Re: Are functions with null derivative locally constant?



In article <eu43qt$idn$1@xxxxxxxxxxxxxxxx>,
lrudolph@xxxxxxxxx (Lee Rudolph) wrote:

The World Wide Wade <aderamey.addw@xxxxxxxxxxx> writes:

In article <56l2oeF29obpuU1@xxxxxxxxxxxxxxxxxx>,
Jose Carlos Santos <jcsantos@xxxxxxxx> wrote:

Hi all:

Let U be a set of complex numbers and let _f_ be a differentiable
function from U into C. My question is: if f' is the null function, then
must _f_ be locally constant? By "locally constant" I mean that, for
each _z_ in U, there is some neighborhood V of _z_ in U such that the
restriction of _f_ to V is constant.

In order to eliminate as much ambiguities as possible, here is the
concept of "differentiable" that I am working with: _f_ is
differentiable if it is differentiable at each point of U and _f_ is
differentiable at a point _z_ in U if _z_ is a non-isolated point of
U and if the limit lim_{w -> z}(f(w) - f(z))/(w - z) exists.

Of course, the answer to the question that I asked above is well-known
to be affirmative if U is assumed to be an open set.

Best regards,

Jose Carlos Santos

No, in fact there is a strictly increasing function on R, differentiable
everywhere, whose derivative = 0 at each rational.

But Jose's question seems to be about complex-differentiability (since
he writes the difference quotient in complex terms). Thus (if I'm
reading him right) _f_ has to be defined on a neighborhood of U
(though _a priori_ only complex-differentiable at every point of U).
Since I got so tangled up when I tried to talk about residues recently,
I'll stop here.

I read it as f : U -> C, U contains no isolated points, and for each z
in U, lim_{w -> z}(f(w) - f(z))/(w - z) exists as w -> z within U. But
there is no problem satisfying the requirement you mention: Let f be the
function on R I cited, and define F(x+iy) = f(x). Now F is defined on
all of C and it's easy to check that F'(z) = 0 for each z in Q x R,
where F'(z) is the full complex derivative. Clearly F is nonconstant in
every neighborhood of a point in Q x R.
.



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