If a function $f$ is analytic in the strip $\mathcal{D}_d = \left\{ z \in \mathbb{C} : |\Im(z)| < d \right\}$, how to show that the Hilbert transform of $f$, which is $\mathcal{H}f(x) = p.v. \int_{\mathbb{R}} \frac{f(t)}{x-t}dt$ ($p.v.$ means Cauchy principal value), is also analytic in this strip?
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5What is the Hilbert transform in this setting? – zhw. Apr 18 '17 at 15:49
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@Yikai I'm a green hand. Thank you! – Claire Fan Apr 21 '17 at 01:50
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@zhw. $\mathcal{H}f(x) = p.v. \int_{\mathbb{R}} \frac{f(t)}{x-t} dt$, where p.v. is the Cauchy principal value. – Claire Fan Apr 21 '17 at 01:51
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@Yikai 谢谢你！我自己也正在看handbook – Claire Fan Apr 29 '17 at 13:18
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1Can you elaborate on your definition of the Hilbert transform? If $f$ is the exponential function, for example, it doesn't seem to be defined. (I suggest editing the additional details into your question. Comments are good for notifying users who have commented, but others expect to see all details in the question itself.) – Joonas Ilmavirta May 01 '17 at 01:32
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One other thing: I warmly recommend going through the site tour. You know have enough reputation to vote up any questions and answers you find helpful. – Joonas Ilmavirta May 01 '17 at 02:22
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@JoonasIlmavirta Thank you! I've added the details in the question. – Claire Fan May 01 '17 at 03:34
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As was already pointed out, this isn't defined, you need some kind of decay of $f$. If we ignore this issue, then one possible argument goes as follows: $f$ holomorphic on the strip is essentially characterized by $|\widehat{f}(k)|\lesssim e^{-d|k|}$ (again, this needs extra assumptions to make it correct), and the Hilbert transform is the multiplier $\textrm{sign}|k|$, so won't disturb this condition. – May 02 '17 at 17:14
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@ChristianRemling Why $f$ holomorphic on the strip is equivalent with its Fourier transform is bounded by $e^{-|d|k}$? – Claire Fan Jun 18 '17 at 14:55
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I have enough reputation that I can see a deleted answer. And that answer seems correct to me.
Of course, there is a problem of defining $\mathcal{H}f(x) = \text{p.v.} \int_{\mathbb{R}} \frac{f(t)}{x-t}\,dt$ for functions that are analytic on $\mathcal{D}_d = \left\{ z \in \mathbb{C} : |\Im(z)| < d \right\}$. But if the function $f$ is sufficiently bounded (e.g. $|\hat f(\zeta)| \le C e^{-(d+\epsilon)|\zeta|}$ for all $\zeta \in \mathbb R$ and some $\epsilon>0$), so that the Hilbert transform is well defined, then one could argue like this: $$ \frac d{dz} \mathcal{H}f(z) = \frac d{dz} \text{p.v.}\int_{\mathbb{R}} \frac{f(z-t)}{t}\,dt = \text{p.v.}\int_{\mathbb{R}} \frac{f'(z-t)}{t}\,dt .$$ Or if you feel squeamish about pulling the derivative inside of the integral, instead use a contour integral: $$ \oint_\gamma \mathcal{H}f(z) \, dz = \text{p.v.}\int_{\mathbb{R}} \frac{\oint_\gamma f(z-t)\, dz}{t}\,dt = 0 $$ and then use Morea's Theorem.
Stephen Montgomery-Smith
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