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Definition: Let $f$ be a real valued function. We say $f$ is $\mathbf{strongly}$ $\mathbf{differentiable}$ at $x = a$ if the following limits exists and is finite:

$$ \lim_{x \to a, y \to a, x \neq y} \frac{ f(x)-f(y)}{x-y} = f^*(a) $$

and we can $f^*(a)$ the strong derivative of $f$ at $a$. Why is this definition of derivative different than the usual one? What is the main crucial point to understand here that makes it different?

1 Answers1

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Here is roughly what every student of analysis should know.

This "strong" derivative was introduced by Peano in 1892 as a "strict derivative". This asks rather more of a function than it merely have an (ordinary) derivative and Peano thought that this was actually better for students and engineers to learn and use. I prefer his terminology since the word "strong" gets rather overused in analysis and interferes with the more popular usages.

PEANO G.: Sur la définition de la dérivée, Mathesis, (2) 2 (1892), 12—14.

For a continuous function $f$ the strict (strong) derivative $f^*(x_0)$ exists at a point if and only if one (and hence all four) of the Dini derivatives $D^+f(x)$, $D_+f(x)$, $D^-f(x)$ or $D_-f(x)$ is continuous at $x_0$.

In particular, if $f'(x)$ is continuous at a point $x_0$ then the strict derivative $f^*(x_0)$ exists and, of course, is equal to the ordinary derivative $f'(x_0)$. If $f'(x)$ exists in some neighborhood of the point $x_0$ then the strict derivative at $x_0$ exists if and only if $f'$ is continuous at $x_0$.

For a bibliography of papers on the subject the ever-reliable Dave Renfro has supplied quite a few in his StackExchange answer.


Here are some considerations if you wish to decide whether you would prefer all your derivatives to be strong ones (as Peano did).

If $f'(x_0)$ exists you can be sure that $f$ is continuous at $x_0$, but it could be discontinuous and quite pathological everywhere else. But if $f^*(x_0)$ exists then you can be certain that $f$ is not only continuous at $x_0$, it is continuous in some neighborhood $(x_0-\delta,x_0+\delta)$. But way more than that: it is even Lipschitz in $(x_0-\delta,x_0+\delta)$.

If $f'(x_0)$ exists then, as already noted there doesn't have to be a derivative at any other point. But if $f^*(x_0)$ exists then there is some neighborhood $(x_0-\delta,x_0+\delta)$ in which $f$ has almost everywhere a derivative and that derivative is continuous at $x_0$ (i.e., continuous relative to the set of points at which it exists).

If you prefer espresso to green tea, a robust merlot to a sauvignon blanc, and a rare steak to a fillet of sole, you would probably like Peano's idea of using strong derivatives in place of their wimpy cousins (the ordinary derivative) when you have to teach the calculus. Especially to engineers.

  • How is it that we know strong differentiation at a point implies a neighborhood of differentiation around that same point? I see that stated here (and in other places) but never with any explanation or justification as to why it is so. –  Mar 12 '19 at 02:16
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    It does not imply differentiability in a neighborhood of the point. It does (easily) imply a Lipschitz condition in some neighborhood though. – B. S. Thomson Mar 13 '19 at 16:49
  • Okay. Your next-to-last paragraph seemed to imply this but maybe I am not reading it properly. As far as being continuous and Lipschitz in some neighborhood... what is the justification behind this? Or can you provide a link to an explanation? I am currently working on a research article on the strong derivative and this is one of the things I am struggling to convince myself of. –  Mar 13 '19 at 20:14
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    Your question (second sentence) is too easy for a hint. That means it follows easily from the definition itself and you need to think about that yourself for a research article. In any event see the link to the Dave Renfro piece: you can use his references for your article. – B. S. Thomson Mar 13 '19 at 20:24