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I believe I have a proof of the claim for functions in $\mathbb R^n$. For convenience, we restate the theorem below.

Theorem: Suppose $f_n - f \to 0$ uniformly and $f’_n - g \to 0$ in $L^\infty$. Then if $f_n$ are differentiable, so is $f$.

Below I outline a sketch of a proof, which I will progressively turn into a proper proof over the course of the next few days. Any preliminary comments are greatly appreciated! (Done, finally!)

We show, as in Pietro Majer's answer that for an everywhere differentiable function $u$ on an convex, open subset $\Omega$ of $\mathbb R^n$ that

$$\text{esssup}_{\Omega} |u'|=\sup_{\Omega} |u'|.$$

The conclusion then follows from the "Limit under the Sign of Derivative" Theorem, which is Theorem (8.6.3) in Dieudonné's Foundations of Modern Analysis.

Since

$$\sup_{\Omega} |u'| \leq \text{Lip}(u),$$

it will suffice to show that $u$ is Lipschitz continuous with Lipschitz constant $\|u'\|_{L^\infty}$.

To see this, let $x, y \in \mathbb R^n$ be arbitrary. By Fubini's theorem, for every $\varepsilon > 0$, there exist points $x_0, y_0$ that are $\varepsilon$-close to $x, y$ respectively such that, denoting by $w$ the unit vector pointing from $x_0$ to $y_0$, we have that $F (t) := (u)(x_0 + tw)$ satisfies $F'(t) \leq \|u'\|_{L^\infty}$ almost everywhere (with respect to one dimensional Lebesgue measure).

In particular since $F$ is also differentiable everywhere, the one dimensional result in Proposition 1 of Pietro Majer's answer applies, and so we have

$$u(x_0) - u(y_0) \leq \|u\|_{L^\infty} |x_0 - y_0|.$$

Consequently by the triangle inequality,

$$|u (x) - u(y)| \leq |\|u\|_{L^\infty}|x_0 - y_0| +|u(x) - u(x_0)|$$ $$ + |u(y) - u (y_0)|$$ $$ \leq |\|u\|_{L^\infty} (|x - y| + 2\varepsilon) +|u(x) - u (x_0)|$$ $$+ |u(y) - u (y_0)|.$$

Now let $\delta > 0$ be arbitrary. Assume that $\varepsilon$ is chosen smaller than $\delta |x - y|$, and small enough such that by continuity,

$$|u(x) - u (x_0)|, |u(y) - u (y_0)| < \delta|x- y|.$$

Then

$$|u (x) - u(y)| \leq (\|u\|_{L^\infty} + 4 \delta)|x - y|,$$

and sending $\delta$ to $0$, we obtain the desired Lipschitz continuity. $\square$.

I believe I have a proof of the claim for functions in $\mathbb R^n$. For convenience, we restate the theorem below.

Theorem: Suppose $f_n - f \to 0$ uniformly and $f’_n - g \to 0$ in $L^\infty$. Then if $f_n$ are differentiable, so is $f$.

Below I outline a sketch of a proof, which I will progressively turn into a proper proof over the course of the next few days. Any preliminary comments are greatly appreciated! (Done, finally!)

We show, as in Pietro Majer's answer that for an everywhere differentiable function $u$ on an convex, open subset $\Omega$ of $\mathbb R^n$ that

$$\text{esssup}_{\Omega} |u'|=\sup_{\Omega} |u'|.$$

The conclusion then follows from the "Limit under the Sign of Derivative" Theorem, which is Theorem (8.6.3) in Dieudonné's Foundations of Modern Analysis.

Since

$$\sup_{\Omega} |u'| \leq \text{Lip}(u),$$

it will suffice to show that $u$ is Lipschitz continuous with Lipschitz constant $\|u'\|_{L^\infty}$.

To see this, let $x, y \in \Omega$ be arbitrary. By Fubini's theorem, for every $\varepsilon > 0$, there exist points $x_0, y_0$ that are $\varepsilon$-close to $x, y$ respectively such that, denoting by $w$ the unit vector pointing from $x_0$ to $y_0$, we have that $F (t) := (u)(x_0 + tw)$ satisfies $F'(t) \leq \|u'\|_{L^\infty}$ almost everywhere (with respect to one dimensional Lebesgue measure).

