5

I have two questions.

My book proves that if $f:\mathbb{C}\rightarrow \mathbb{C}$ is a holomorphic function, then it satisfies the Cauchy-Riemann equations, and if we look at the function as $F: \mathbb{R}^2\rightarrow \mathbb{R}^2$, then this function is differentiable. The point is that we are looking at conditions when we can go from complex differentiability(holomorphic), to differentiability when we look at the function in terms of real variables only ,that is differentiability in multivariate calculus.

question 1:

The book also has the converse of the above[picture], and in the converse result they write $ f = u(x,y)+i*v(x,y)$, where $z=x+i*y$, now in the converse result they have that the u and v must be continuously differentiable, why not only differentiable? Do you guys see in the proof where they need to use that it is continuously differentiable and it would fail if they were only differentible?

question 2:

A related question is: If we have a function $h: \mathbb{R}^2\rightarrow \mathbb{R}^2 $. And we know that the partial derivatives exist, but they are not continous, will it may be that h is not differentiable?

enter image description here

user119615
  • 10,176
  • 5
  • 44
  • 112
  • How is "holomorphic" defined in that book? Is it only the existence of the derivative on $\Omega$, or is continuity of the derivative included in the definition? – Daniel Fischer Aug 18 '14 at 15:01
  • It's been a while since I've taken complex analysis, but aren't holomorphic functions analytic (meaning they would be infinitely differentiable)? So the first derivatives would have to be continuous – Silynn Aug 18 '14 at 15:03
  • @Silynn Yes, but the cool thing is that you don't need to demand that. It follows from the mere existence of the derivative on an open set. – Daniel Fischer Aug 18 '14 at 15:05
  • @DanielFischer From what I see the existence of the derivative is the only thing that is required. – user119615 Aug 18 '14 at 15:06
  • @Silynn I think you are mixing up the derivative when we look at f as a complex function and when we look at it as a real function. The point is that it tries to find conditions when we can go from one to another. – user119615 Aug 18 '14 at 15:07

2 Answers2

9

The continuous differentiability of $u$ and $v$ is not required for the proof, for the representation

$$u(x+h_1,y+h_2) - u(x,y) = \frac{\partial u}{\partial x}(x,y) \cdot h_1 + \frac{\partial u}{\partial y}(x,y)\cdot h_2 + \lvert h\rvert\cdot \psi_1(h)\tag{1}$$

and the analogous for $v$, the differentiability of $u$ (and $v$) in $(x,y)$ [resp. $x+iy$] is enough. From these representations and the Cauchy-Riemann equations (in the single point $(x,y)$) follows the complex differentiability of $f$ in $x+iy$.

That shall be the case for all $x+iy \in \Omega$, hence $f$ is holomorphic on $\Omega$.

Since every holomorphic function is analytic, nothing is lost by imposing stricter conditions than the proof requires on $u$ and $v$, but the proof only uses the differentiability in each point.

Concerning question 2, yes, there are functions where the partial derivatives exist everywhere that are not differentiable at least in some points.

The mere existence of the partial derivatives does not imply a representation of the type $(1)$, hence the existence of the partial derivatives plus the Cauchy-Riemann equations does not suffice to deduce the complex differentiability of $f$ in $x+iy$.

