Background: Let us suppose $f\colon \text{dom}(f)\subseteq \mathbb{R} \rightarrow \mathbb{R}$ is a function with $a \in \text{int}(\text{dom}(f))$.

Note: (condition (2.) is due to Carathéodory's in (Funktionentheorie,Erster Band,p.121,1950), thanks to Tony Piccolo for this reference)

Claim: The following two conditions are equivalent:

  1. $f$ is differentiable at $a$ with $f'(a)=\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}=N$
  2. There exists a function $\phi$ which is continuous at $a$ with $\lim_{x \rightarrow a}\phi(x)=\phi(a)=N$ and $f(x)=f(a)+\phi(x)(x-a)$ for $x \neq a$.

Let me establish the equivalence above since it is important towards understanding my question.

Proof: To show $(1.) \Rightarrow (2.)$ suppose $f'(a)=N$ exists and define, for $x \neq a$, $$ \phi(x) = \begin{cases} \frac{f(x)-f(a)}{x-a} & x \neq a \\ N & x=a \end{cases}$$ Observe that, by construction, $f(x)=f(a)+\phi(x)(x-a)$ for $x \neq a$. Continuing, the continuity of $\phi$ at $a$ naturally follows from the differentiability of $f$ at $a$: $$ \lim_{x \rightarrow a} \phi(x) = \lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}=N = \phi(a). $$ To show $(2.) \Rightarrow (1.)$ suppose condition (2.) holds true. Observe that for $x \neq a$ we may solve the given $f(x)=f(a)+\phi(x)(x-a)$ for $\phi(x) = \frac{f(x)-f(a)}{x-a}$. Hence: $$ \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a} = \lim_{x \rightarrow a} \phi(x) = N $$ We thus identify $f$ is differentiable at $a$ and $f'(a)=N$ as desired. $\Box$

With the result above settled, we can prove that differentiability of $f$ at $a$ implied continuity of $f$ at $a$ as follows: use (2.) to calculate: \begin{align} \lim_{x \rightarrow a}f(x) &= \lim_{x \rightarrow a} \left[ f(a)+\phi(x)(x-a) \right] \\ &= \lim_{x \rightarrow a} f(a)+\lim_{x \rightarrow a}\phi(x)\lim_{x \rightarrow a}(x-a) \\ &=f(a)+f'(a)(a-a) \\ &=f(a). \end{align} Therefore $f$ is continuous at $a$.

Question: is the argument above that differentiable implies continuous circular? Here I mean for us to assume that the Claim is proved as I argued.

I talked myself into thinking it the other day, but now that I come back, I can't see the flaw. Thanks in advance for your insights.

  • $\begingroup$ Your condition 2. is the Carathéodory's formulation (Funktionentheorie,Erster Band,p.121,1950). $\endgroup$ – Tony Piccolo Sep 8 '13 at 23:05
  • $\begingroup$ @TonyPiccolo thanks for the reference! I read this result in a real analysis text which I have since given to a student, I was aware of this, but leary of including the details as I was uncertain. I will edit the post to give credit where credit is due. $\endgroup$ – James S. Cook Sep 9 '13 at 0:50

It looks just fine to me! Nice work.

  • $\begingroup$ thanks for looking over it for me. I do appreciate it. $\endgroup$ – James S. Cook Sep 9 '13 at 1:10

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