# what is f prime?

currently taking Measure and Integration course, which seems to have a different definition of f'.

$$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$

but in folland's book, it seems to be defined as

$$f'(x)=\lim_{r\to 0} \frac{f(x+r)-f(x-r)}{m(B(x,r))}$$

i was just wondering if these 2 definitions are really the same thing. thanks in advance

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Which of Folland's books? Real Analysis? And where in the book? – Hans Lundmark Apr 10 '11 at 17:24
it's just like $$\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}$$ which is the derivative, if the derivative exists – yoyo Apr 10 '11 at 18:22

I'm assuming that $m(B(x,r))$ means $r$, because I can't make the question sensible any other way.
For the first claim, note that $$\frac{f(x+r/2) - f(x-r/2)}{r} = \frac{1}{2} \left( \frac{f(x+r/2) - f(x)}{r/2} + \frac{f(x) - f(x-r/2)}{r/2} \right)$$ and use the standard result that, if $\lim_{t \to 0} g(t)$ and $\lim_{t \to 0} h(t)$ exist, then $\lim_{t \to 0} g(t) + h(t)$ exists and is the sum of the previous limits.
For the second claim, let $f(x) = |x|$. Then the derivative, defined in the usual sense, does not exist, but Folland's limit is $0$.
I think $m(B(x,r))$ is supposed to be the measure of the open ball of radius $r$ centered at $x$, which would be...$2r$? – REDace0 Apr 10 '11 at 15:54