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Consider the Cauchy integral of the function $f\in L^2(\mathbb{R})$, that is, $F(z)=\frac{1}{2\pi i}\int^{\infty}_{-\infty}\frac{f(t)}{t-z}dt$

  1. $F$ is holomorphic in the open upper half-plane
  2. $F$ has a continuous extension to closed upper half-plane
  3. $F(z)=\int ^{\infty}_{0}\hat f(\xi)e^{2\pi iz\xi}d\xi$ on open upper half plane

Could someone help me establish this or give me a hint? Thanks a lot.

jk001
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  • 'Cauchy integral' --> is this like an actual thing? like in cauchy's integral/differential formula of $f(z) = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(t)}{t-z} dt$ (for suitable $\gamma$ and $f$) if $\gamma$ is instead replaced by $\mathbb R$ in the RHS and then (the equation is no longer true) and the RHS is defined as the *Cauchy integral* $F$ of $f$ (and it exists for $f \in L^2(\mathbb R)$) ? So what, $F$ is defined on the whole of $\mathbb C$? or what? – BCLC Aug 10 '21 at 01:24
  • what if $\hat{f}$ ? Where did you get this? – BCLC Aug 10 '21 at 01:32
  • I guess cauchy integral is not meant like cauchy sum as in 1, 2 or 3 but like 4, 5, 6, 7, 8, – BCLC Aug 10 '21 at 01:41
  • 9 or 10? – BCLC Aug 10 '21 at 01:41
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    @BCLC The adjective "Cauchy" here is being used to describe the Cauchy kernel $\frac{1}{\zeta-z}$. The integral here is a Lebesgue integral of an $L^1$ function: if $z=x+iy$ and $|y|>\delta>0$, then by the Cauchy-Schwarz inequality $|F(z)|\leq \frac{1}{2\pi}|f|{L^2}\cdot\sqrt{\int{\Bbb{R}}\frac{,dt}{|t-z|^2}}\leq\frac{|f|{L^2}}{2\pi}\sqrt{\int{\Bbb{R}}\frac{ds}{s^2+\delta^2}}<\infty.$ In other words, the integrand is in $L^1$ and for each $\delta>0$, $F$ is a bounded function on ${z\in\Bbb{C},:,|\text{im}(z)|>\delta}$ – peek-a-boo Aug 10 '21 at 01:57
  • thanks @peek-a-boo why do you change from $t$ to $s$? – BCLC Aug 10 '21 at 01:59
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    @BCLC if you do it fully, you'll get (inside square roots of course) $\int\frac{dt}{(t-x)^2+y^2}\leq \int\frac{dt}{(t-x)^2+\delta^2}=\int\frac{ds}{s^2+\delta^2}$, by a change of variables – peek-a-boo Aug 10 '21 at 02:00
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    @BCLC Sorry for the lack of clarity. I borrowed the term from Stein & Shakarchi's Functional analysis page 61.. I am not sure if this term is original. And by $\hat f$ I meant Fourier transform of the function $f\in L^2$ – jk001 Aug 10 '21 at 02:26

1 Answers1

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Everything but point 2 is chasing definitions: For the first, write the difference quotient $$ \dfrac{F(z)-F(z_0)}{z-z_0}= \dfrac{1}{2\pi i}\int_{-\infty}^\infty \dfrac{f(t)}{z-z_0}\left( \dfrac{1}{t-z}- \dfrac{1}{t-z_0}\right)\, dt, $$ and the quantity in parentheses is controlled by $$ \dfrac{|z_0-z|}{|z-t||z_0-t|} \lesssim_{z_0} \dfrac{|z-z_0|}{1+t^2}, $$ since $z_0$ is fixed and $z,z_0$ are both in the upper half space. Now apply the dominated convergence theorem to conclude.

The third one is a bit lengthy if you don't know you Fourier transform identities, but the basic idea is that for fixed $z=x+iy$, we can write $$ F(z)= \dfrac{-1}{2\pi i}\int_{-\infty}^\infty \dfrac{f(t)}{x-t +iy}\, dt =: f* g_y(x), $$ where $g_y(s):=\frac{1}{2\pi i(s+iy)}$, and $*$ denotes convolution in one dimension. Then by properties of convolution and Fourier transform we have $$ f*g_y(x)= (\hat{f} \hat{g}_y)^\vee (x)= \int_{\mathbb{R}} e^{2\pi i x\xi} \hat{f}(\xi) \hat{g}_y (\xi)\, d\xi, $$ and so it remains to show that $\hat{g}_y$ has the right form. For this, write $$ \hat{g}_y(\xi)= \int_{\mathbb{R}} \dfrac{e^{-2\pi i x\xi}}{x+iy}\, dx = \int_{\mathbb{R}} e^{-2\pi i x\xi} \dfrac{ x-iy}{x^2+y^2}\, dx. $$ Now separate these into two Fourier transforms, and recall that the FT of $\frac{1}{\pi}\frac{y}{x^2+y^2}$ is given by $e^{-2\pi y|\xi|}$. I leave the rest of this part to you.

