Given $f \in L^1$, prove that $$F(z) = \frac{1}{2\pi i} \int^\infty_{-\infty} \frac{f(t)}{t-z}\,dt$$ is an analytic function and $$F'(z)=\frac{1}{2\pi i} \int^\infty_{-\infty} \frac{f(t)}{(t-z)^2}\,dt.$$

My attempt: It is sufficient to prove its analytic by showing that the derivative of $F(z)$ exists. (Any flaws with this argument?)

\begin{align} F'(z)&=\lim_{\Delta z_n\to 0} \frac{F(z+\Delta z_n) - F(z)}{\Delta z_n}\\ &=\lim_{\Delta z_n\to 0} \frac{1}{\Delta z_n} \frac{1}{2\pi i} \left[\int^\infty_{-\infty} \left( \frac{f(t)}{t-(z+\Delta z_n)}-\frac{f(t)}{t-z} \right) dt \right]\\ \mbox{(a bit algebra)} &= \frac{1}{2\pi i} \lim_{\Delta z_n\to 0} \int^\infty_{-\infty} \frac{f(t)}{(t-z)(t-z-\Delta z_n)} \, dt \\ \end{align}

I came so close to justify the exchange of the limit with the integral (Hence, DCT). I somehow cannot find a appropriate $g\in L^1$ to bound the integrand.

  • $\begingroup$ For $z \notin\mathbb{R}$, choose $|\Delta z_n| < |\Im z|/2$. Then $|t-z-\Delta z_n| \ge |t-z| - |\Delta z_n| \ge |t-z|-|\Im z|/2$. $\endgroup$ – DisintegratingByParts Dec 6 '15 at 3:55

If $z$ is not real, you can always bound the denominator away from zero, and so you can use $f(t)/(t-z+\delta)^2$ for an appropriate (and small enough) $\delta>0$.

But when $t$ is real I see problems: for instance let $f$ be any $L^1$ function that is constant on some interval around $2$. Then $$ \int_{-\infty}^\infty\frac{f(t)}{t-2} $$ doesn't exists (it blows up at $2$).

Are you really supposed to show that $F$ is analytic on all of the complex plane?

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  • $\begingroup$ How about if I restricted the domain to $\mathbb{R}\setminus \{z\}$? $\endgroup$ – Baffin Chu Dec 5 '15 at 18:18
  • $\begingroup$ Yes, that would work. Then you can use the distance from $z$ to the real axis to bound $t-z$ away from zero. $\endgroup$ – Martin Argerami Dec 5 '15 at 19:02
  • $\begingroup$ From your hint, I think the correct one should be $$ g(t) = f(t) / (t-z-\delta)^2 ,$$ since it needs to satisfy |(the ingetrand)|$\le g(t)$, am I right? $\endgroup$ – Baffin Chu Dec 6 '15 at 3:45
  • $\begingroup$ Yes, with $\delta=\text {Im}z/2$. $\endgroup$ – Martin Argerami Dec 6 '15 at 4:45

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