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Assume $f(z)$ is a complex function of a complex variable $z$. The definition of $f$ possessing a derivative at $z$ is the existence of the limit $$\lim\limits_{h\to0}\frac{f(z+h)-f(z)}{h}.$$ See, e.g., $\S$1.1 of "Complex Analysis (3rd Ed.)" by Lars V. Ahlfors.

A multivariate vector-valued function $\mathbf{f}$ mapping from $\Bbb R^n\to\Bbb R^m$ is differentiable at $\bf{x}$ if there is a linear transformation $A$ of $\Bbb R^n$ into $\Bbb R^m$ such that $$\lim\limits_{{\bf{h}}\to\bf{0}}\frac{|{\bf{f}}({\bf{x}}+{\bf{h}})-{\bf{f}}({\bf{x}})-A{\bf{h}}|}{|{\bf{h}}|}=0,$$ which is equivalent to $$\lim\limits_{{\bf{h}}\to\bf{0}}\frac{|{\bf{r}}({\bf{h}})|}{|{\bf{h}}|}=0,$$ where ${\bf{r}}({\bf{h}})={\bf{f}}({\bf{x}}+{\bf{h}})-{\bf{f}}({\bf{x}})-A{\bf{h}}$. See, e.g., "DIFFERENTIATION" Section in Chapter 9 of Rudin's "Principles of Mathematical Analysis (3rd Ed.)".

I've been being tempted to think of complex functions of a complex variable $f(z)=u(z)+iv(z)$ as a $\Bbb R^2\to\Bbb R^2$ function $\mathbf{f}(\mathbf{x})$ where $\mathbf{x}=(x,y)$ corresponds to $z$ and $\mathbf{f}=(u,v)$, and subsequently the condition that $f(z)$ possesses derivative at $z$ (first paragraph) is equivalent to (i.e., can imply each other) $\mathbf{f}$ being differentiable at $\mathbf{x}$ with $n=m=2$ (second paragraph). But today, when I thought more about this, I found myself standing in front of an obvious contradiction that arises from the following theorem (P26 of Ahlfors' "Complex Analysis (3rd Ed.)"):

If $u(x,y)$ and $v(x,y)$ have continuous first-order partial derivatives which satisfy the Cauchy-Riemann differential equations, then $f(z) = u(z) + iv(z)$ is analytic with continuous derivative $f'(z)$, and conversely.

If the above equivalence claim were correct, the continuity of first-order partial derivatives will guarantee differentiability of $\mathbf{f}$ (by, e.g., Theorem 9.21 on P219 of Rudin's text) and in turn that $f$ possesses derivative. Then the Cauchy-Riemann differential equations would be useless in the proof. Since the Ahlfors' text is so classical, I'm sure this condition is indispensable (I guess it is needed in line 11 and 12 of P26 in Ahlfors' text, right?). As a result, the equivalence claim in the above paragraph must be wrong. Then I tried to figure out why on earth they are not equivalent. Later I came to a point that (1) $\Rightarrow$ (2) while (2) $\not\Rightarrow$ (1), where (1) stands for $f$ possessing derivative and (2) for the corresponding $\Bbb R^2\to\Bbb R^2$ function $\mathbf{f}$ being differentiable. That is, (1) is stronger than (2). I guess the strongerness comes from the fact that $\Bbb C$ defines a multiplication operation, which is not available in $\Bbb R^2$. The contradiction can be solved by this relation: the continuity of first-order partial derivatives establishes (2), but we need additional conditions (the Cauchy-Riemann differential equations) to arrive at a stronger conclusion (1).

To show the above relation, First, I gave a proof of (1) $\Rightarrow$ (2): Assume the derivative is a complex number $\alpha$ and we define linear transformation according to complex number multiplication: if ${\bf{h}}=(x,y)$, $A{\bf{h}}\cong\alpha(x+iy)$ (an isometry between $\Bbb R^2$ and the complex plane). Since $\lim\limits_{z\to0}|f(z)|=0$ iff $\lim\limits_{z\to0}f(z)=0$, we have $$\begin{eqnarray} \lim\limits_{{\bf{h}}\to\bf{0}}\frac{|{\bf{f}}({\bf{x}}+{\bf{h}})-{\bf{f}}({\bf{x}})-A{\bf{h}}|}{|{\bf{h}}|}&=&\lim\limits_{h\to0}\frac{|f(x+h)-f(x)-\alpha h|}{|h|}\\ &=&\lim\limits_{h\to0}\frac{f(x+h)-f(x)-\alpha h}{h}\\ &=&\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}-\alpha\\ &=&\alpha-\alpha=0 \end{eqnarray}$$ For (2) $\not\Rightarrow$ (1): From the equivalent formulation of differentiability, we have $f(z+h)-f(z)=A{\bf{h}}+r(h)$ where $r(h)=o(h)$. If we can write $A{\bf{h}}\cong\alpha h$ for some complex number $\alpha$ as in the last paragraph, then the derivative of $f$ is immediate. But the thing is, we can't (at least I can't). $A{\bf{h}}/h$ is not necessarily a constant complex number for a given general linear transformation $A$. That's why (2) does not imply (1).

Now comes my question: 1) Is it true that possessing derivative for a complex function of a complex variable is not equivalent to its corresponding $\Bbb R^2\to\Bbb R^2$ function being differentiable? 2) Is my relation that (1) $\Rightarrow$ (2) while (2) $\not\Rightarrow$ (1) a correct argument to explain this inequivalence? 3) Is there anything wrong (or not rigorous) throughout my argument? Thank you.

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    $\begingroup$ All I'd say is that to be sure about the non-implication you need an explicit counterexample: something like $f(z)=\bar{z}$ at the origin. ( A great book, Ahlfors!) $\endgroup$ Jan 11 at 11:33
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    $\begingroup$ Differentiability means existence of a suitably good linear approximation. For the real-variable case, this means an $\mathbb R$-linear map $\mathbb R^2\to\mathbb R^2$. For the complex-variable case, it means a $\mathbb C$-linear map $\mathbb C\to\mathbb C$. And these are not the same, because the latter requires the map to respect multiplication by non-real scalars like $i$. (In this sense, your difficulty is not with analysis but with linear algebra.) $\endgroup$ Jan 11 at 18:19
  • $\begingroup$ @ancientmathematician: thank you for the counterexample of $f(z)=\bar{z}$ that is not analytic, which clearly confirms my conjecture that being analytic is stronger than being differentiable. $\endgroup$ Jan 12 at 4:09
  • $\begingroup$ @Andreas Blass: I studied the concept of differentiability in Rudin's book, so I don't quite understand what you mean by "real-variable case" (for multivariate function, each of the variables is assumed to be real), "R-linear map" (for vector-value function, each component of the vector is a real) and how differentiability can be used in complex-variable case (I studied derivative of a complex function of a complex variable in Ahlfors' book, where there is no concept of "differentiability"). $\endgroup$ Jan 12 at 4:16
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Update of my longer answer:

$ $ 1) Yes, 2) Yes, 3) No.

Concerning "complexifying": This was a word invented by me describing a thought process that is common, and that you do all the time in your question.

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  • $\begingroup$ I'm sorry I'm at a loss about your answer. Can you please be more specific about how the answer is related to the three questions I asked at the end of the original question? In particular, what do you mean by "complexify" (can you explain it in math as opposed to natural language)? $\endgroup$ Jan 12 at 3:54
  • $\begingroup$ Thank you for the editing. $\endgroup$ Jan 12 at 10:57

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