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Let $\{f_n\}$ a sequence of function differentiable at $x_0$

We have equidifferentiability at $x_0$, if

$\lim_{h \to 0} \max_n \left| \frac{f_n(x_0+h) - f(x_0)}{h}- f'_n(x_0)\right| = 0$

Are the two following statement true?

(1?) The family is equidifferentiable at $x_0$ iff the derivatives $f'_n(x_0)$ are equicontinuous at $x_0$.

(2?) If the derivatives $f'_n(x_0)$ are uniformly bounded, then $f_n(x_0)$ is equicontinuous at $x_0$

The result would be motivated by similar results when studying a single function $f$ over multiple $x$,

(1') A function $f$ is uniformly differentiable iff its derivative is uniformly continuous.

(2') If the derivative of $f$ is bounded, then $f$ is uniformly continuous.

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  • $\begingroup$ The if part follows directly from the fundamental theorem of calculus and continuity. The only if part looks non trivial to me. $\endgroup$
    – user251257
    May 31, 2018 at 0:01
  • $\begingroup$ I don't think the "only if" part is true. The functions may not even be differentiable in a neighborhood of $x_0$, much less have $f'_n(x_0)$ be equicontinuous. $\endgroup$ May 31, 2018 at 2:17

1 Answer 1

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For (1), for the if part,

A. We will prove the stronger statement for $\mathcal{A}$ an arbritrary set:

If $f'_n(x)$ is equicontinuous at every $x \in \mathcal{A}$ uniformly in $x$, then $f_n(x)$ is equidifferentiable at every $x \in \mathcal{A}$ uniformly in $x$.

Proof. From the uniformly equicontinuity of $f'_n(x)$, for any $\epsilon > 0$, there is a $\delta > 0$ such that for any $x \in \mathcal{B}$, $|x - y| < \delta$ implies $f'_n(y)$ exists for all $n$, and

$|f_n'(y) - f'_n(x)| < \epsilon$, if $|y - x| < \delta$

Note that from the mean value theorem,

$\frac{ f_n(y) - f_n(x) }{ y - x } = f'_n(\tilde{x})$

where $|\tilde{x}-x| < \delta$. Therefore, for any $(x,y)$ with $x \in \mathcal{A}$ and $|x - y| < \delta$,

$| \frac{ f_n(y) - f_n(x) }{ y-x } - f'_n(x)| = |f'_n(\tilde{x}) - f'_n(x)| < \epsilon$

which proves the statement. Taking $\mathcal{A} = \{x_0\}$ proves the original statement.

B. The "only" part is not true in general; just consider a family of functions $f_n = f$ for every $n$ with $f$ only differentiable at $x_0$, in which case the derivative cannot be continuous at $x_0$ but the $\{f_n\}$ are equidifferentiable.

However, we will prove the following statement,

If $f_n(x)$ is uniformly equidifferentiable on $\mathcal{A}$, then $f_n'(x)$ is uniformly equicontinuous on $\mathcal{A}$

Proof. From the uniform equidifferentiability on $\mathcal{A}$, there exist a $\delta > 0$ such that for $|x - y| < \delta$ and $x$ and $y$ in said set,

$\Big| \frac{f_n(x)-f_n(y)}{x-y}-f'_n(x) \Big|\lt\epsilon/2$

$\Big| \frac{f_n(x)-f_n(y)}{x-y}-f'_n(y) \Big|\lt\epsilon/2$

Hence, with the triangle inequality

$|f'_n(x)-f'_n(y)| \le \Big| \frac{f_n(x)-f_n(y)}{x-y}-f'_n(x) \Big| + \Big| \frac{f_n(x)-f_n(y)}{x-y}-f'_n(y) \Big| < \epsilon$

for $x$ and $y$ in $\mathcal{A}$, which proves uniform equicontinuity on $\mathcal{A}$.

Note that a sufficient condition for equicontinuity of $f'_x(x)$ at $x_0$ is uniform equidifferentiability of $f(x)$ on a ball centered at $x_0$.

For (2)

It isn't true in general. Take $f_n(x) = n x^2$. It has derivative zero at $x_0 = 0$, but we have,

$f_n(x_0+\delta) - f_n(x_0) = n\delta^2$

which goes to infinity as $n \to \infty$ for any $\delta \ne 0$.

However, we will prove the following statement:

If derivatives $f'_n(x)$ are uniformly bounded on a convex set $\mathcal{C}$, then $f_n(x)$ is uniformly equicontinuous on $\mathcal{C}$.

Proof. Let $M$ be the bound. We have, for $x$ and $y$ in $\mathcal{C}$,

$|f_n(y) - f_n(x)| \le |y - x| M$

which means $f_n$ is equi-Lipschitz on $\mathcal{C}$.

Therefore, for $x$ and $y$ in $\mathcal{C}$, $|x - y| < M^{-1}\epsilon$ implies $|f_n(y) - f_n(x)| < \epsilon$, which proves the result.

Hence, if the derivatives are bounded on a ball centered at $x_0$, $f_n(x)$ is equicontinuous at $x_0$.

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  • $\begingroup$ Nice${}{}{}{}{}$ $\endgroup$
    – GFauxPas
    May 31, 2018 at 4:15

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