In particular since $F$ is also differentiable everywhere, the one dimensional result in Proposition 1 of Pietro Majer's answer applies, and so we have

$$u(x_0) - u(y_0) \leq \|u\|_{L^\infty} |x_0 - y_0|.$$

Consequently by the triangle inequality,

$$|u (x) - u(y)| \leq |\|u\|_{L^\infty}|x_0 - y_0| +|u(x) - u(x_0)|$$ $$ + |u(y) - u (y_0)|$$ $$ \leq |\|u\|_{L^\infty} (|x - y| + 2\varepsilon) +|u(x) - u (x_0)|$$ $$+ |u(y) - u (y_0)|.$$

Now let $\delta > 0$ be arbitrary. Assume that $\varepsilon$ is chosen smaller than $\delta |x - y|$, and small enough such that by continuity,

$$|u(x) - u (x_0)|, |u(y) - u (y_0)| < \delta|x- y|.$$

Then

$$|u (x) - u(y)| \leq (\|u\|_{L^\infty} + 4 \delta)|x - y|,$$

and sending $\delta$ to $0$, we obtain the desired Lipschitz continuity. $\square$.

I believe I have a proof of the claim for functions in $\mathbb R^n$. For convenience, we restate the theorem below.

Theorem: Suppose $f_n - f \to 0$ uniformly and $f’_n - g \to 0$ in $L^\infty$. Then if $f_n$ are differentiable, so is $f$.

Below I outline a sketch of a proof, which I will progressively turn into a proper proof over the course of the next few days. Any preliminary comments are greatly appreciated! (Done, finally!)

We show, as in Pietro Majer's answer that for an everywhere differentiable function $u$ on an convex, open subset $\Omega$ of $\mathbb R^n$ that

$$\text{esssup}_{\Omega} |u'|=\sup_{\Omega} |u'|.$$

The conclusion then follows from the "Limit under the Sign of Derivative" Theorem, which is Theorem (8.6.3) in Dieudonné's Foundations of Modern Analysis.

Since

$$\sup_{\Omega} |u'| \leq \text{Lip}(u),$$

it will suffice to show that $u$ is Lipschitz continuous with Lipschitz constant $\|u'\|_{L^\infty}$.

To see this, let $x, y \in \mathbb R^n$ be arbitrary. By Fubini's theorem, for every $\varepsilon > 0$, there exist points $x_0, y_0$ that are $\varepsilon$-close to $x, y$ respectively such that, denoting by $w$ the unit vector pointing from $x_0$ to $y_0$, we have that $F (t) := (u)(x_0 + tw)$ satisfies $F'(t) \leq \|u'\|_{L^\infty}$ almost everywhere (with respect to one dimensional Lebesgue measure).

In particular since $F$ is also differentiable everywhere, the one dimensional result in Proposition 1 of Pietro Majer's answer applies, and so we have

$$u(x_0) - u(y_0) \leq \|u\|_{L^\infty} |x_0 - y_0|.$$

Consequently by the triangle inequality,

$$|u (x) - u(y)| \leq |\|u\|_{L^\infty}|x_0 - y_0| +|u(x) - u(x_0)|$$ $$ + |u(y) - u (y_0)|$$ $$ \leq |\|u\|_{L^\infty} (|x - y| + 2\varepsilon) +|u(x) - u (x_0)|$$ $$+ |u(y) - u (y_0)|.$$

Now let $\delta > 0$ be arbitrary. Assume that $\varepsilon$ is chosen smaller than $\delta |x - y|$, and small enough such that by continuity,

$$|u(x) - u (x_0)|, |u(y) - u (y_0)| < \delta|x- y|.$$

Then

$$|u (x) - u(y)| \leq (\|u\|_{L^\infty} + 4 \delta)|x - y|,$$

and sending $\delta$ to $0$, we obtain the desired Lipschitz continuity. $\square$.

I believe I have a proof of the claim for functions in $\mathbb R^n$. For convenience, we restate the theorem below.

