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?
Best Answer
"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 $y>1/n$.