The following argument is used as a lemma in my recent paper with D. Azagra and A. Cappello (work in progress).

The answer is yes and it does not require modification of Theorem 1 since following stronger form of Theorem 2 follows from Theorem 1:

Theorem. If $f:\mathbb{R}^n\to\mathbb{R}$ is convex, then for almost all $x\in\mathbb{R}^n$ where $f$ is differentiable, we have $$ \lim_{y\to x}\sup_{\sigma_y\in\partial f(y)}\frac{|\sigma_y-Df(x)-D^2f(x)(y-x)|}{|y-x|}=0. $$

If we want to prove differentiability of $\partial f$ at $x$, we cn always assume that $x=0$ by placing the origin at $x$.

If $f$ is twice differentiable say, at $0$ as in Theorem 1, then we have $$ f(x)=f(0)+Df(0)x+\frac{1}{2}x^TD^2f(0)x+R(x)= f(0)+Df(0)x+\langle Ax,x\rangle +R(x), $$ where $A=\frac{1}{2}D^2f(0)$ and $R(x)=0(|x|^2)$. Note that $$ a(r):=\sup_{0<|x|\leq 2r}\frac{|R(x)|}{|x|^2}\to 0 \qquad \text{as $r\to 0^+$.} $$ Moreover, $$ |R(x)|\leq a\Big(\frac{|x|}{2}\Big)\, |x|^2\leq a(|x|)|x|^2. $$ It suffices to prove:

Proposition. If a convex function $f:\mathbb{R}^n\to\mathbb{R}$ is twice differentiable at $0$ as in Theorem 1, then $$ \lim_{x\to 0}\frac{\sigma_x-Df(0)-D^2f(0)x}{|x|}=0, $$ whenever $\sigma_x\in\partial f(x)$.

Proof. For points $x,y\neq 0$, we have $$ f(x)=f(0)+Df(0)x+\langle Ax,x\rangle +R(x), \quad f(y)=f(0)+Df(0)y+\langle Ay,y\rangle +R(y). $$ Since $f(x)+\langle \sigma_x,y-x\rangle\leq f(y)$, we have $$ \langle\sigma_x,y-x\rangle \leq f(y)-f(x)= Df(0)(y-x)+\langle A(x+y),y-x\rangle+R(y)-R(x). $$ We used here the fact that $A$ is symmetric and hence $\langle Ax,y\rangle=\langle Ay,x\rangle$. Let $$ y=x+w, \quad \text{where} \quad w=\sqrt{a(|x|)}\,|x|z,\ |z|=1. $$ Then $$ \langle\sigma_x,w\rangle\leq Df(0)w+ \langle A(2x+w),w\rangle +R(y)-R(x), $$ $$ \langle\sigma_x-Df(0)-2Ax,w\rangle\leq \langle Aw,w\rangle +R(y)-R(x). $$ If $|x|$ is sufficiently small, then $a(|x|)\leq 1$ and hence $|w|\leq |x|$, so $|y|\leq 2|x|$. Therefore, $$ |R(y)|\leq a\Big(\frac{|y|}{2}\Big)\, |y|^2\leq 4a(|x|)|x|^2, \qquad |R(y)-R(x)|\leq 5a(|x|)|x|^2. $$ Taking the supremum over all $z$ with $|z|=1$ we get $$ |\sigma_x-Df(0)-2Ax|\sqrt{a(|x|)}|x|\leq |A|a(|x|)|x|^2+5a(|x|)|x|^2, $$ and hence $$ \frac{|\sigma_x-Df(0)-2Ax|}{|x|}\leq (|A|+5)\sqrt{a(|x|)}\to 0 \quad \text{as $x\to 0$.} $$ Since $2A=D^2f(0)$, the result follows.

The following argument is used in my recent paper with D. Azagra and A. Cappello (work in progress).

The answer is yes. I asked about whether the proof of Theorem 1 can be modified so to cover Theorem 2. In fact the following stronger version of Theorem 2 is a direct cosequence of Theorem 1.

Theorem. If $f:\mathbb{R}^n\to\mathbb{R}$ is convex, and it is twice differentiable at $x$, then $$ \lim_{y\to x}\sup_{\sigma_y\in\partial f(y)}\frac{|\sigma_y-Df(x)-D^2f(x)(y-x)|}{|y-x|}=0. $$

We can always assume that $x=0$ by placing the origin at $x$. If $f$ is twice differentiable at $0$ as in Theorem 1, then we have $$ f(x)=f(0)+Df(0)x+\frac{1}{2}x^TD^2f(0)x+R(x)= f(0)+Df(0)x+\langle Ax,x\rangle +R(x), $$ where $A=\frac{1}{2}D^2f(0)$ and $R(x)=0(|x|^2)$. Note that $$ a(r):=\sup_{0<|x|\leq 2r}\frac{|R(x)|}{|x|^2}\to 0 \qquad \text{as $r\to 0^+$.} $$ Moreover, $$ |R(x)|\leq a\Big(\frac{|x|}{2}\Big)\, |x|^2\leq a(|x|)|x|^2. $$ It suffices to prove:

Proposition. If a convex function $f:\mathbb{R}^n\to\mathbb{R}$ is twice differentiable at $0$ as in Theorem 1, then $$ \lim_{x\to 0}\frac{\sigma_x-Df(0)-D^2f(0)x}{|x|}=0, $$ whenever $\sigma_x\in\partial f(x)$.

