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I was trying the last few days to generalize the Paley-Wiener theorem in a quite obvious direction... or so I thought. The original Paley-Wiener theorem talks about functions and distributions with compact support and - thinking of the Laplace transform - I though this could be generalized to distributions supported in a pointed, closed, convex cone $K$ with non-empty interior (think positive octant in $\mathbb{R}^n$).

Given such a cone $K$ the set $K^\vee := \{y\in\mathbb{R}^n \mid \langle y,\cdot\rangle \text{bounded from above on}\, K\}$ is a pointed, closed, convex cone with non-empty interior. I then proved that for any $s\in\mathbb{C}^n$ with $\Im(s)\in int(K^\vee)$ the function $x\mapsto e^{-i\langle s,x\rangle}$ is Schwartz on $K$ in the sense that $x^\alpha \partial^\beta e^{-i\langle s,x\rangle}$ is bounded on (an open neighbourhood of) $K$ for all $\alpha,\beta$. The set of seminorms $\{\|x^\alpha \partial^\beta f\|_\infty \mid \alpha,\beta\in\mathbb{N}^n\}$ turns $\mathcal{S}(K)$ into a locally convex vector space.

This is therefore a function $\mathbb{R}^n+int(K^\vee)i \to \mathcal{S}(K)$ and my hope was that this is a holomorphic function so that composing with a distribution $u: \mathcal{S}(K)\to\mathbb{C}$ would be the holomorphic function $s\mapsto \int u(x) e^{-i\langle s,x\rangle} dx$ whose properties would then be the subject of a Paley-Wiener-type theorem.

After several days of calculating derivatives of various functions and failing to find the right estimates, I'm now starting to doubt whether this is even true. So my question is:

Is $\mathbb{R}^n+int(K^\vee)i \to \mathcal{S}(K), s\mapsto(x\mapsto e^{-i\langle s,x\rangle})$ holomorphic? And if yes, where can I find a proof?

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  • $\begingroup$ The Fourier/Laplace transform of a tempered distribution $u \in S'(\mathbb{R}^n)$ is subject to the Paley-Wiener theorem at least when you look at $f_{a,s_0}(z) = \langle u, e^{-i \langle s_0+z a,x \rangle}\rangle$ where $z \in \mathbb{C}$ and $a,s_0 \in \mathbb{R}^n$ $\endgroup$ – reuns Oct 18 '16 at 18:19
  • $\begingroup$ And if $\text{support}(u) \subset K$ then $f_{a,s_0}(z)$ is well-defined (and analytic in $z$) for $a,s_0,z$ in some correctly chosen region $\endgroup$ – reuns Oct 18 '16 at 18:22
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$E:=x\mapsto(s\mapsto e^{-i\langle{s,x}\rangle})$ is indeed holomorphic.

Step 1: Define the support function $h_K(y) := \sup_{x\in K} \langle{x,y}\rangle$ for all $y\in K^\vee$. $h_K$ is convex and therefore continuous in $int(K^\vee)$.

Step 2: For any $s$ with $\Im(s)\in int(K^\vee)$ the bound $$|x^\alpha \partial^\beta e^{i\langle{s,x}\rangle}| \leq |s^\beta| \alpha! \delta^{-|\alpha|} e^C \tag{1}$$ holds, where $\delta$ is so small that a $\delta$-neighbourhood of $\Im(s)$ is still in $int(K^\vee)$ and $h_K\leq C$ on that neighbourhood. This is a consequence of $|r|^k \leq k! e^{|r|}$ for all real numbers r.

Step 3: Using Taylor expansion we will prove that the obvious choice for the differential $d_s E=e^{-i\langle{s,x}\rangle}\langle{\cdot,x}\rangle$ works. We will show that $$\lVert{x^\alpha \partial^\beta (E(s+h)-E(s)-d_sE(h))}\rVert_{L^\infty(K)} \in O(h^2)$$ for $h\to 0$. First assume $\alpha=0$ for simplicity. Then $\partial^\beta (E(is+ih)-E(is)-d_s E(ih))$ simplifies to

$ ... = \partial^\beta (e^{\langle{s+h,x}\rangle}-e^{\langle{s,x}\rangle} - e^{\langle{s,x}\rangle}\langle{h,x}\rangle) \\ = (s+h)^\beta e^{\langle{s+h,x}\rangle} - s^\beta e^{\langle{s,x}\rangle} - \sum_{\beta_1+\beta_2=\beta} s^{\beta_1} e^{\langle{s,x}\rangle} \underbrace{\partial^{\beta_2} \langle{h,x}\rangle}_{=0 \, \text{if}\,|{\beta_2}|\geq 2} \\ = (s+h)^\beta e^{\langle{s+h,x}\rangle} - s^\beta e^{\langle{s,x}\rangle} - \left(s^\beta e^{\langle{s,x}\rangle} \langle{h,x}\rangle + \sum_{j=1}^n \beta_j s^{\beta-e_j} e^{\langle{s,x}\rangle} h_j \right)$

Now we apply Taylor expansion to $f(s):=s\mapsto s^\beta e^{\langle{s,x}\rangle}$ (for every $x$ this is an analytic function!) and find that the expression equals