Theorem: Suppose $f_n - f \to 0$ uniformly and $f’_n - g \to 0$ in $L^\infty$. Then if $f_n$ are differentiable, so is $f$.

Below I outline a sketch of a proof, which I will progressively turn into a proper proof over the course of the next few days. Any preliminary comments are greatly appreciated! (Done, finally!)

We show, as in Pietro Majer's answer that for an everywhere differentiable function $u$ on an convex, open subset $\Omega$ of $\mathbb R^n$ that

$$\text{esssup}_{\Omega} |u'|=\sup_{\Omega} |u'|.$$

The conclusion then follows from the "Limit under the Sign of Derivative" Theorem, which is Theorem (8.6.3) in Dieudonné's Foundations of Modern Analysis.

Since

$$\sup_{\Omega} |u'| \leq \text{Lip}(u),$$

it will suffice to show that $u$ is Lipschitz continuous with Lipschitz constant $\|u'\|_{L^\infty}$.

To see this, let $x, y \in \mathbb R^n$ be arbitrary. By Fubini's theorem, for every $\varepsilon > 0$, there exist points $x_0, y_0$ that are $\varepsilon$-close to $x, y$ respectively such that, denoting by $w$ the unit vector pointing from $x_0$ to $y_0$, we have that $F (t) := (u)(x_0 + tw)$ satisfies $F'(t) \leq \|u'\|_{L^\infty}$ almost everywhere (with respect to one dimensional Lebesgue measure).

In particular since $F$ is also differentiable everywhere, the one dimensional result in Proposition 1 of Pietro Majer's answer applies, and so we have

$$u(x_0) - u(y_0) \leq \|u\|_{L^\infty} |x_0 - y_0|.$$

Consequently by the triangle inequality,

$$|u (x) - u(y)| \leq |\|u\|_{L^\infty}|x_0 - y_0| +|u(x) - u(x_0)|$$ $$ + |u(y) - u (y_0)|$$ $$ \leq |\|u\|_{L^\infty} (|x - y| + 2\varepsilon) +|u(x) - u (x_0)|$$ $$+ |u(y) - u (y_0)|.$$

Now let $\delta > 0$ be arbitrary. Assume that $\varepsilon$ is chosen smaller than $\delta |x - y|$, and small enough such that by continuity,

$$|u(x) - u (x_0)|, |u(y) - u (y_0)| < \delta|x- y|.$$

Then

$$|u (x) - u(y)| \leq (\|u\|_{L^\infty} + 4 \delta)|x - y|,$$

and sending $\delta$ to $0$, we obtain the desired Lipschitz continuity. $\square$.

I believe I have a proof of the claim for functions in $\mathbb R^n$. For convenience, we restate the theorem below.

Theorem: Suppose $f_n - f \to 0$ uniformly and $f’_n - g \to 0$ in $L^\infty$. Then if $f_n$ are differentiable, so is $f$.

Below I outline a sketch of a proof, which I will progressively turn into a proper proof over the course of the next few days. Any preliminary comments are greatly appreciated! (Done, finally!)

We show, as in Pietro Majer's answer that for an everywhere differentiable function $u$ on an convex, open subset $\Omega$ of $\mathbb R^n$ that

$$\text{esssup}_{\Omega} |u'|=\sup_{\Omega} |u'|.$$

The conclusion then follows from the "Limit under the Sign of Derivative" Theorem, which is Theorem (8.6.3) in Dieudonné's Foundations of Modern Analysis.

Since

$$\sup_{\Omega} |u'| \leq \text{Lip}(u),$$

it will suffice to show that $u$ is Lipschitz continuous with Lipschitz constant $\|u'\|_{L^\infty}$.

To see this, let $x, y \in \mathbb R^n$ be arbitrary. By Fubini's theorem, for every $\varepsilon > 0$, there exist points $x_0, y_0$ that are $\varepsilon$-close to $x, y$ respectively such that, denoting by $w$ the unit vector pointing from $x_0$ to $y_0$, we have that $F (t) := (f_i - f_j)(x_0 + tw)$ satisfies $F'(t) \leq \|u'\|_{L^\infty}$ almost everywhere (with respect to one dimensional Lebesgue measure).

