On a take home exam, I gave a function of the type: and asked the students to explain why such a function was continuous everywhere but not analytic anywhere.

This really isn’t hard but that got me to thinking: if is analytic at and NON CONSTANT, is ever analytic? Before you laugh, remember that in calculus class, is differentiable wherever .

Ok, go ahead and laugh; after playing around with the Cauchy-Riemann equations at bit, I found that there was a much easier way, if is analytic on some open neighborhood of a real number.

Since is analytic at , real, write and then compose with and substitute into the series. Now if this composition is analytic, pull out the Cauchy-Riemann equations for the composed function and it is now very easy to see that on some open disk which then implies by the Cauchy-Riemann equations that as well which means that the function is constant.

So, what if is NOT on the real axis?

Again, we write and we use to denote the partials of these functions with respect to the first and second variables respectively. Now . Now turn to the Cauchy-Riemann equations and calculate:

Insert into the Cauchy-Riemann equations:

From this and from the assumption that we obtain after a little bit of algebra:

This leads to which implies either that is zero which leads to the rest of the partials being zero (by C-R), or this means that which is absurd.

So must have been constant.

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[…] realize that what I did in the previous post was, well, lame. The setting: let be continuous but non-analytic in some disk in the complex […]

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