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In a problem I've been doing it is stated to show that even though the function is differentiable along the coordinate axes, it is nowhere analytic. By definition, a function is differentiable if the following limit exists and is finite $$\lim _{\Delta z \to0 }\frac{f\left ( z+\Delta z \right )-f\left ( z \right )}{\Delta z}.$$

Along the x-axis I got something like $f^{'}\left ( z \right )=3x^{2}+3x$(it's important that it exists). Similar analysis holds for the y-axis. The Cauchy-Riemann equations yield the following system : $$x^{2}+y^{2}=y^{2}+x^{2}$$ $$xy=-xy$$ That means the the necessary conditions are met only along the lines $x=0$ or $y=0$. The partials of $u$ and $v$ are elementary functions and as such are continuous on their whole domain(which includes the axes). I have, thus proven that the function IS differentiable along the axes, which is contrary to the problem's statement. This seemed a bit odd, so I tried doing this analysis by definition, namely, after proving differentiability(the difference between analyticity and differentiabilty is subtle) I used the following two definitions:

Definition 1. A function is analytic in a point if it is differentiable in SOME neighborhood of that point.

This obviously holds true for every point on the axes.

Definition 2. A function is analytic in some region if it is analytic in every point of that region.

This intuitively also holds true. One point where this might break down is the fact that the axes might not even be considered regions(which are open here, in the topological sense.). By all definitions I have read and searched for, a line is to be considered an open region. Please, feel free to prove me wrong, as I have been thinking this over for a long period.

Henno Brandsma
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Emir Šemšić
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1 Answers1

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In order to be analytic at a point $z_0$, it must be complex-differentiable at every point in some open disk centered at $z_0$. (This is your definition $1$).

You've shown using the Cauchy-Riemann equations that this open disk condition doesn't hold at any $z_0$, including $z_0$ on the coordinate axes. Hence $f(z)$ is analytic nowhere.

Zarrax
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  • Actually he seems to have shown that the C-R equations do hold at real and purely imaginary $z_0$, and the function is therefore complex differentiable on the axes. However, that is not enough to fit a neighborhood, so it is still not complex analytic. – hmakholm left over Monica Nov 01 '15 at 13:47
  • @HenningMakholm A neighborhood is ought to be a disk? If so, then it is obvious that it is not analytic. The definition which confused me is the one which stated the it can be an arbitrary neighborhood, which by analogy to the one dimensional real-case I took to be a line segment – Emir Šemšić Nov 01 '15 at 13:54
  • @HenningMakholm When I say "this condition" I mean being complex-differentiable at every point in some open disk centered at z_0. – Zarrax Nov 01 '15 at 13:57
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    @Shemafied: A neighborhood of $x$ is an open set that contains $x$ (or, for some authors, any superset of an open set that contains $x$). And the correct analogy to the real case is: An open set is one that contains a ball around each of its elements. – hmakholm left over Monica Nov 01 '15 at 13:57
  • Thank you for the clarification. I am just being super pedant about this subject. – Emir Šemšić Nov 01 '15 at 14:07
  • @Shemafied.In complex analysis, it is important to understand that "open" ,"closed"and "nbhd" usually mean with respect to the topology of the complex plane,unless stated otherwise. In topology a nbhd of a point means an open set containing that point. In the complex plane a set is open iff it covers some open disc about each of its points. Your Q looks like it was designed to test your knowledge of this. – DanielWainfleet Nov 01 '15 at 16:14
  • @user254665 Indeed, if one is not careful enough, they end up making false assumptions. – Emir Šemšić Nov 01 '15 at 16:16