Can a function be differentiable but not strongly differentiable (Knuth)?

Question

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?

Answer 1

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).

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 $.