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Synopsis

Note that there is an identical question that has no answers based off the same exercise. I hope that I can be a bit more lucky if I provide some more definitions.

The exercise is from Tao's Analysis I and asks simply to prove the chain rule, which he gives as

Let $X,Y$ be subsets of $\mathbb{R}$, let $x_0 \in X$ be a limit point of $X$, and let $y_0 \in Y$ be a limit point of $Y$. Let $f: X \to Y$ be a function such that $f(x_0) = y_0$ and such that $f$ is differentiable at $x_0$. Suppose that $g: Y \to \mathbb{R}$ is a function which is differentiable at $y_0$. The nthe function $g \circ f:X \to \mathbb{R}$ is differentiable at $x_0$, and $$(g \circ f)'(x_0) = g'(y_0)f'(x_0).$$

He recommends utilizing a solution involving a rigorous version of Newton's Approximation we proved earlier in order to avoid the proof which must take into account the case of $f'(x_0) = 0$. He defines Newton's Approximation as such:

Let $X$ be a subset of $\mathbb{R}$, let $x_0 \in X$ be a limit point of $X$, let $f:X \to \mathbb{R}$ be a function, and let $L$ be a real number. Then the following statements are equivalent

  1. $f$ is differentiable at $x_0$ on $X$ with derivative $L$.
  2. For every $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $x \in X$ and $|x-x_0| \leq \delta$, we have $$|f(x)-(f(x_0)+L(x-x_0))| \leq \epsilon|x-x_0|.$$

I am always sure to add these definitions because Tao defines things in non-standard ways and also uses notation that is a little different from other texts.

Attempt

Let $\epsilon > 0$. Note that our domain is $X$, so $g$ is restricted to $f(X)$. Hence, there exists a $\delta > 0$ such that if $|f(x) - f(x_0)| < \delta$ then $$|g(f(x)) - (g(f(x_0))+g'(y_0)(f(x)-f(x_0)))| \leq \epsilon |f(x)-f(x_0)|.$$ But I have no idea where to go from here. Note that I must show $$|g(f(x)) - (g(f(x_0))+g'(y_0)f'(x_0)(x-x_0)))| \leq \epsilon |x-x_0|$$ and I'm evidently somehow close, but how do I get $f'(x_0)$ into this? It seems like all I'm missing is one piece of the puzzle but I don't know where to find it. I can fix the right hand side pretty easily through the fact that $f$ must be continuous, but all together I'm a bit stuck on how to get that pesky $f'(x_0)$ term in this. Note that I've already tried somehow connecting Newton's approximation for $f$ into it, but I couldn't find a way. I'd appreciate any hints. Thank you.

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    $\begingroup$ Write it as $$\Bigl(g(f(x)) - \bigl(g(f(x_0)) + g'(y_0)\bigl(f(x) - f(x_0)\bigr)\bigr)\Bigr) + g'(y_0)\cdot \bigl(f(x) - f(x_0) - f'(x_0)(x - x_0)\bigr)\,.$$ For the first, you have a start, then estimate $\lvert f(x) - f(x_0)\rvert$ by a multiple of $\lvert x - x_0\rvert$ using the differentiability of $f$ at $x_0$. $\endgroup$ Commented Jan 14, 2021 at 22:24
  • $\begingroup$ Sorry for the late response. I got it. Thank you! $\endgroup$ Commented Jan 16, 2021 at 16:41
  • $\begingroup$ Since your question is about how to get the $f'(x_0)$ into the estimate, you may also want to take a look at this previous answer of mine, which is "essentially" the main idea behind the chain rule. The tough part though, of course, is making sure all the leftover terms are "small enough". $\endgroup$ Commented Jan 23, 2021 at 10:01

1 Answer 1

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This is just to follow through with the suggestion of Daniel Fischer:

Let $\varphi:=|f'(x_0)|+|g'(f(x_0))|\geq{0}$, and let $\varepsilon>0$ be arbitrary. We have that $\varphi^2<\varphi^2+4\varepsilon$, so by Proposition $6.7.3$ in the book, $\varphi<\sqrt{\varphi^2+4\varepsilon}$, thus $0<\frac{-\varphi+\sqrt{\varphi^2+4\varepsilon}}{2}:=\psi$. Since $\psi>0$ and $f$ is differentiable at $x_0$, there exists a $\delta>0$ such that for all $x\in{X}:|x-x_0|\leq{\delta}$, $|f(x)-f(x_0)-f'(x_0)(x-x_0)|\leq{\psi{|x-x_0|}}\implies{|f(x)-f(x_0)|\leq{(\psi+|f'(x_0)|)|x-x_0|}}$. Since $g$ is differentiable at $f(x_0)$, it follows that there exists a $\delta'>0$ such that for all $y\in{Y}:|y-f(x_0)|\leq{\delta'}$, $|g(y)-g(f(x_0))-g'(f(x_0))(y-f(x_0))|\leq{\psi{|y-f(x_0)|}}$. Finally, since $f$ is continuous at $x_0$, there exists a $\delta''>0$ such that for all $x\in{X}:|x-x_0|\leq{\delta''}$, $|f(x)-f(x_0)|\leq{\delta'}$.

Thus, let $x\in{X}:|x-x_0|\leq{\min(\delta,\delta'')}$. Then we have that: $$|g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))|\leq{\psi{|f(x)-f(x_0)|}}\leq{\psi(\psi+|f'(x_0)|)|x-x_0|}$$

Now observe: $$|g(f(x))-g(f(x_0))-g'(f(x_0))f'(x_0)(x-x_0)|=|g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))+g'(f(x_0))(f(x)-f(x_0)-f'(x_0)(x-x_0))|\leq{|g(f(x))-g(f(x_0))-g'(f(x_0))(f(x)-f(x_0))|+|g'(f(x_0))(f(x)-f(x_0)-f'(x_0)(x-x_0))|}\leq{\psi(\psi+|f'(x_0)|)|x-x_0|+|g'(f(x_0))|\psi{|x-x_0|}}=\psi(\psi+|f'(x_0)|+|g'(f(x_0))|)|x-x_0|$$ But it is a straightforward matter to check that $\psi(\psi+|f'(x_0)|+|g'(f(x_0))|)=\varepsilon$, as desired.

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