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In Spivak calculus proof of the chain rule, which is essentially the same as this proof, Spivak mentioned that

the continuity of $\phi(h)$ is the crux of the whole proof

I'm not sure I see why it is the crux move in this proof, any help?

Community
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Loai Ghoraba
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2 Answers2

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I believe Spivak is alluding to the following result-

(Carathéodory formulation of the derivative) A function $f : \mathbb R \to \mathbb R$ is differentiable at $x\in\mathbb R$ iff there exists a function $\Phi:\mathbb R\to \mathbb R$ (depending on $x$) continuous at $h=0$ such that (at least near $h=0$) $$ f(x+h) = f(x) + \Phi(h)h.$$ In addition, $f'(x) = \Phi(0)$.

The equivalence is not hard to show at all, but it allows you an alternative proof of all the differentiation rules that is very mechanical. Say we write $\Phi_{f,x}(h) = \Phi_f(h)$ to make the dependence on $x,f$ explicit: $$f(x+h) = f(x) + \Phi_f(h)h,$$ $$g(x+h) = g(x) + \Phi_g(h)h.$$ It is immediate from mere algebraic properties of continuous functions (without any $\epsilon-\delta$ management) $$ (f+g)(x+h) = (f+g)(x) + \underbrace{(\Phi_{f}+\Phi_g)(h)}_{=\Phi_{f+g}(h)}h,$$ $$ (fg)(x+h) = f(x)g(x) + f(x)\Phi_g(h)h + \Phi_f(h)g(x)h + \Phi_f(h) f(x)\Phi_g(h)h^2\\ = (fg)(x) + \Big(\underbrace{f(x)\Phi_g(h) + \Phi_f(h)g(x) + \Phi_f(h) f(x)\Phi_g(h)h}_{=\Phi_{fg}(h)}\Big) h,$$ $$ \frac{1}{f(x+h)}=\frac1{f(x)} + \frac{f(x) - f(x+h)}{f(x+h)f(x)} = \frac1{f(x)} + \underbrace{\frac{-\Phi_f(h)}{f(x+h)f(x)}}_{=\Phi_{1/f}(h)} h \quad (\text{if }f(x)\neq 0),$$ and finally chain rule (this time using $\Phi_f = \Phi_{f,g(x)}$ instead of $\Phi_{f,x}$) $$ f\circ g(x+h) = f\circ g(x) + \Phi_f(g(x+h)-g(x))(g(x+h)-g(x)) \\ =f\circ g(x) + \underbrace{\Phi_f(\Phi_g(h)h)\Phi_g(h)}_{= \Phi_{f\circ g}(h)}h. $$ In this way, each rule for derivatives follows with no effort from the corresponding rule for continuous functions.

(PS there is more discussion about the Carathéodory derivative, in particular higher dimensional analogues here)

Calvin Khor
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If $\Phi$ was not continuous at 0, you wouldn't be able to write $\lim_{h \to 0} \left( \Phi(h) \cdot \frac{\Delta_h}{h} \right) = \lim_{h \to 0} \left( \Phi(h) \right) \cdot \lim_{h \to 0} \left( \frac{\Delta_h}{h}\right)$ because you wouldn't know whether $\lim_{h \to 0} \left( \Phi(h) \right)$ even exists. The lemma fortunately proves that the limit does exist, and therefore that $\Phi$ is continuous.

Edit: whether it is the "crux" of the proof I am not sure either, but it is certainly necessary for the proof to be valid.

Lashoun
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  • what I see is that the way $\phi$ is defined makes the limit exists(as proved in the lemma) and hence can used. The limit is equal to the function value (at 0), and hence continuous, great, however I don't see where we benefited from the fact that the limit is equal to the function value after that. – Loai Ghoraba Feb 12 '19 at 20:57
  • We benefited from the continuity because we could then prove that the limit of the left expression exists, hence proving that fog is differentiable. – Lashoun Feb 12 '19 at 23:19