# Equidifferentiable functions

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?

• "if" part follows from lagrange theorem: $f_n(x)-f_n(y)=f_n'(\theta)(x-y)$ for certain $\theta$ between $x$ and $y$, and $|f_n'(x)-f_n'(\theta)|<\varepsilon$ provided that $x$ and $\theta$ are close enough Jun 4 at 10:05

"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 $$y>1/n$$.
• The "right" notion of equidifferentiable for this is perhaps the uniform version, where one requires $\delta$ to be independent of $x$ also. At least I think functions as in your example will not be equidifferentiable in this stronger sense, when one can take $x$ to be at the bump of $f_n'$ consistently. Jun 4 at 16:30
• @Christian With such notion, simply interchange $x$ and $y$ and subtract Jun 4 at 16:37
• Sorry, I don't understand your comment. If (to make this concrete) $f_n'$ goes from $0$ to $1$ and back linearly on an interval of length $1/n^2$, centered at $c_n$, then $2n^2(f_n(c_n)-f_n(c_n-1/2n^2))=1/2$, while $f_n'(c_n)=1$, so this sequence is not equidifferentiable in the stronger (uniform) sense. Jun 4 at 17:35
• @Christian I mean that the answer is positive with such understanding, because you may take two inequalities for pairs $(x, y)$ and $(y, x)$ and subtract them (or sum up?) to get $|f_n'(x)-f_n'(y)|<2\varepsilon$ Jun 4 at 18:38