The interesting part 2 you can do as follows: Fix $t_0\in \mathbb{R}$, we want to show that $\lim_{z\to t_0} F(z)$ exists. Fix $\varepsilon>0$ and break up the integral as $$ F(z)= \int_{|t-t_0|\leq \varepsilon} + \int_{|t-t_0|>\varepsilon}=:I_\varepsilon(z)+ II_\varepsilon(z). $$ For the second term, you can use dominated convergence to show that $\lim_{z\to t_0} II_{\varepsilon}(z)$ exists (basically for this one, we've cut-off the singularity that appears when $z$ tends to the real line).

From here on out, I'll make the assumption that $f\in L^2\cap C^\alpha(\mathbb{R})$ for some $\alpha>0$, meaning $\sup_{t\neq s}\frac{|f(t)-f(s)|}{|t-s|^\alpha}=:[f]_\alpha <\infty$ (You can probably get away with less, but a continuous extension can't be expected for a general $L^2$ function, unless the problem means something else...).

Notice first that $$ I_\varepsilon(z) = \dfrac{1}{2\pi i} \int_{|w-t_0|\leq \varepsilon} \dfrac{f(w)-f(t_0)}{w-z}\, dw + \dfrac{f(t_0)}{2\pi i} \int_{|z-t_0|\leq \varepsilon} \dfrac{dw}{w-z}=: I'_\varepsilon +I''_\varepsilon. $$ For the first of these terms you can use dominated convergence, since the integrand is dominated (see (1) below) by $|w-t_0|^{\alpha-1}$, which is integrable for $w\in \mathbb{R}$.

To treat $I''_\varepsilon$, consider the curve $\Gamma$ given by the lower half-circle: $|w-t_0|=\varepsilon$ and $\text{Im}(w)\leq 0$, together with $w\in\mathbb{R}$, $|w-t_0|\leq \varepsilon$ travelled clock-wise.

Then $$ I''_\varepsilon(z)= \dfrac{1}{2\pi i}\int_{\Gamma} \dfrac{dw}{w-z} - \dfrac{1}{2\pi i} \int_{\substack{|w|=\varepsilon\\ \text{Im}(w)\leq 0}} \dfrac{dw}{w-z} =- \dfrac{1}{2\pi i} \int_{\substack{|w|=\varepsilon\\ \text{Im}(w)\leq 0}} \dfrac{dw}{w-z}, $$ where the last equality follows from Cauchy's theorem, since the function $1/(z-w)$, for fixed $z$ in the upper half space, is holomorphic for $w$ in the lower half space.

Now the important thing to notice is that the singularity for the last integral is on the lower half of the circle $|w|=\varepsilon$, so that we may apply dominated convergence again to get that the limit exists. Combining all of the above, we see that the limit for $F$ exists.

Sidenotes: You can actually get a formula for the boundary values of $F$ by examining the limit $I_\varepsilon$ (or more specifically $I''_\varepsilon$) as $\varepsilon\to 0$ (notice that $II_\varepsilon$ tends to the Hilbert transform of $f$).

You can also extend these result to $L^2$ functions (without the continuity assumption), in the sense that the limits are attained nontangentially instead of cotninuously.

(1) Added: On reread, it seems that to dominate like I say you do, you need to restrict $w$ to be in a region like $\{ w: |\text{Re}(w)-t_0|\leq A\cdot \text{Im}(w)\}$ for a given constant $A>0$ (convince yourself these are cones). The reason for this is that if $z$ approaches the real axis way too fast, while being still relatively far away from $t_0$, then the smallness of the numerator $|f(w)-f(t_0)|$ is not helping at all to control the denominator, which blows up like $|w-z|^{-1}$.

That said, I do believe that, at least for $C^\alpha$ data, one should be able to upgrade this nontangential convergence to a full continuous extension; however, for now I'm not seeing it clearly.

Jose27
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