complex analysis - Fourier transform of meromorphic function

Suppose that I have a function $f(z)$ which is meromorphic on the entire complex plane, meaning holomorphic everywhere except for a discrete set of poles.

I then take a vertical slice of this function, which we can denote $f(s+it)$ (for constant $s$), and then take the Fourier transform of it, denoted $\mathcal{F}\{f(s+it)\}(\omega)$. Let's assume the function decays quickly enough on this slice for the Fourier transform to be well-defined without distributions having to get involved.

My two questions:

1. Does the result always also admit a meromorphic continuation to the entire complex plane? (I believe yes, this is the Laplace transform)


2. If so, does the meromorphic continuation depend on which slice we choose? (Much less sure about this)

Answer

[Note: I use the unitary FT $\mathcal{F}u(\xi) = \sqrt{2\pi}^{-n/2} \int_{\mathbb{R}^n} u(x)e^{-ix\xi} dx$. Also my answer will be 90 degrees rotated compared to your question and consider horizontal slices, not vertical ones, just because it doesn't introduce more $i$'s and $(-1)$'s than absolutely necessary.]

The answer to your first question is "no". Wlog you can assume that the slice you're asking about is the slice at zero. An easy counterexample would be $f(z) = \frac{1}{z}$. Its Fourier transform is the sign function which does not extend meromorphically to the complex plane because it is locally constant.

If you don't want to take a slice through one of the poles, take $f(z) = \frac{1}{1+z^2}$. Its FT is $\sqrt{\tfrac{\pi}{2}} e^{-|x|}$ which does not extend because of the non-differentialibity at zero.

It is not even true if you remove all poles and consider only entire functions. The Fourier transform of any $C^\infty$-function with compact support $\phi$ will give you a counterexample. By the Paley-Wiener theorem $\mathcal{F}\phi$ extends to an entire function $F$ that decays rapidly along any horizontal slice, namely $$F(\zeta) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} f(x) e^{-ix\zeta}dx$$
But now the FT of all slices $F_b := a\mapsto F(a+ib)$ are functions with compact support so that none of them extend to a holomorphic function.

And this behaviour is to be expected! Always remember that the Fourier transform exchanges decay at infinity with smoothness. You asked about a very smooth functions (analytic) with very modest decay properties (just enough that the FT exists). Therefore its Fourier transform will decay rapidly towards infinity, but will have modest smoothness properties. In particular it would be way to much to ask for analyticity.

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Now for your second question: Even if you are in the nice case, the functions you get from Fourier transforming two different slices are not equal to each other. And again, the Paley-Wiener theorem gives you the reason why: With the notations above $F_b$ is the Fourier transform of $x\mapsto e^{bx} f(x)$ so that $\mathcal{F}^{-1}(F_b) = e^{-bx}f(-x)$. So even if both $\mathcal{F}(F_{b_1})$ and $\mathcal{F}(F_{b_2})$ extend to meromorphic functions, those extensions will differ by a factor of $e^{(b_1-b_2)x}$.

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However: There is still hope! The Paley-Wiener theorem generalises very well and gives a very good characterisation of holomorphic functions whose slices are Fourier transforms of something. The gist of it is that the FT exchanges exponential decay in certain directions for analytic continuations in "dual directions" just like it exchanges polynomial decay for real differentiability.

This works as follows:

Start with a (generalised) function $f: \mathbb{R}^n\to\mathbb{C}$ and define its points of exponential decay:
$$I_f := \{b\in\mathbb{R}^n \mid e^{\langle b,x\rangle}f(x) \in \mathcal{S}'(\mathbb{R}^n) \}$$
Some comments:

• This set might be empty (think of $f(x)=e^{\|x\|^2}$).


• $f$ has a Fourier transform iff $0\in I_f$.


• $I_f$ is always convex. In the one-dimensional case this means that $I_f$ is an interval (possibly empty or degenerate).