In particular since $F$ is also differentiable everywhere, the one dimensional result in Proposition 1 of Pietro Majer's answer applies, and so we have

$$u(x_0) - u(y_0) \leq \|u\|_{L^\infty} |x_0 - y_0|.$$

Consequently by the triangle inequality,

$$|u (x) - u(y)| \leq |\|u\|_{L^\infty}|x_0 - y_0| +|(f_i - f_j)(x) - (f_i - f_j )(x_0)|$$ $$ + |(f_i - f_j)(y) - (f_i - f_j) (y_0)|$$ $$ \leq |\|f'_i - f'_j\|_{L^\infty} (|x - y| + 2\varepsilon) +|(f_i - f_j)(x) - (f_i - f_j) (x_0)|$$ $$+ |(f_i - f_j)(y) - (f_i - f_j) (y_0)|.$$

Now let $\delta > 0$ be arbitrary. Assume that $\varepsilon$ is chosen smaller than $\delta |x - y|$, and small enough such that by continuity,

$$|(f_i - f_j)(x) - (f_i - f_j) (x_0)|, |(f_i - f_j)(y) - (f_i - f_j) (y_0)| < \delta|x- y|.$$

Then

$$|(f_i - f_j) (x) - (f_i - f_j)(y)| \leq (\|f'_i - f'_j\|_{L^\infty} + 4 \delta)|x - y|,$$

and sending $\delta$ to $0$, we obtain the desired Lipschitz continuity. $\square$.

I believe I have a proof of the claim for functions in $\mathbb R^n$. For convenience, we restate the theorem below.

Theorem: Suppose $f_n - f \to 0$ uniformly and $f’_n - g \to 0$ in $L^\infty$. Then if $f_n$ are differentiable, so is $f$.

Below I outline a sketch of a proof, which I will progressively turn into a proper proof over the course of the next few days. Any preliminary comments are greatly appreciated! (Done, finally!)

We show, as in Pietro Majer's answer that for an everywhere differentiable function $u$ on an convex, open subset $\Omega$ of $\mathbb R^n$ that

$$\text{esssup}_{\Omega} |u'|=\sup_{\Omega} |u'|.$$

The conclusion then follows from the "Limit under the Sign of Derivative" Theorem, which is Theorem (8.6.3) in Dieudonné's Foundations of Modern Analysis.

Since

$$\sup_{\Omega} |u'| \leq \text{Lip}(u),$$

it will suffice to show that $u$ is Lipschitz continuous with Lipschitz constant $\|u'\|_{L^\infty}$.

To see this, let $x, y \in \mathbb R^n$ be arbitrary. By Fubini's theorem, for every $\varepsilon > 0$, there exist points $x_0, y_0$ that are $\varepsilon$-close to $x, y$ respectively such that, denoting by $w$ the unit vector pointing from $x_0$ to $y_0$, we have that $F (t) := (u)(x_0 + tw)$ satisfies $F'(t) \leq \|u'\|_{L^\infty}$ almost everywhere (with respect to one dimensional Lebesgue measure).

In particular since $F$ is also differentiable everywhere, the one dimensional result in Proposition 1 of Pietro Majer's answer applies, and so we have

$$u(x_0) - u(y_0) \leq \|u\|_{L^\infty} |x_0 - y_0|.$$

Consequently by the triangle inequality,

$$|u (x) - u(y)| \leq |\|u\|_{L^\infty}|x_0 - y_0| +|u(x) - u(x_0)|$$ $$ + |u(y) - u (y_0)|$$ $$ \leq |\|u\|_{L^\infty} (|x - y| + 2\varepsilon) +|u(x) - u (x_0)|$$ $$+ |u(y) - u (y_0)|.$$

Now let $\delta > 0$ be arbitrary. Assume that $\varepsilon$ is chosen smaller than $\delta |x - y|$, and small enough such that by continuity,

$$|u(x) - u (x_0)|, |u(y) - u (y_0)| < \delta|x- y|.$$

Then

$$|u (x) - u(y)| \leq (\|u\|_{L^\infty} + 4 \delta)|x - y|,$$

and sending $\delta$ to $0$, we obtain the desired Lipschitz continuity. $\square$.