# Analysis Question-Schwarz Derivative

1. Nov 21, 2009

1. The problem statement, all variables and given/known data
Def. f is Schwarz Differentiable at a pt c in its domain if

lim(h->0) [f(c+h)-f(c-h)]/2h exists as a finite limit.

1.)Prove or disprove: f is differentiable at c => f is Schwarz Differentiable at c
2.)Prove or disprove: f is Schwarz Differentiable at c => f is differentiable at c

2. Relevant equations

a function f is differentiable at c if lim(x->c) [f(x)-f(c)]/[x-c] exists

3. The attempt at a solution

My suspicion, from picturing each derivative, is that 1 is true and 2 is false. To prove 1, I've tried to set h=|x-c| and evaluate the derivative for x>c, and x<c and then use some linear combination of those limits to derive the schwarzian derivative, but I keep running into problems with extra terms.

Last edited: Nov 21, 2009
2. Nov 21, 2009

### snipez90

$$f'(c) = \lim_{h \rightarrow 0}\frac{f(c + h) - f(c)}{h} = \lim_{h \rightarrow 0}\frac{f(c - h) - f(c)}{-h} = \lim_{h \rightarrow 0}\frac{f(c) - f(c + h) }{h}$$

3. Nov 21, 2009

Big Help. Thank you snipez but, as the third part of the equality, did you mean to write f(c-h) instead of f(c+h)? As in

f'(c)=~=~=lim(h->0)[f(c)-f(c-h)]/h ?

I believe that's correct, and gives me what I need to proceed to part 1.

For part two I'm trying to think of a function that will provide me with a counterexample. Maybe f=|x| a x=0?

Last edited: Nov 21, 2009
4. Nov 21, 2009

### Dick

If you think that's a good counterexample, why don't you show us why it is?

5. Nov 21, 2009

Sorry, I still had some figuring to do on my own.

Clearly, for f=|x|, f is not differentiable at c=0.

However, fs(c)=lim(h->0)[f(c+h)-f(c-h)]/2h=0, as |x| is symmetric about the y-axis. That is, f(c+h)-f(c-h) is always zero.

Seems right to me. What do you think?

6. Nov 22, 2009

### HallsofIvy

Staff Emeritus
I think that's excellent.

(I would write it specifically as fs(0)= lim(h->0) (|h|- |-h|)/2h= 0, however.)