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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?

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    $\begingroup$ "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 $\endgroup$
    – Fedor Petrov
    Commented Jun 4, 2022 at 10:05

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"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 $y>1/n$.

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    $\begingroup$ 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. $\endgroup$
    – Christian Remling
    Commented Jun 4, 2022 at 16:30
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    $\begingroup$ @Christian With such notion, simply interchange $x$ and $y$ and subtract $\endgroup$
    – Fedor Petrov
    Commented Jun 4, 2022 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. $\endgroup$
    – Christian Remling
    Commented Jun 4, 2022 at 17:35
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    $\begingroup$ @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 $|f_n'(x)-f_n'(y)|<2\varepsilon$ $\endgroup$
    – Fedor Petrov
    Commented Jun 4, 2022 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
    Commented Jun 5, 2022 at 0:28

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