5
$\begingroup$

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?

$\endgroup$
1
  • 5
    $\begingroup$ "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 $\endgroup$ Jun 4 at 10:05

1 Answer 1

10
$\begingroup$

"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 $y>1/n$.

$\endgroup$
5
  • 2
    $\begingroup$ 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. $\endgroup$ Jun 4 at 16:30
  • 1
    $\begingroup$ @Christian With such notion, simply interchange $x$ and $y$ and subtract $\endgroup$ Jun 4 at 16:37
  • $\begingroup$ 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. $\endgroup$ Jun 4 at 17:35
  • 2
    $\begingroup$ @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 $|f_n'(x)-f_n'(y)|<2\varepsilon$ $\endgroup$ Jun 4 at 18:38
  • $\begingroup$ Ah, I did not expect the uniform/non uniform distinction to matter. Interesting that for equicontinuity, non uniform equicontinuity implies uniform, but not for equidifferentiability! @ChristianRemling $\endgroup$
    – Nate River
    Jun 5 at 0:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.