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Donald Knuth defined $f$ is strongly differentiable at $x$ if $$f(x+\epsilon) = f(x) + \epsilon f'(x) + \mathcal{O}(\epsilon^2)$$ for sufficiently small $\epsilon$.

What differentiable functions are not strongly differentiable? Clearly any analytic function is strongly differentiable by Taylor's Theorem with Remainder. But even the non-analytic smooth functions would seem to have remainders that are subquadratic, making them strongly differentiable. It would seem to me that to not be strongly differentiable requires a remainder that is supraquadratic but sublinear.

What is an example of a non strongly differentiable function? What is the remainder of the Taylor series? Can the remainder of a smooth function be supraquadratic?

Or perhaps the only examples are non smooth functions that still have a single derivative, perhaps constructable via integration?

SRobertJames
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For things like this, I like functions of the form $$ f ( x ) = \cases { x ^ n \sin ( 1 / x ) & $ x \ne 0 $ \\ 0 & $ x = 0 $. } $$ If $ n \leq 1 $, then this isn't even differentiable at $ 0 $. If $ 1 < n < 2 $, then it's differentiable but not strongly differentiable. If $ n = 2 $, then it's strongly differentiable but not continuously differentiable. And if $ n > 2 $, then it's continuously differentiable.

Notice that because $ f ( 0 ) $ and $ f ' ( 0 ) $ are both $ 0 $, the remainder at $ 0 $ is $ f ( \epsilon ) = \epsilon ^ n \sin ( 1 / \epsilon ) $, so wanting this to be supraquadratic but sublinear is exactly asking for $ 1 < n < 2 $.

Toby Bartels
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  • Neat. Can this happen on an interval (not just an isolated point)? – SRobertJames Dec 17 '23 at 01:47
  • @SRobertJames : No, because if a function is differentiable on an interval, then it must be continuously differentiable on a dense subset of that interval. See https://math.stackexchange.com/questions/112067/how-discontinuous-can-a-derivative-be for a characterization of how often a derivative must be continuous. (Since strong differentiability is weaker than continuous differentiability, that may have to happen even more often; I don't know any results about that.) – Toby Bartels Dec 17 '23 at 15:42
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    @SRobertJames there are two competing definitions of "strong" differentiability. Assuming (after two traslations) that $f(0)=0,$ $f$ is strong differentiable, according to Knuth [K], if $f(h)−h=O(h^2)$ whereas Cartan's definition [C] asserts that $f(h)−h$ is $\varepsilon$-lipschitzian on some (small enough) neighbourhood of zero. The link of this answer assumes [C] and proves that $\mathscr{C}^1$ implies [C]. However, the function $f(x)=|x|^{3/2}$ (given in the comments) is $\mathscr{C}^1$ and therefore [C] but it is not [K]. – William M. Jun 19 '24 at 23:57
  • ...Also, the function of this answer, namely $f(x)=x^2 \sin \frac{1}{x}$ will not satisfy [C] even though it satisfies [K]. To see they do not satisfy [C] observe we must prove that the relation "for every $\varepsilon > 0$ there exists $\delta > 0$ such that $|f(x)−f(y)|=|x^2 \sin \frac{1}{x} − y^2 \sin \frac{1}{y}| \leq \varepsilon |x−y|$ for close all $|x-y| \leq \delta$" does not hold; that is, we will show there exists some $\varepsilon > 0$ such that for all $n \in \mathbf{N}$ there are $x_n, y_n$ such that $|x_n-y_n| \to 0$ and $\frac{f(x_n)-f(y_n)}{x_n-y_n} \geq \varepsilon.$... – William M. Jun 20 '24 at 00:52
  • ....Now consider sequences $x_n^{-1} = 2\pi n + \frac{\pi}{2}$ and $y_n^{-1} = 2\pi n + \pi + \frac{\pi}{2}.$ Note that $\sin x_n^{-1}=1$ and $\sin y_n^{-1} = -1$ for all $n$ and both $x_n, y_n \to 0.$ Then, $$ \frac{f(x_n)-f(y_n)}{x_n-y_n} = \frac{x_n^2+y_n^2}{x_n-y_n} = \frac{(2\pi n + a)^{-2}+(2\pi n + b)^{-2}}{(2\pi n + a)^{-1} - (2\pi n + b)^{-1}} = \frac{(2\pi n + a)^2 + (2\pi n + b)^2}{(b-a) (2\pi n + a)(2\pi n + b)}, $$ where $a = \frac{\pi}{2}$ and $b = \pi + \frac{\pi}{2},$ so $b-a = \pi.$.... – William M. Jun 20 '24 at 00:52
  • In the above fraction only the dominant $n^2$ terms survive as $n \to \infty,$ leading to $\frac{f(x_n)-f(y_n)}{x_n-y_n} \to \frac{8 \pi^2}{\pi (2\pi)(2\pi)} = \frac{2}{\pi}.$ (The conclusion is that $f$ satisfies [K] but not [C]. And since above we showed [C] does not imply [K], it turns out [C] and [K] are completely independent, logically, from one another. They are just different definitions altogether.) – William M. Jun 20 '24 at 00:53
  • @WilliamM. : Yes, you're right. (I kept meaning to come back to this and check whether the two are really the same, but I never got to it.) I removed the now irrelevant link to the other question, but I put it here so that people will understand what your comments are referring to: – Toby Bartels Jun 23 '24 at 02:49
  • ‘You're right that smooth functions won't help here; in fact any continuously differentiable function is strongly differentiable (I think that this is an equivalent notion).’ – Toby Bartels Jun 23 '24 at 02:50