Let $f\_n: [0, 1] \to \mathbb R$ be a sequence of continuously differentiable functions. We say that the sequence $f\_n$ is *equidifferentiable* if for every $x \in [0, 1]$ and every $\varepsilon > 0$, there exists a $\delta > 0$ such that for all $n \in \mathbb N$,


$$\frac{|f\_n (x) - f\_n (y) - f\_n’(x)(x-y)|}{|x-y|} < \varepsilon$$


for all $y$ with $|x - y| < \delta$.


**Question:** Given a sequence $f\_n$ of continuously differentiable functions, is it true that $f\_n$ are equidifferentiable if and only if the sequence $f’\_n$ is equicontinuous?



Equidifferentiable functions
"If" part follows from Lagrange theorem: $f\_n(x)−f\_n(y)=f\_n'(\theta)(x−y)$ for certain $θ$ between $x$ and $y$, and $|f\_n'(x)−f\_n'(\theta)|<\varepsilon$ provided that $x$ and $\theta$ are close enough.


"Only if" part does not hold in general. Let $f\_n'$ be supported on $[1/n,1/n+1/n^2]$ and vary on this segment from 0 to 1. Then for all $x\ne 0$ the claim is obvious, since $f\_n$ are locally constant at $x$ for all large enough $n$. For $x=0$ the inequality reads as $|f\_n(y)-f\_n(0)|<\varepsilon y$, this also holds for large enough $n$, since $f\_n(y)-f\_n(0)=0$ for $y\leqslant 1/n$ and $0\leqslant f\_n(y)-f\_n(0)\leqslant 1/n^2\leqslant y/n$ for $y>1/n$.


