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A similar question was asked before (however, there were a few issues with the definitions and answer given, as I pointed out over there): Can a function be differentiable but not strongly differentiable (Knuth)?

I am looking for an example of a strongly differentiable function that is not continuously differentiable. To clarify, a function $f:(a,b) \to \mathbf{R}$ is said to be strongly differentiable at $x$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that $h \mapsto f(x+h)-f(x)-f'(x) h$ is $\varepsilon$-lipschitzian on $|h| < \delta.$ The same definition applies if $f:\mathrm{A} \to \mathrm{W}$ is a function between the open set $\mathrm{A}$ of a normed space with values in some normed space $\mathrm{W}.$

In the linked post, the family of functions $f_\alpha(x) = x^\alpha \sin x^{-1}$ can be shown to be differentiable for $1 < \alpha \leq 2$ but not strongly differentiable, and they can be shown to be continuously differentiable for $2 < \alpha$ (and therefore, strongly differentiable as well since $\mathscr{C}^1$ implies strong differentiability: Show that a function is strongly differentiable if it is continuously differentiable.). But I suppose the concept of strong differentiability is weaker than differentiability with continuity, but I cannot find nor construct any example.

NOTE: the accepted answer provides a reference where they show that Strong Differentiable and Continuously Differentiable are the same concept. (Prop. 2.56 of here)

William M.
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  • I've never thought about strongly differentiable functions, but my first try would be $$ f_\alpha = \begin{cases} x^\alpha \sin(1/x) & x \leq 0 \ x + x^\alpha \sin(1/x) & x > 0 \end{cases} $$ for $2<\alpha$ to try to shove a unit discontinuity into $Df_\alpha$ at $x=0$. – Eric Towers Jun 20 '24 at 16:00
  • @EricTowers very interesting intuition, I definitely like it, but $f_\alpha'(0)$ is required to exist. Making your example differentiable in an obvious way (e.g. $x^\beta + x^\alpha \sin(1/x)$ forces $\beta > 1$ and $f_\alpha$ to be $\mathscr{C}^1$). – William M. Jun 20 '24 at 16:16
  • Besides what others have said, the remarks and references in this MSE answer may be of interest. – Dave L. Renfro Jun 20 '24 at 20:25

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When a function $f\colon W\to Y$ is differentiable, the continuous differentiability is equivalent to the strict differentiability which means that $$\lim_{\substack{x',x''\to x\\ x'\neq x''}} \frac{f(x')-f(x'')-Df(x)(x'-x'')}{\|x'-x''\|} = 0$$ for all $x\in W$. (see, Proposition 2.56 on this book).

Suppose $f$ is strongly differentiable at $x$, then for $\epsilon>0$, there is $\delta$ such that $\phi_{x}:h\mapsto f(x+h)-f(x)-Df(x)h$ is $\epsilon$-Lipschitz on $B(x,\delta)$, then taking $x',x''\in B(x,\delta)$ with $x'\neq x''$ we have $$\|\varphi_x(x'-x)-\varphi_x(x''-x)\|\leq \epsilon\|x'-x''\|$$ but $\varphi_x(x'-x)-\varphi_x(x''-x) = f(x')-f(x'')-Df(x)(x'-x'')$. Then we have $$\frac{\| f(x')-f(x'')-Df(x)(x'-x'')\|}{\|x'-x''\|}\leq \epsilon,\forall x',x''\in B(x,\delta), x'\neq x''$$ it proves that $f$ is strictly differentiable therefore it is continuously differentiable.

juancodmw
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  • I check the book, thanks for the reference. Indeed, strong differentiability and differentiability with continuity are the same concept. The proof you wrote, I think is circular (you start with P and conclude P). – William M. Jun 20 '24 at 17:08
  • why do you say that the proof is circular? I mean, I started by supposing the strong differentiability of $f$ and I concluded the strict differentiability of $f$ – juancodmw Jun 20 '24 at 18:21
  • Okay, so you meant strict and strong differentiability to be two things. They are trivially equivalent. So you prove something trivial and not asked, but the resourced you shared answers my original question. – William M. Jun 20 '24 at 18:31