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Here is the pertinent section: enter image description here It seems in the statement of the theorem he has the unnecessary assumption "suppose $f$ is continuous on $[a,b]$". I can't spot where it is used (obviously $f$ is continuous at $x$, but perhaps it's possible that $\forall \epsilon >0 \exists y \in (x-\epsilon, x+\epsilon)$ s.t. $f$ is not continuous @ y, or something to that effect). This assumption is not stated in Wolfram Mathworld or Wikipedia's statement of the chain rule either. So is it necessary? Is it possible to have $f$ differentiable at $x$ but not continuous on any (non-trivial, we have to take $a\neq b$ for the derivative to be defined) interval, and then chain rule doesn't hold?

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  • $\begingroup$ The last sentence says "...by the continuity of $f$..." $\endgroup$ – Randall Jul 15 at 13:42
  • $\begingroup$ Continuity of $f$ @ $x$, which is implied by differentiability of $f$ @ $x$. $\endgroup$ – Keefer Rowan Jul 15 at 13:45
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    $\begingroup$ Hmmm, maybe. Let me think. $\endgroup$ – Randall Jul 15 at 13:46
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I think it is unnecessary too. Every statement of the chain rule I've seen runs like:

Suppose that $f$ and $g$ are defined on open sets (of the real line, or more generally a Banach space) such that $g \circ f$ makes sense. Suppose also that $f$ is differentiable at $x$, and $g$ is differentiable at $f(x)$. Then $g \circ f$ is differentiable at $x$ and \begin{align} (g \circ f)'(x) = g'(f(x)) \cdot f'(x) \end{align} (In single variable calculus, $\cdot$ is multiplication of real numbers; in higher dimensions, we might replace it with a composition $\circ$)

So, we only need the domains of the functions to be nice, so that compositions make sense etc, and for the functions to be differentiable at the respective points. We do not need any further continuity assumptions on the entire domain.

Also, in the proof (I only gave it a quick read), it seems like Rudin only uses continuity of $f$ at the particular point $x$ (rather than the entire interval), which as you mentioned follows from the assumption that $f$ is differentiable at $x$.


By the way, here's an example of a pretty badly behaved function: let $f: \Bbb{R} \to \Bbb{R}$ be defined by \begin{align} f(x) = \begin{cases} x^n & \text{if $x$ is irrational} \\ 0 & \text{if $x$ is rational} \end{cases} \end{align} where $n \geq 2$. Then, $f$ is differentiable at $0$ with $f'(0) = 0$, but if $x \neq 0$, then $f$ is not even continuous at $x$. (This example is taken from Spivak's Calculus, page $413$, 3rd Edition).

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    $\begingroup$ Huh, you'd think that'd be correcting in a book as well known as PMA. $\endgroup$ – Keefer Rowan Jul 15 at 14:36

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