Complex analysis: I have to find sequence of C^inf functions that

QIsReluctant
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Homework Statement


... if fj are holomorphic on an open set U and fj \stackrel{uniformly}{\rightarrow} f on compact subsets of U then δ/δz(fj) \stackrel{uniformly}{\rightarrow} δ/δz(f) on compact subsets of U. Give an example to show that if the word "holomorphic" is replaced by "infinitely differentiable" then the result is false.

Homework Equations


The above.


The Attempt at a Solution


I've used the disk D(0, 1) and all of the obvious choices: |z|, \overline{z}, etc. None of them work. It seems that their derivatives exhibit similar behaviour to the functions themselves, i.e., converge uniformly on that disk. I've considered some wackier functions like ln(z) but finding the real and imaginary parts of the function and doing partial derivatives is, shall we put it mildly, a chore. Can anyone hrlp me please?
 
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Think about the real function ##f_n(x)=e^{-\frac{1}{nx}}## for x>0 and ##f_n(x)=0## for x<=0. Can you prove that's infinitely differentiable? What the limit function like as n->infinity?
 
Thanks for the response, Dick.

The problem is that in complex analysis "infinitely differentiable" means something different: if a function is of the form f(z) = u + iv then the two "parts" have to have continuous partial derivatives w/r to x (real part) and y (imaginary part).

Unfortunately your example takes a while to put into the above format, and I fear that all my work will be for nothing ...
 
Apologies if you know all of that, by the way, and I'm just not seeing what's obvious!
 
QIsReluctant said:
Thanks for the response, Dick.

The problem is that in complex analysis "infinitely differentiable" means something different: if a function is of the form f(z) = u + iv then the two "parts" have to have continuous partial derivatives w/r to x (real part) and y (imaginary part).

Unfortunately your example takes a while to put into the above format, and I fear that all my work will be for nothing ...

I guess I don't see the problem. Just extend the real function to the complex plane. Define u to be my ##f_n## and v to be 0.
 
... I am officially an idiot. Thank you.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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