Is it absolutely true that $\frac{d}{dk}\lim_{x\to k}f(x)=0 \implies f'(x)=0$?

Question

I thought of the following setup where there is a limit inside a derivative $$\frac{d}{dk}\lim_{x\to k}f(x)=0$$

Since the derivative turns out to zero no matter what $k$ is, I think it's necessary that $f(x)$ is a constant function, but proving it is different. If $f(x)$ is assumed to be continuous, then

$$\frac{d}{dk}\lim_{x\to k}f(x)=\frac{d}{dk}f(k)=f'(k)=0$$

Then I can just relabel to say $f'(x)=0$. My question is how to reach this conclusion without the continuous assumption. Is there a stronger argument that can establish $\frac{d}{dk}\lim_{x\to k}f(x)=0 \implies f'(x)=0$?

Answer 1

Let $g(k):=\lim_{x\to k}f(x)$. Since you want $g$ to be differentiable everywhere, it has an antiderivative.

So by integrating both sides you get $g(k)=\lim_{x\to k}f(x)=c$ for some constant $c$.

But as has been shown already,$g$ being constant doesn't mean that $f$ is constant, as the limit irons away all isolated points. That is, if we define $M:= \{x\mid f(x) \neq c\}$, then if $M$ has no accumulation point, then every converging sequence $(x_n)_{n\in\mathbb N} \subseteq R$ can only contain a finite amount of elements of $M$, and therefore we have $\lim_{n\to\infty} f(x_n)=c$.

That said, even if $M$ has an accumulation point, it still can work out, e.g. if $$f(x) = \begin{cases} x, &\text{if}& \frac 1x \in \mathbb N \\ 0, &\text{else}& \end{cases}$$.

However, $M$ mustn't be uncountable: If it were, it had an accumulation point $x$.Therefore $M_L:= \{x\mid f(x) < c\}$ or $M_U:= \{x\mid f(x) > c\}$ has to be uncountable. Let wlog $M_L$ be uncountable. Rewrite it as $$ M_L=\big( M_L\cap f^{-1}((-\infty,c-1) ) \big)\cup\bigcup_{n=0}^\infty \big(M_L\cap f^{-1}( [c-\frac 1{2^n},c-\frac 1{2^{n+1}}) )\big) $$ Since the right hand side is a disjunct union of countably many elements, at least one of it has to be uncountable. Let thus for some $n$ the part $P:=\big(M_L\cap f^{-1}( [c-\frac 1{2^n},c-\frac 1{2^{n+1}}) )\big)$ be uncountable.

Then this part has an accumulation point, and therefore a converging sequence $(x_n)_{n\in\mathbb N} \subseteq P$. But $\lim_{n\to\infty} f(x_n) \le c-\frac 1{2^{n+1}}<c$.

Answer 2

Technically, you are right, and the derivative isn't $0$ everywhere if the function is discontinuous. A function is differentiable only if it's continuous by the definition of a derivative with limits (Differentiability). Differentiability implies continuity. Say your function is continuous in an interval and discontinuous at a point, then the derivative doesn't exist at the point of discontinuity and is $0$ everywhere else. As an example, let's consider a piecewise defined function $\operatorname{sgn}(x)$, the sign function: $$\operatorname{sgn}(x) = \begin{cases} 0, & \text{if } x=0\\ 1, & \text{if } x>0\\ -1, & \text{if } x<0 \end{cases}$$ Then $$\operatorname{sgn'}(x) = \begin{cases} 0, & \text{if } x=\mathbb{R}-{0}\\ \text{Doesn't exist}, & \text{if } x=0\\ \end{cases}$$ This can be understood by taking the intervals $(0, \infty)$, $(-\infty, 0)$ and ${0}$ separately for sgn(x) (the tangent for a single point doesn't exist).

The exact same argument can be applied for your case.