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

traditionally,

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

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
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up vote 7 down vote accepted

I'm assuming that $m(B(x,r))$ means $r$, because I can't make the question sensible any other way.

If the first limit exists then so does the second and they are equal. However, the second limit can exist while the first one doesn't.

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 $0$.

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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
3  
I thought of that, but then the formula is off by a factor of 2, right? –  David Speyer Apr 10 '11 at 17:15
    
sorry i fixed it –  jack Apr 10 '11 at 17:34
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