$ = \sum_{|{\gamma}|\geq 2} \frac{1}{\gamma!} \partial_s^\gamma f(s) h^\gamma \\ = \sum_\gamma \sum_{\substack{\gamma_1+\gamma_2=\gamma \\ \gamma_1\leq\beta}} \frac{1}{\gamma!} \binom{\gamma}{\gamma_1,\gamma_2} \frac{\beta!}{(\beta-\gamma_1)!} s^{\beta-\gamma_1} x^{\gamma_2} e^{\langle{s,x}\rangle} h^\gamma \\ = \sum_{\substack{|{\gamma_1}|+|{\gamma_2}|\geq 2 \\ \gamma_1\leq\beta}} \frac{\beta!}{\gamma_1!\gamma_2!(\beta-\gamma_1)!} s^{\beta-\gamma_1} x^{\gamma_2} e^{\langle{s,x}\rangle} h^{\gamma_1+\gamma_2} \\ = \sum_{\substack{\gamma_1+\gamma_2\geq 2 \\ \gamma_1\leq\beta}} \binom{\beta}{\gamma_1} \frac{1}{\gamma_2!} s^{\beta-\gamma_1} h^{\gamma_1} x^{\gamma_2} e^{\langle{s,x}\rangle} h^{\gamma_2} \\ = \sum_{\substack{\gamma_1+\gamma_2\geq 2}} \binom{\beta}{\gamma_1} s^{\beta-\gamma_1} h^{\gamma_1} \cdot \frac{1}{\gamma_2!} x^{\gamma_2} e^{\langle{s,x}\rangle} h^{\gamma_2}$

Multiplying by $x^\alpha$ we find \begin{equation} x^\alpha \partial^\beta(...) = \sum_{\substack{\gamma_1+\gamma_2\geq 2}} \binom{\beta}{\gamma_1} s^{\beta-\gamma_1} h^{\gamma_1} \cdot \frac{1}{\gamma_2!} x^{\gamma_2+\alpha} e^{\langle{s,x}\rangle} h^{\gamma_2} \tag{2} \end{equation} Notice that this is a power series in $h$ missing the lowest powers of $h$. The full power series would be $$\sum_{\gamma_1,\gamma_2} \binom{\beta}{\gamma_1} s^{\beta-\gamma_1} h^{\gamma_1} \cdot \frac{1}{\gamma_2!} x^{\gamma_2+\alpha} e^{\langle{s,x}\rangle} h^{\gamma_2} = \sum_{\gamma_2} (s+h)^\beta \cdot \frac{1}{\gamma_2!} x^{\gamma_2+\alpha} e^{\langle{s,x}\rangle} h^{\gamma_2}$$ which has positive convergence radius uniformly in $x$ as we will now prove: $$|{\frac{1}{\gamma_2!} x^{\gamma_2+\alpha} e^{\langle{s,x}\rangle}}| \overset{(1)}{\leq} \frac{1}{\gamma_2!} (\gamma_2+\alpha)! \delta^{-|{\gamma}_2| - |{\alpha}|} e^C = \frac{(\gamma_2+\alpha)!}{\gamma_2!} \delta^{-|{\gamma_2}|} \cdot \delta^{-|{\alpha}|} e^C$$ Now $\frac{(g+a)!}{g!}$ is a polynomial in $g$ of degree $a$. Therefore the power series converges whenever $\lVert{h}\rVert_\infty<\delta$ holds and this is independent of $x$.

In particular: The power series (2) is in $O(\lVert{{h}^2}\rVert)$ for $h\to 0$ uniformly in $x$. This shows that $E$ is indeed holomorphic as a map $\mathbb{R}^n+int(K^\vee)i \to \mathcal{S}(K)$.

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  • $\begingroup$ Very unclear to me. What did you want to show or check ? $\endgroup$ – reuns Oct 18 '16 at 17:14
  • $\begingroup$ Since the "what" is very clearly stated in the above question as well as the first and the last sentence of this answer, let me ask instead what exactly you find unclear about this proof? $\endgroup$ – Johannes Hahn Oct 18 '16 at 17:18
  • $\begingroup$ What do you want to show ? wiki/Paley–Wiener_theorem is very clear : we look at the analyticity of $f(\xi) = \int_{-\infty}^\infty F(x) e^{-2i \pi \xi x} dx$ when $f \in L^2$, and at the converse : if $f(\xi)$ is analytic and decreasing fast enough, then $F \in L^2([0,\infty))$ $\endgroup$ – reuns Oct 18 '16 at 17:29
  • $\begingroup$ Did you read the question or the wikipedia article? The classical Paley-(Schwartz-)Wiener theorem deals with compactly supported functions(distributions). It doesn't say anything about functions with support in some convex cone. The wikipedia article only mentions the one-dimensional case and at best implicitly hints at the more general form for cones (in fact that's one of the sources that gave me the idea of such a generalisation in the first place) and it definitely doesn't provide a proof of (an intermediate step of a proof of) the result which is what this question is about. $\endgroup$ – Johannes Hahn Oct 18 '16 at 18:03
  • $\begingroup$ Come on... I'm just asking you to say in 2 lines what you want to show !!!!! $\endgroup$ – reuns Oct 18 '16 at 18:04

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