Intuition behind the following: If $\lim_{x \to c} f'(x)=L$ and $L$ is finite, then $f’$ is continuous at $c$.

real-analysissolution-verification

Theorem: Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. If $\lim_{x \to c} f'(x)=L$ and $L$ is finite, then $f'$ is continuous at $c$.

Proof:

let $c \in (a,b)$

$\lim_{x \to c} f'(x)=L \implies \forall \epsilon > 0 \hspace{.4cm} \exists \delta_1>0 \hspace{.2cm}$ s.t. $\hspace{.2cm} |x-c|<\delta_1 \implies |f'(x) – L|<\epsilon$

$f$ is differentiable at $c$ $\implies \forall \epsilon > 0 \hspace{.4cm} \exists \delta_2>0 \hspace{.2cm}$ s.t. $\hspace{.2cm} |x-c|<\delta_2 \implies |\frac{f(x)-f(c)}{x-c} – f'(c)|<\epsilon$

let $\epsilon>0$. then, $\epsilon/2>0$. for this particular $\epsilon/2$, let $\delta_1,\delta_2$ be determined by the above. let $\delta= \min \{\delta_1,\delta_2\}$.

let $\beta \in (c-\delta, \delta+c)$ such that $\beta > c$.

by M.V.T. there exists $\alpha \in (c,\beta)$ such that $\frac{f(\beta)-f(c)}{\beta-c}=f'(\alpha)$. thus, $|\beta-c|<\delta \le \delta_2$ implies $|f'(\alpha)-f'(c)|<\epsilon/2$

$|\alpha-c|<\delta$ implies $|f'(\alpha)-L|<\epsilon/2$

by triangle inequality, $|f'(c) -L|\le|f'(\alpha)-f'(c)|+|f'(\alpha)-L|<\epsilon$

so for all $\epsilon>0, \hspace{.2cm}$ $|f'(c) -L|<\epsilon$. hence, $f'(c)=L$

therefore, $f'$ is continuous at $c$.

Firstly, is this proof correct? If not, please explain why and ignore what is written below. Otherwise, please address the following:

I was able to successfully prove the statement, but I am not really sure that I understand what I have written. I know this may seem like an odd question, but can you explain my proof to me? Can you provide intuition, perhaps in the form of pictures, that helps me grasp this at a conceptual level? Moreover, what is the intuition behind this additional assumption $\lim_{x \to c} f'(x)=L$ in the statement of the theorem? I was able to find a counter example, but, again, I am not sure I intuitively grasp the need for this assumption.

Best Answer

Intuition? In this case the key to the intuition is having an understanding of what properties a derivative must have. This is not a small topic. It is not an obvious problem to state necessary and sufficient conditions in order that a function should be the derivative of some other function on an interval.

But there are two baby steps that we should all know. (i) A derivative $f'$ must belong to the first class of Baire, and (ii) A derivative $f'$ is necessarily a Darboux function (i.e., it satisfies the intermediate value theorem just like a continuous function would).

It is the Darboux property that you need here. This is not always taught in elementary courses although it should be.

See, for example, Section 7.9 The Darboux Property of the Derivative in the reference [1] below if you have been missing this. It is a simple consequence of the mean-value theorem. Indeed your proof essentially just uses the mean-value theorem instead of directly using the Darboux property.

Now it is a simpler task to prove your theorem. Just prove this instead:

Theorem. Let $g:(a,b)\to \mathbb R$ be a Darboux function. Then $g$ is continuous at a point $c$ if and only if $\lim_{x\to > c}\,\, g(x)$ exists.

Proof. Easy.

Corollary. Let $f:(a,b)\to \mathbb R$ be a differentiable function. Then $f'$ is continuous at a point $c$ if and only if $\lim_{x\to c} f'(x)$ exists.

Proof. Immediate since $f'$ is a Darboux function.

REFERENCE: [1] https://classicalrealanalysis.info/documents/TBB-AllChapters-Landscape.pdf

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