Let $f(t,z)$ be a bounded (say by a constant $M$) continuous function on $\mathbb{R}_t \times \mathcal{U}$ where $\mathcal{U}$ is an open neighborhood of $0 \in \mathbb{C}_z$. Moreover, for each fixed $t \in \mathbb{R}_t$, assume that $f$ is analytic in $\mathcal{U}$.
Questions
- Is $\partial f /\partial z$ continuous on $\mathbb{R}_t \times \mathcal{U}$? How do I show this?
- For $r > 0$, let $D_r(a)$ be a closed disk of radius $r$, centered at $a$. Is there an $r > 0$ for which $\frac{\partial f}{\partial z} (t,z)$ is bounded on $\mathbb{R}_t \times D_r(0)$?
My Attempt: Fix an acceptable $r >0$ so that $D_r(0) \subset \mathcal{U}$. Let $0 < r' < r$. Then $D_{r'}(0) \subset D_r(0)$. If we take an $a \in D_{r'}(0)$, then $D_{r-r'}(a) \subset D_r(0)$. We can use Cauchy's estimate to get $$ \left| \frac{\partial f}{\partial z} (t,a) \right| \leq \frac{M}{r-r'} \text{ on } \mathbb{R}_t. $$ Since $a$ is arbitrary, we have $$ \left| \frac{\partial f}{\partial z} (t,z) \right| \leq \frac{M}{r-r'} \text{ on } \mathbb{R}_t \times D_{r'}(0). $$
Am I on the right track here? Am I missing something?
Thanks for the help!
You answer question 2 correctly, the Cauchy estimates show that $\frac{\partial f}{\partial z}$ is uniformly bounded on $\mathbb{R}\times D_r(0)$ for all $r > 0$ such that $\overline{D_r(0)} \subset \mathcal{U}$.
To see the continuity of $\frac{\partial f}{\partial z}$, fix a $z_0\in\mathcal{U}$. Then for $r > 0$ small enough that $\overline{D_r(z_0)} \subset \mathcal{U}$, the Cauchy integral formula for the derivative gives you
$$\frac{\partial f}{\partial z}(t,z) = \frac{1}{2\pi i} \int_{\lvert \zeta-z_0\rvert = r} \frac{f(t,\zeta)}{(\zeta - z)^2}\,d\zeta\tag{1}$$
for all $(t,z) \in \mathbb{R}\times D_r(z_0)$. The integrand
$$h(t,z,\zeta) = \frac{f(t,\zeta)}{(\zeta-z)^2}$$
is continuous on $\mathbb{R}\times D_r(z_0)$ (and uniformly bounded on $\mathbb{R}\times D_\rho(z_0)$ for all $\rho \in (0,r)$), so the integral $(1)$ is continuous. The exact argument for the continuity depends on whether you consider the integral as a Lebesgue integral, a Riemann integral, or some other type of integral. For the Lebesgue integral, the usual argument would be the dominated convergence theorem, for the Riemann integral, one would probably argue with the uniform continuity of continuous functions on compact spaces, which means that the integrands in
$$\frac{\partial f}{\partial z} (t_n,z_n) = \frac{1}{2\pi i}\int_{\lvert\zeta - z_0\rvert = r} \frac{f(t_n,\zeta)}{(\zeta - z_n)^2}\,d\zeta$$
converge uniformly to $\frac{f(t,z)}{(\zeta-z)^2}$ on the circle $\lvert \zeta-z_0\rvert = r$ as $(t_n,z_n) \to (t,z)$.