# Unnecessary assumption in Rudin PMA Theorem 5.5

Here is the pertinent section: 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?

• The last sentence says "...by the continuity of $f$..." – Randall Jul 15 at 13:42
• Continuity of $f$ @ $x$, which is implied by differentiability of $f$ @ $x$. – Keefer Rowan Jul 15 at 13:45
• Hmmm, maybe. Let me think. – Randall Jul 15 at 13:46

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 $$\circ$$)
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).