4
$\begingroup$

Let $f$ be holomorphic in $D(0;1)$ and define $k$ by $k(z)=\overline{f(z)}$. Prove that $k$ is differentiable at $a\in D(0;1)$ if and only if $f'(a)=0$.

What I tried was first, assuming $k$ is differentiable and letting $f=u+iv$ we have (first when $h \in \mathbb{R}$)

$$k'(z)= \lim_{h \to 0} \frac{u(x+h,y)-u(x,y)}{h} -i\frac{v(x+h,y)-v(x,y)}{h} = u_x -iv_x$$ and when $h=ik, \ k\in \mathbb{R}$

$$k'(z)=\lim_{k \to 0} \frac{u(x,y+k)-u(x,y)}{ik} -\frac{v(x,y+k)-v(x,y)}{h} = \frac{1}{i}u_y -v_y$$

And equating real and imaginary parts, we get that

$$u_x=-v_y, \; u_y=v_x$$ Since $f$ is holomorphic, it satisfies the Cauchy-Riemann equations and thus

$$u_x=v_y, \; u_y=-v_x$$ so $$f'(a)=-f'(a)$$ and then $f'(a)=0$. I don't know if this works, so please correct me if I'm wrong. Besides that, I'm stuck in proving the other implication. So far I did

$$0=f'(a)=\lim_{h\to 0} \frac{f(a+h)-f(h)}{h}=\overline{\lim_{h\to 0}\frac{f(a+h)-f(h)}{h}}=\lim_{h\to 0}\frac{\overline{f(a+h)} -\overline{f(h)}}{\overline{h}}=\overline{f'(a)}=k'(a)$$

But again, I'm not sure if this is right. Any help will be highly appreciate, and thanks in advance!

$\endgroup$
3
$\begingroup$

Yes, both directions of your proof are essentially correct, but some steps could be better justified. To see them done out in a bit more detail:

For all $a \in D(0,1)$, $f$ is holomorphic, so the Cauchy-Riemann equations hold, i.e.

$$u_x=v_y, \ \ \ \ u_y=-v_x$$

Assuming $k'(a)=0$, you have shown that the following equations must hold at $a$:

$$u_x=-v_y, \ \ \ \ u_y=v_x$$

Combining all of these, we get the following:

$$u_x=-u_x, \ \ \ \ u_y=-u_y, \ \ \ \ v_x=-v_x, \ \ \ \ v_y=-v_y$$

Now, $u_x=-u_x$ implies that $u$ is constant in $x$, $u_y=-u_y$ implies that $u$ is constant in $y$, etc. Hence, these equations imply $f'(a)=0$.

Now, assume $f'(a)=0$. Then we certainly have $f'(a)=\overline{f'(a)}$. We also know that complex conjugation is additive, multiplicative, and that the complex conjugate of a continuous function is continuous. Hence, we have

$$0=\overline{\lim_{h\to 0}\frac{f(a+h)-f(h)}{h}}=\lim_{h\to 0}\overline{\frac{1}{h}(f(a+h)-f(h))}=\lim_{h\to 0}\frac{\overline{f(a+h)} -\overline{f(h)}}{\overline{h}}$$

Now, this last limit does evaluate exactly to $k'(a)$. To see what happens in detail, write $h=re^{i\theta}$. Then, we have

$$\lim_{h\to 0}\frac{\overline{f(a+h)} -\overline{f(h)}}{\overline{h}} = \lim_{r\to 0}\frac{\overline{f(a+re^{i\theta})} -\overline{f(h)}}{re^{-i\theta}} = e^{2i \theta}\lim_{r\to 0}\frac{\overline{f(a+re^{i\theta})} -\overline{f(h)}}{re^{i\theta}}=e^{2i \theta}k'(a)$$

Hence, we have $e^{2i \theta}k'(a)=0$. Note that, for all $\theta \in [0,2\pi]$, this implies $k'(a)=0$. Hence, $k'(a)=0$.

$\endgroup$
  • $\begingroup$ But the first set of the Cauchy-Riemann equations holds for $f$, while the second set holds for $k$, which are two different functions. So I don't see why you need to equate the Cauchy-Riemann equations for two completely different functions. $\endgroup$ – sequence Apr 6 '17 at 8:38
4
$\begingroup$

$\displaystyle \lim_{h\to0} |\dfrac{f(z+h)-f(z)}{h}|=0 \Leftrightarrow \displaystyle \lim_{h\to0} |\dfrac{\overline{f(z+h)-f(z)}}{h}|=0$

But the expression $\dfrac{\overline{f(z+h)-f(z)}}{h}= \dfrac{\overline{f(z+h)-f(z)}}{\overline{h}}\times\dfrac{\overline{h}}{h}$ does not tend to a limit if the first half of it tends to $w=\overline{f'(a)}\ne0$, because the second half can be made to have any complex value of unit length.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.