Analysis Problem of the Day 58

Today’s problem appeared as Problem 11 on the UCLA Fall 2022 Analysis Qual:

Problem 11: For $f \in C^1(\mathbb{R}) \cap L^1(\mathbb{R}),$ define

$u(z) = \frac{1}{2\pi i}\int_{-\infty}^\infty \frac{f(t)}{t-z} dt, \quad |\text{Im }z| \not =0.$

a) Show that $u$ is holomorphic on $\mathbb{C} \setminus \mathbb{R}$ and $u(z) \to 0$ as $|\text{Im }z| \to 0.$

b) Show that the limit $\lim_{y \to 0} u(z) - u(\bar{z}), z = x+iy$ exists and compute it.

Solution: a) To show that a function is holomorphic, we apply Morera’s theorem. Indeed, the integral is finite by Hölder, since $\frac{1}{t-z}$ is bounded for $z=x+iy$ for fixed nonzero $y$ and $f$ is integrable. Thus, $u(z)$ is well-defined. Next, note that $u$ is continuous in $z$ by dominated convergence theorem, since $\left|\frac{f(t)}{t-z_n}\right| \leq \frac{|f(t)|}{\inf_{z_n \in B(z,\epsilon)} |y_n|} \in L^1, z_n = x_n+iy_n$ for $\epsilon < \text{Im }z$ as $z_n \to z,$ since $f$ is integrable. Finally, the Cauchy integral for $u$ evaluates by Fubini to

$\oint_\Delta u(z) dz = \int_{-\infty}^\infty \frac{1}{2\pi i}\oint_\Delta \frac{1}{t-z} dz f(t) dt = 0$

since $z \to \frac{1}{t-z}$ is holomorphic on any triangle avoiding the real line (and so the inner integral evaluates to 0 by Cauchy’s theorem). It follows that $u$ is continuous on $\mathbb{C} \setminus \mathbb{R}$ and every contour integral over a triangle evaluates to 0. Thus, by Morera’s theorem, $u$ is holomorphic on $\mathbb{C} \setminus \mathbb{R}.$

b) Notice by definition of $u$ that

$u(\bar{z}) = \frac{1}{2\pi i} \overline{\int_{-\infty}^\infty \frac{f(t)}{t-z}}= - \overline{u(z)},$

so the expression to evaluate becomes

$u(z)-u(\bar{z}) = 2 \text{Re }u(z) = \int_{-\infty}^\infty \frac{f(t)y}{\pi((t-x)^2+y^2)} dt, \quad z=x+yi.$

Now, notice that $\frac{y}{\pi(t^2+y^2)}$ is an approximation to the identity as $y \to 0,$ i.e. it converges weak- $^*$ to the Dirac delta measure at 0. Thus, the above integral expression converges to $\int_{-\infty}^\infty f(t) \delta(t-x)dt = f(x).$

Remark: This problem can be generalized to the so-called Sokhotski-Plemelj formulae, which state that given a continuous function defined on a closed simple curve in the complex plane, the Cauchy integrals as one approaches the curve approach half of the principal value of the Cauchy integral at the point plus/minus half of the value of the function at the point.

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