• If $b\in I_f$ and $f$ has support contained in a closed, convex $K\subseteq\mathbb{R}^n$, then $b+K^\vee \subseteq I_f$, where $K^\vee := \{b \mid \exists C\in\mathbb{R} \forall x\in K: \langle b,x\rangle \leq C\}$ is the dual of $K$. In particular: If $f$ has compact support, then $I_f=\mathbb{R}^n$; If $f$ has support in $[0,\infty)$ and $I_f\neq\emptyset$, then $I_f = (-\infty,b]$ or $I_f=(-\infty,b)$ for some $b\in\mathbb{R}\cup\{+\infty\}$.

Now consider the set $\Omega_f := \mathbb{R}^n + i I_f \subseteq\mathbb{C}^n$. In the one-dimensional case think of a horizontal "strip" or a half-plane.

The Fourier-Laplace-transform of $f$ is the function $F:\Omega_f\to\mathbb{C}, a+ib \mapsto \mathcal{F}(e^{bx}f(x))(a)$.

One can prove:

1. $F$ is holomorphic in the interior of $\Omega_f$. In particular: This really is a function, even if $f$ is a proper distribution. (Also the expression $\mathcal{F}(e^{bx}f(x))(a)$ really makes sense: The Fourier transform is an analytic function and thus can be evaluated at $a$.)


2. $F$ grows at most polynomially along horizontal stripes and the degree of the polynomials does not vary too wildly between stripes. More precisely: There are locally bounded functions $C,N: I_f\to \mathbb{R}$ such that
   $$|F(a+ib)| \leq C(b) (1+|a|)^{N(b)}$$


3. Conversely: If $\Omega=\mathbb{R}^n+ i I$ for some open convex subset $I\subseteq\mathbb{R}^n$ and $F: \Omega\to\mathbb{C}$ is a holomorphic function with the same kind growth restriction, then there is a unique distribution $f$ such that $I\subseteq I_f$ and $F$ is the Fourier-Laplace-transform of $f$.

There is a variety of different growth conditions that could be imposed instead and they will correspond to better behaved $f$. For example: $f\in L^2$ with support in $[0,\infty)$ corresponds to $F$ being holomorphic on the lower half-plane with the growth condition $\exists C\forall b<0: \int_\mathbb{R} |F(a+bi)|^2 da < C$.

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Now for the meromorphic case: Assume that the slices with poles on them are not dense, i.e. $\{b \mid \exists a: \text{pole at } a+bi\}\subseteq\mathbb{R}$ is discrete.

For simplicity we then restrict our attention to $b=0$ so that we can split $F$ into a upper and lower half: $F_\pm := F_{|\mathbb{R}\pm i(0,\varepsilon)}$. Furthermore let $a_n$ be the poles of $F$ and $p_n= \sum_{k=0}^{d_n} c_{nk} (z-a_n)^{-(k+1)}$ be the principal parts of $F$ at $a_n$.

If $F$ is nice enough at infinity, then we can apply the Paley-Wiener theorem from above to $F_+$ and $F_-$ and obtain two distributions $u_{\pm}$ such that $F_b = \mathcal{F}(e^{bx} u_{sgn(b)}(x))$ for all $b\neq 0$. We think of $u_\pm$ as $u_\pm = \lim\limits_{b\to 0, \pm b >0} \mathcal{F}^{-1}(F_b)$, although strictly speaking this limit might not exist, at least not as a distribution, although it is a hyperfunction. (I am not sure if stronger growth restrictions on $F$ make this limit exist in $\mathcal{D}'$, but I think so)

One can show that $F_{+0}$ and $F_{-0}$ differ exactly by a distribution localised at $\{a_n \mid n\}$, namely:

$$F_{+0} - F_{-0} = 2\pi i \sum_{n} \sum_{k=1}^{d_n} \frac{c_{nk}(-1)^k}{k!} \delta^{(k)}(x-a_n)$$
so that
$$u_+ - u_- = 2\pi i\sum_n \sum_{k=1}^{d_n} \frac{c_{nk}}{k!} (-ix)^k e^{ia_n x}$$

Of course there is nothing special about the zero slice: Everytime you cross a slice with a pole on it, you will get an additional summand of this form.