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)
Best Answer
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.