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This must be something really easy/standard, but for some reason I can't see what's wrong with this argument.

Let $f : \mathbb{C} \rightarrow \mathbb{C}$ be any complex function, and write $f = u + iv$, with $u, v : \mathbb{C} \rightarrow \mathbb{R}$.

I know that if $f$ is $C^1$ in the real sense and $f$ satisfies the Cauchy-Riemann equations, then $f$ is holomorphic. My definition of holomorphic here is that the limit $$\lim_{\zeta \rightarrow 0} \frac{f(z + \zeta) - f(z)}{\zeta}$$ exists and is a complex number. This is not hard to prove, as I outline below, but I'm having trouble seeing where in this proof I'm actually using the continuity of the derivative of $f$. Here's the argument:

Proof: Let $z_0 = x_0 + iy_0 \in \mathbb{C}$, and let $A = u_x(x_0, y_0)$, $B = u_y(x_0, y_0)$, $C = v_x(x_0, y_0)$ and $D = v_y(x_0, y_0)$.

Since $u$ is differentiable at $(x_0, y_0)$, we can write

$$\epsilon_1(h, k) = u(x_0 + h, y_0 + k) - u(x_0, y_0) - Ah - Bk,$$ where $\epsilon_1(h, k)/\|(h, k)\|$ goes to zero as $(h, k) \rightarrow 0$. Now, for a moment there I thought this was the part where you need continuity of the partials $u_x$ and $u_y$, but note that they're being computed at a fixed point $(x_0, y_0)$, and the fact that $\epsilon_1$ goes to zero above is just the definition of differentiability, isn't it?

If we accept this and do the same expansion for $v$, then it's easy to see that the aforementioned limit will exist and will be a complex number, because of the Cauchy-Riemann equations.

So, again, my question is:

Where in the above proof are we using that $f$ is $C^1$, and not just differentiable?

Thanks!

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A necessary and sufficient condition for $f(z)$ to be differentiable (as a complex function) is that the functions $u$ and $v$ are differentiable (as functions of two real variables) and satisfy the C.R. equations. You do not need $C^1$. –  PAD Oct 8 '12 at 22:36
    
@PantelisDamianou I know, but that is Looman-Menchoff's theorem, right? And it is quite hard to prove, so I really doubt that my argument above is establishing the same result that easily. –  student Oct 8 '12 at 22:40
    
This is not Looman-Menchoff. L-M says that if you have a continuous function whose partial derivatives exist and satisfy C-R, then it is in fact holomorphic. You're assuming that f is differentiable (as a function $\mathbb{R}^2 \to \mathbb{R}^2$). This is stronger than just assuming the existence of partial derivatives. –  mrf Oct 9 '12 at 9:43
    
In other words, there is nothing wrong with your proof, but maybe you are confused about the word differentiable. For functions of one real variable, it means the same as having a derivative. Not so for functions of several variables. –  mrf Oct 9 '12 at 10:33
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up vote 1 down vote accepted

EDIT:I misunderstood your definition.

I'ld like to point out When is a function satisfying the Cauchy-Riemann equations holomorphic? instead.


Example Consider the function$$f(z)=e^{-\frac{1}{z^4}} \ \ \forall z \neq 0$$ $$f(0)=0$$

This function satisfies cauchy Riemann equation but is not holomorphic in any open neighbourhood of $0$.

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I think you misunderstood my definition; I call a function holomorphic if that limit exists everywhere. Note that I'm not fixing $z$ (maybe this part was not clear, but this is what I mean). Moreover, I don't claim that satisfying Cauchy-Riemann is a sufficient condition for holomorphy. On the other hand, you seem to be claiming that yourself in the 2nd sentence in your 2nd paragraph. Finally, your function $f$ is holomorphic (just check the $\mathbb{C}$-limit). –  student Oct 8 '12 at 22:19
    
No I wrote, complex differentiability $\iff$ Cauchy Riemann. It is true. Complex differentiability does not imply analyticity. Also the function is right example. Consider $z$ tending to zero with $arg (z) = \frac {\pi}{4}$. –  TheJoker Oct 8 '12 at 22:30
    
Okay, you're right about the function not being differentiable at $z = 0$. But still, if you look for instance Rudin's (Real & Complex Analysis) definition of holomorphic, it only requires that the $\mathbb{C}$-limit exists, and this is the same as being analytic. –  student Oct 8 '12 at 22:38
    
yes, sorry. I'll edit my post. Thanks. –  TheJoker Oct 8 '12 at 22:43
    
Thanks for the link, which contains relevant information. The only problem is that my question is really about my proof, not about $\mathbb{C}$-differentiability or conditions implying analiticity/holomorphy (I changed the title). By the way, this seems to be the same proof outlined there, but unfortunately nobody there said anything about it. –  student Oct 8 '12 at 23:29
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