# Showing that a Fourier transform is holomorphic

Let $f\in \mathcal{S}(\mathbb{R})$ be a Schwartz function. I would like to show that the Fourier function $$F(z)=\int_{\mathbb{R}}f(t)e^{itz}\, dt$$ is an entire function.

Here is my approach:

1. Write $z=x+iy$ and note that $itz=it(x+iy)=-ty+itx$. Thus, $$F(z)=\int_{\mathbb{R}}f(t)e^{-ty}e^{itx}\, dt$$ which shows that $F(z)$ exists for each $z$, because $f(t)e^{-ty}$ is also a Schwartz function and $F(z)$ is written as the (real) Fourier transform of a Schwartz function which is convergent.
2. Use Morera's theorem (together with Cauchy integral formula and Fubini's theorem) to conclude.

Of course, to complete the second step I need to check the continuity of $F$ which probably requires the dominated convergence theorem.

My questions is whether there is an alternative/neater way of proving that $F$ is entire. (For instance, can Morera's theorem be avoided by an application of the dominated convergence theorem or one of its cousins to differentiate under the integral sign?)

• $f(t) e^{-ty}$ is in general not a Schwartz function. Schwartz functions can have slower than exponential decay. Commented Jun 23, 2015 at 15:00
• @DanielFischer Thanks for the correction! I need to show that $f(t)e^{-ty}$ is $L^1$, right?
– EPS
Commented Jun 23, 2015 at 15:05
• That would suffice, but in general it isn't in $L^1$. Commented Jun 23, 2015 at 15:05
• People have pointed out why the obvious proof doesn't work. Someone should point out that the result is in fact easily seen to be false. For example: We know that if $f$ is a Schwarz function then $\hat f$ can be any Schwarz function, right? In particular $\hat f$ can be non-zero but with compact support. The restriction of a holomorphic function to the line cannot behave that way. Commented Jun 23, 2015 at 17:10
• You can certainly do it by differentiating under the integral sign. Some professor said to me a long time ago: MVT fails for complex-valued functions, but all the important consequences of MVT still hold. For example $|f'|\le M$ on $[a,b]$ still implies $|(f(b)-f(a))/(b-a)|\le M$. Commented Jun 24, 2015 at 4:42

For this answer, I will suppose that $$x \mapsto f(x)e^{a|x|}$$ is $$L^1$$ for every $$a \geq 0$$.

The theorem of holomorphy under the integral sign is :

Theorem :

Let $$(E,\mu)$$ be a measured space, $$U$$ an open set of $$\mathbb{C}$$ and $$f$$ a function from $$E \times U$$ to $$\mathbb{C}$$. We suppose that :

• $$x \mapsto f(x,z)$$ is measurable on $$E$$ for every $$z\in U$$
• $$z \mapsto f(x,z)$$ is holomorphic on $$U$$ for $$\mu$$-almost every $$x \in E$$
• there exists $$g \in L^1(E)$$ such that for every $$z\in U$$ and $$\mu$$-almost every $$x \in E$$, $$|f(x,z)|\leq|g(x)|$$

Then, $$F : z \mapsto \int_E f(x,z) \text{d}\mu(x)$$ is holomorphic on $$U$$ and for every $$k \in \mathbb{N}$$ : $$F^{(k)}(z)=\int_E \dfrac{\partial^k f}{\partial z^k}(x,z) \text{d}\mu(x)$$ and these integrals are well defined and finite.

To come back to the initial problem, let's consider $$E=\mathbb{R}$$ with the Lebesgue measure and $$f(x,z)=f(x)e^{izx}$$. Fix $$a>0$$ and define $$U_a=\lbrace z \in \mathbb{C} \; | \; -a<\text{Im}(z).

For every $$x \in \mathbb{R}$$ and $$z \in U_a$$, we have : $$|f(x,z)|=|f(x)e^{izx}|=|f(x)e^{-\text{Im}(z)x}|\leq |f(x)|e^{a|x|}$$

Thus, we can apply the theorem with the control function $$g_a(x)=f(x)e^{a|x|}$$, which is $$L^1$$ by hypothesis. Hence, $$F$$ is holomorphic on any open band $$U_a$$ and thus, is holomorphic on $$\mathbb{C}$$.

Edit : I'll add a little remark on the theorem. The great difference between the theorem of holomorphy under the integral and the one of differentiation under the integral is that here, we only have to find a control function $$g$$ on the initial function $$f$$, not on its derivative, to have the existence and the expression of $$F$$ and its derivatives (this is due to the Cauchy inequalities, which allows to control the derivatives of $$f$$ just by controling $$f$$).

• This is a nice theorem! Thanks for writing up your comment. Do you happen to have a reference for it? It seems to me that the classical differentiation under the integral sign theorem would also do the job in this case because $|\frac{\partial}{\partial z}f(x, z)|$ is also bounded by an $L^1$ function as you have argued above. Does that sound OK to you?
– EPS
Commented Jun 24, 2015 at 4:16
• Well yes, the classical theorem would work with the help of Cauchy inequalities, but doing that, you would actually just prove the holomorphic theorem (I guess in the end it's just a matter of what you want to admit and the details you want to give). I do have references on the subject, but unfortunately, there in french ; so maybe the best is that I let the english mathematicians answer this question. Commented Jun 24, 2015 at 8:12
• I don't see why we need the Cauchy inequalities. Checking that $|\frac{\partial}{\partial z}f(x, z)|$ is bounded by an $L^1$ function uniformly in $U_a$ (using your notation) we can apply the classical differentiation under the integral sign which proves that $F(z)$ is differentiable. That's all we wanted to show, isn't it? Can you clarify why Cauchy inequalities enter this argument, because what I've sketched doesn't seem to have any gaps? Thanks
– EPS
Commented Jun 24, 2015 at 15:11
• In your case indeed, you can do it without the use of Cauchy inequalities : the derivative simply multiply with $ix$ and it is not difficult to see that this is still $L^1$ with our hypothesis. Sorry if misdirect you on this point. What I wanted to say was that, for holomorphic function, you don't even have to ask yourself this kind of question, as the boundedness of derivatives will come from the one of $f$ with cauchy inequality. So in a sense, you're doing too much work. Yet, I understand that you might prefer to work with the classical theorem if you don't have reference on this one. Commented Jun 24, 2015 at 15:24
• Hi Sylvain, could you please share the title of the French reference you mentioned? Thank you so much! Commented Jul 12, 2020 at 5:34