightarrow C$ such that $\phi(n) \rightarrow 0$ as $|n|\rightarrow\infty$, and $\phi$ is not the Fourier transform of some function $f\in L^1([-\pi,\pi])$?

Rudin proves in Real and Complex Analysis that such a $\phi$ exists, but the proof isn't constructive.

Answer 1

Following function is an example:

$$f(x)=\sum_{n=2}^\infty \frac{\sin nx}{\ln n}$$

For the proof, see here.

It is based on the fact that for a $2 \pi$-periodic function $g$, Lebesgue-integrable on $[0,2 \pi]$, the sum $$\sum_{n=1}^\infty \frac{c_n-c_{-n}}{n}$$ is convergent where $(c_n)_{n \in \mathbb Z}$ are the complex Fourier coefficients of $g$: $$c_n = \frac{1}{2 \pi} \int_0^{2 \pi} g(t)e^{-ikt} \ dt.$$

Answer 2

I'm pretty sure that $$\sum n^{-1/3}e^{i3^nt}$$gives an example.

The point being that lacunary series behave like independent random variables; the lacunary-series version of Khintchine's inequality shows that if a lacunary series is in $L^1$ it must be in $L^2$.

Ah, right. The point is this:

Lemma If $a_n$ is bounded then there exists a complex measure $\mu$ on $\Bbb T$ such that $\hat\mu(3^n)=a_n$ for $n=1,2,\dots$ and $||\mu||\le c\sup|a_n|$.

That's a standard application of "Riesz products". Suppose the lemma is true for a second, and let's see how the main result follows:

Theorem If $f=\sum a_ne^{i3^nt}$ is in $L^1$ then $f\in L^2$.

Proof: The lemma shows that for any sequence of plus and minus signs there exists $\mu$ so that if $g=\sum\pm a_ne^{i3^nt}$ then $g=f*\mu$. Hence $||g||_1\le c||f||_1$, and hence $||g||_1\sim||f||_1$, since we also have $f=g*\mu$.

So Fubini and the probabilistic Khintchine inequality show that $$||f||_1\sim\Bbb E||g||_1\sim\left(\sum|a_n|^2\right)^{1/2};$$in particular if $||f||_1$ is finite then $f\in L^2$.

Now the lemma follows from this:

Lemma' If $-1/2\le a_n\le 1/2$ there exists a probability measure $\mu$ on $\Bbb T$ with $\hat\mu(3^n)=a_n$, $n=1,2,\dots$.

Proof: Define $$\mu_N=\prod_{n=1}^N(1+2a_n\cos(3^nt)).$$

Note that $\mu_N\ge0$. Now, a given integer has at most one representation in the form $\sum\epsilon_j3^j$, where each $\epsilon_j$ is $-1,0,$ or $1$; it follows that $$\hat\mu_N(0)=1$$and $$\hat\mu_N(3^n)=a_n\quad(n=1,2,\dots,N).$$

In particular $||\mu_N||_1=1$; let $\mu$ be a weak-star accumulation point of $\mu_N$.