Here's a little snippet of the proof in page 20 of Spivak's Calculus on Manifolds (please ignore the first sentence):
Where equation $(5)$ is $\displaystyle\lim_{y\to b}\frac{|\psi(y)|}{|y-b|}=0$, the function $f:\mathbb R^n\to\mathbb R^m$ is differentiable at $a\in\mathbb R^n$ and $b=f(a)$. I am aware that this is a result of the continuity of $f$ at $a$ but what worries me is the implicit assumption that $0<|f(x)-b|$, while nowhere in the proof exactly declares such property of $f$. On the other hand, $\frac{|\psi(y)|}{|y-b|}$ is clearly not continuous at $b$ to allow $|f(x)-b|=0$ in the condition. How do we know that there exists a deleted $\delta$-neighborhood of $a$ small enough such that $f(x)\ne b$ everywhere?
My initial thoughts were that Spivak intended to prove the chain rule for functions $f$ such that all points $x$ sufficiently close to $a$ yield $f(x)\ne b$; since otherwise, the derivative of $f$ is zero and the chain rule trivially follows. Though, I find this to be a slightly wild assumption as it raises another question: is it necessary that the derivative of $f$ is zero if we fail to find a sufficiently small enough deleted $\delta$-neighborhood of $a$, all of whose points $x$ yield $f(x)\ne b$? I tried to prove this but to no avail. I would highly appreciate any help.
Best Answer
Since $\lim\limits_{y \to b}\dfrac{\lVert\psi(y)\rVert}{\lVert y-b\rVert} = 0$, it means that for every $\epsilon > 0$, there is a $\delta > 0$ such that for all $y\in \Bbb{R}^n$, if $0 < \lVert y-b\rVert < \delta$ then \begin{align} \left|\dfrac{\lVert\psi(y)\rVert}{\lVert y-b\rVert} - 0 \right| < \epsilon, \end{align} or equivalently, \begin{align} \lVert \psi(y)\rVert < \epsilon \cdot \lVert y-b\rVert \end{align} By definition, it is true that $\psi(b) = 0$. So, we can say that:
Here, it is necessary to keep the last inequality $\leq$ a weak one, in case $y=b$. So, sure, Spivak should have put a weak inequality as well. But I don't think the rest of the proof requires a strict inequality anywhere, because ultimately, all you wanna show is that there is a constant $K \geq 0$, such that given any $\epsilon > 0$, there is a $\delta' > 0$ such that if $|x-a| < \delta'$ then \begin{align} \lVert \psi(f(x))\rVert & \leq \epsilon K \lVert x-a\rVert. \end{align}