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I am trying to understand why the differentiability of a complex function $f(z) = u(x, y) + iv(x, y)$ at a point $z_0 = a + ib$ implies that the real part $u(x, y)$ and the imaginary part $v(x, y)$ are continuously differentiable at $(a, b)$.

Here’s what I know so far:

  • I am assuming that $f(z)$ is differentiable at $z_0 = a + ib$ but I am not assuming that $f(z)$ is analytic at $z_0$.
  • For $f(z)$ to be differentiable at $z_0$, the Cauchy-Riemann equations must hold at $(a, b)$, and the partial derivatives of $u(x, y)$ and $v(x, y)$ must exist.
  • However, I don’t fully understand why these partial derivatives must also be continuous at $(a, b)$.

Could someone explain:

  1. Why differentiability in the complex sense ensures continuous differentiability of $u$ and $v$?
  2. If possible, how can this be proved rigorously, given that $f(z)$ is only assumed to be differentiable at $z_0$, not analytic?

Any clarification or pointers to relevant theorems would be greatly appreciated!

Sartaj Ansari
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  • Most textbooks follow the same approach. They use Cauchy’s integral formula to prove that complex differentiability of $f$ (in a domain, not just at a point) implies analyticity. (There is also an integration-free proof by Whyburn, but my impression is that it makes use of some topological notions and facts that few people except topologists have heard of.) Continuous differentiability of $u$ and $v$ then follows. – user1551 Jan 26 '25 at 13:00
  • If a function f(z)=u+iv is analytic at a point imply partial derivatives of u and v are continuous at that point question is if f is not analytic but differentiable at a point then partial derivatives of u and v are continuous – Sartaj Ansari Jan 26 '25 at 18:04
  • I think I have made my point clear. If $f$ is differentiable in a domain, it is necessarily analytic in that domain and hence $u$ and $v$ are continuously differentiable. If $f$ is only differentiable at a point, Weber’s answer below shows that $u$ and $v$ are not necessarily continuously differentiable. I am well aware of your assumption. I don’t understand why you keep repeating it. Which part of my comments do you not understand? – user1551 Jan 26 '25 at 18:30

1 Answers1

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Let $u(x, y) = \begin{cases} x^2 \sin(1/x) &\mbox{if } x \neq 0 \\ 0 & \mbox{if } x=0 \end{cases}$, $v(x, y) = 0$ and $(a, b) = (0, 0)$. Then

  1. $\frac{\partial u}{\partial x}(a, b) = \frac{\partial u}{\partial y}(a, b) = \frac{\partial v}{\partial x}(a, b) = \frac{\partial v}{\partial y}(a, b) = 0$,
  2. the partial derivatives $\frac{\partial u}{\partial x}(x, y), \frac{\partial u}{\partial y}(x, y), \frac{\partial v}{\partial x}(x, y), \frac{\partial v}{\partial y}(x, y)$ exist for all $(x, y) \in \mathbb{R}^2$, but
  3. $\frac{\partial u}{\partial x}$ is discontinuous in $(a,b)$, cf. Mark McClure's answer to "What are examples of functions with 'very' discontinuous derivative?".

So the presented statement "complex differentiability at a single point implies partial derivatives' continuity in that point" is not true. Correct is only "complex differentiability in a neighborhood of a point implies partial derivatives' continuity in that point", for reasons given e.g. in user1551's comment.

Toralf Weber
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