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We define the difference quotient of a function $f$ at $x$ in the direction $l$ as $$\triangle^h_l f(x) = \frac{f(x+he_l)-f(x)}{h}$$ where $e_l$ is the unit vector in the direction $l$.

The book I am reading claims the following : Let $f \in C^{1,\alpha}(\Omega)$ (a $C^1$ function also in the Hölder space, but I think all the claim is using is $C^1$). Then $$\triangle^h_l f(x) = \frac{1}{h}\int_0^1{\frac{d}{dt}f(x+the_l)} = \int_0^1{D_lf(x+the_l)}$$ where $D_l$ is the partial with respect to $x_l$. The first equality comes from $C^1$ is easy to see. How did they get the second equality? You need to take a limit for this, and we certainly didn't do that here. Am I missing something obvious?

(The reference being used is Gilbarg-Trudinger p. 110)

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From what I see, it is simply the chain rule. (I am not sure if I understand your question correctly.) –  Tunococ Sep 11 '12 at 18:11
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Nice nickname! :-) I hear Euler is always partying with his good friend Elvis Presley. –  Giuseppe Negro Sep 11 '12 at 19:23

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up vote 1 down vote accepted

Both equalities come from the fundamental theorem of calculus.

Let $\phi(t) = f(x + t h e_l)$. Then $\phi'(t) = D f(x+ t h e_l) (h e_l) = h D f(x+ t h e_l) e_l= h D_l f(x+the_l) $.

Hence $\frac{f(x+he_l)-f(x)}{h} = \int_0^1 \frac{\phi'(t)}{h} dt = \int_0^1 D_l f(x+the_l) dt$.

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Thanks! ~~~~~~~~ –  Euler....IS_ALIVE Sep 11 '12 at 19:09
    
You're welcome! –  copper.hat Sep 11 '12 at 19:44

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