Daniel Fischer
  • 206,697
  • You say that the existence of partial derivatives does not imply representation as (1), does continuous partial derivatives imply it? I do not understand what you mean when you first say it is not required in the proof, but last say that we can't get the representation? – user119615 Aug 18 '14 at 15:30
  • Yes. If you look at the proof, you see that the continuity of the derivatives of $u$ and $v$ is nowhere even mentioned except in the statement of the theorem. What is used is the differentiability, which is the existence of a representation $(1)$. Only if the definition of holomorphicity in that book includes the continuity of the derivative, we would need something to obtain the continuity and not only the existence of $f'$ on $\Omega$. – Daniel Fischer Aug 18 '14 at 15:32
  • @DanielFischer What about this post? It seems to list some counterexamples for why $\mathcal{C}^1$ is needed. – angryavian Aug 18 '14 at 15:34
  • Continuous partial derivatives imply continuous differentiability, so that is enough. Note that differentiability is a strictly stronger condition than partial differentiability. But if you add continuity, both coincide. – Daniel Fischer Aug 18 '14 at 15:34
  • So what you are saying is that for continuous partial derivatives is a consequence of f beeing holomorphic, so in order for f to even be holomorphic they need to be continuous? The thing is I don't see how holomorphicity would require continuous partial derivatives? I mean a holomorphic function was defined as a function where the complex derivative existed, it didn't say anything about the continuity of the partial derivatives of the components, is this something that follows? – user119615 Aug 18 '14 at 15:38
  • Oh no I see, since my book does not include conditions of a continuous complex derivative I can assume that the conditions in the proof are too strict? – user119615 Aug 18 '14 at 15:41
  • @angryavian I can't see much at the JSTOR link of the first answer there. The Looman-Menchoff theorem linked in the second answer is even stronger, it states that continuity of $f$, existence of the partial derivatives and the Cauchy-Riemann equations imply holomorphicity. Here, we have differentiability (which is total differentiability or Fréchet differentiability), which is stronger than only continuity plus partial differentiability. – Daniel Fischer Aug 18 '14 at 15:42
  • @user119615 It is a theorem that every holomorphic function is analytic, in particular infinitely differentiable. So the partial derivatives of the real and imaginary part of a holomorphic function are necessarily also infinitely differentiable (even real-analytic). Thus by requiring $u$ and $v$ to be continuously differentiable you don't miss any holomorphic functions. But as I said, the proof does not require that assumption. It works with plain Fréchet differentiability, the continuity of the (partial) derivatives is a consequence. – Daniel Fischer Aug 18 '14 at 15:46
  • Sorry I am still confused about your last statement in your main post. You talk about the existence of representation (1), do you mean that we CANT write it as (1) if we delete continuous? It seems you say that we can remove continuous because we do not use it in the proof, but you also say that only existence of partial differentiability does not gurantee the representation we need to use in the proof? It seems that yous ay the opposite thing in the comments and in the last section of your main post? – user119615 Aug 18 '14 at 15:51
  • You say this in the last section "The mere existence of the partial derivatives does not imply a representation of the type (1), hence the existence of the partial derivatives plus the Cauchy-Riemann equations does not suffice to deduce the complex differentiability of f in x+iy." But you also say that we can remove the word continuous from the proof?, doesn't that mean that indeed if the partial derivatives exist and the cauchy riemann equations are satisfied the it is complex differentiable(holomorphic). – user119615 Aug 18 '14 at 15:53
  • I think I know what I misunderstood now: We have three situations, partial derivatives exist, differentiability and continuous differentiability, you mean that differentiability is enough, but existence of partial derivatives is too weak, but continuous differentiability is too strong? – user119615 Aug 18 '14 at 15:58
  • @user119615 Yes. The mere existence of the partial derivatives is too weak for the proof (but, if $f$ is assumed continuous, not for the conclusion). Plain differentiability is enough for the proof. Hence also everything stronger is enough. But continuous (partial) differentiability is more than is required. – Daniel Fischer Aug 18 '14 at 16:03
  • @DanielFischer Than you very much for your time!, you are extremely knowledgeable. When I had multivariate calculus I never learned why the theory worked etc., we just did the calculations and never tought why it worked etc., I need to go back and study that. – user119615 Aug 18 '14 at 16:06
2

\begin{align*} &u(x_0+h_1,y_0+h_2)-u(x_0,y_0)\\ &= h_1 \frac{u(x_0+h_1,y_0+h_2)-u(x_0,y_0+h_2)}{h_1}+h_2\frac{u(x_0,y_0+h_2)-u(x_0,y_0)}{h_2}\\ &= h_1 \left(\frac{\partial u}{\partial x}(x_0,y_0+h_2) + \varphi_1(h_1)\right) + h_2 \left(\frac{\partial u}{\partial y}(x_0,y_0) + \varphi_2(h_2)\right) & \text{where $\lim\limits_{h_i \to 0}\varphi_i(h_i)=0$}\\ &= h_1 \left(\frac{\partial u}{\partial x}(x_0,y_0) + \varphi_1(h_1)+\widetilde{\varphi}_2(h_2)\right) + h_2 \left(\frac{\partial u}{\partial y}(x_0,y_0) + \varphi_2(h_2)\right) & \text{$\lim\limits_{h_2 \to 0}\widetilde{\varphi}(h_2)\to 0$ by continuity of $\partial u / \partial x$}\\ \end{align*}

angryavian
  • 89,882
  • Wouldn't those two function tend to zero regardless of the continuity of $\partial u/ \partial x$? – user119615 Aug 18 '14 at 15:26
  • Yes for $\varphi_1$ and $\varphi_2$. However, I applied the continuity of $u_x$ to show that $\widetilde{\varphi_2}$ (the difference between $u_x(x_0,y_0+h_2)$ and $u_x(x_0,y_0)$) is small. However, Daniel seems to say that continuous differentiability is not needed so maybe I have made a mistake... – angryavian Aug 18 '14 at 15:30
  • Shouldn't $\varphi_1(h_1)$ be $\varphi_1(h_1,h_2)$, which tends to $0$ as $h_1 \to 0$ with $h_2$ fixed? It would then need to be shown that $\varphi_1(h_1,h_2) \to 0$ as $(h_1,h_2) \to 0$. – Antonio Vargas Feb 14 '17 at 21:02