Proof. For points $x,y\neq 0$, we have $$ f(x)=f(0)+Df(0)x+\langle Ax,x\rangle +R(x), \quad f(y)=f(0)+Df(0)y+\langle Ay,y\rangle +R(y). $$ Since $f(x)+\langle \sigma_x,y-x\rangle\leq f(y)$, we have $$ \langle\sigma_x,y-x\rangle \leq f(y)-f(x)= Df(0)(y-x)+\langle A(x+y),y-x\rangle+R(y)-R(x). $$ We used here the fact that $A$ is symmetric and hence $\langle Ax,y\rangle=\langle Ay,x\rangle$. Let $$ y=x+w, \quad \text{where} \quad w=\sqrt{a(|x|)}\,|x|z,\ |z|=1. $$ Then $$ \langle\sigma_x,w\rangle\leq Df(0)w+ \langle A(2x+w),w\rangle +R(y)-R(x), $$ $$ \langle\sigma_x-Df(0)-2Ax,w\rangle\leq \langle Aw,w\rangle +R(y)-R(x). $$ If $|x|$ is sufficiently small, then $a(|x|)\leq 1$ and hence $|w|\leq |x|$, so $|y|\leq 2|x|$. Therefore, $$ |R(y)|\leq a\Big(\frac{|y|}{2}\Big)\, |y|^2\leq 4a(|x|)|x|^2, \qquad |R(y)-R(x)|\leq 5a(|x|)|x|^2. $$ Taking the supremum over all $z$ with $|z|=1$ we get $$ |\sigma_x-Df(0)-2Ax|\sqrt{a(|x|)}|x|\leq |A|a(|x|)|x|^2+5a(|x|)|x|^2, $$ and hence $$ \frac{|\sigma_x-Df(0)-2Ax|}{|x|}\leq (|A|+5)\sqrt{a(|x|)}\to 0 \quad \text{as $x\to 0$.} $$ Since $2A=D^2f(0)$, the result follows.

The answer is yes and it does not require modification of Theorem 1 since following stronger form of Theorem 2 follows from Theorem 1:

Theorem. If $f:\mathbb{R}^n\to\mathbb{R}$ is convex, then for almost all $x\in\mathbb{R}^n$ where $f$ is differentiable, we have $$ \lim_{y\to x}\sup_{\sigma_y\in\partial f(y)}\frac{|\sigma_y-Df(x)-D^2f(x)(y-x)|}{|y-x|}=0. $$

If $f$ is twice differentiable say, at $0$ as in Theorem 1, then we have $$ f(x)=f(0)+Df(0)x+\frac{1}{2}x^TD^2f(0)x+R(x)= f(0)+Df(0)x+\langle Ax,x\rangle +R(x), $$ where $A=\frac{1}{2}D^2f(0)$ and $R(x)=0(|x|^2)$. Note that $$ a(r):=\sup_{0<|x|\leq 2r}\frac{|R(x)|}{|x|^2}\to 0 \qquad \text{as $r\to 0^+$.} $$ Moreover, $$ |R(x)|\leq a\Big(\frac{|x|}{2}\Big)\, |x|^2\leq a(|x|)|x|^2. $$ It suffices to prove:

Proposition. If a convex function $f:\mathbb{R}^n\to\mathbb{R}$ is twice differentiable at $0$ as in Theorem 1, then $$ \lim_{x\to 0}\frac{\sigma_x-Df(0)-D^2f(0)x}{|x|}=0, $$ whenever $\sigma_x\in\partial f(x)$.

Proof. For points $x,y\neq 0$, we have $$ f(x)=f(0)+Df(0)x+\langle Ax,x\rangle +R(x), \quad f(y)=f(0)+Df(0)y+\langle Ay,y\rangle +R(y). $$ Since $f(x)+\langle \sigma_x,y-x\rangle\leq f(y)$, we have $$ \langle\sigma_x,y-x\rangle \leq f(y)-f(x)= Df(0)(y-x)+\langle A(x+y),y-x\rangle+R(y)-R(x). $$ We used here the fact that $A$ is symmetric and hence $\langle Ax,y\rangle=\langle Ay,x\rangle$. Let $$ y=x+w, \quad \text{where} \quad w=\sqrt{a(|x|)}\,|x|z,\ |z|=1. $$ Then $$ \langle\sigma_x,w\rangle\leq Df(0)w+ \langle A(2x+w),w\rangle +R(y)-R(x), $$ $$ \langle\sigma_x-Df(0)-2Ax,w\rangle\leq \langle Aw,w\rangle +R(y)-R(x). $$ If $|x|$ is sufficiently small, then $a(|x|)\leq 1$ and hence $|w|\leq |x|$, so $|y|\leq 2|x|$. Therefore, $$ |R(y)|\leq a\Big(\frac{|y|}{2}\Big)\, |y|^2\leq 4a(|x|)|x|^2, \qquad |R(y)-R(x)|\leq 5a(|x|)|x|^2. $$ Taking the supremum over all $z$ with $|z|=1$ we get $$ |\sigma_x-Df(0)-2Ax|\sqrt{a(|x|)}|x|\leq |A|a(|x|)|x|^2+5a(|x|)|x|^2, $$ and hence $$ \frac{|\sigma_x-Df(0)-2Ax|}{|x|}\leq (|A|+5)\sqrt{a(|x|)}\to 0 \quad \text{as $x\to 0$.} $$ Since $2A=D^2f(0)$, the result follows.

The following argument is used as a lemma in my recent paper with D. Azagra and A. Cappello (work in progress).

The answer is yes and it does not require modification of Theorem 1 since following stronger form of Theorem 2 follows from Theorem 1:

Theorem. If $f:\mathbb{R}^n\to\mathbb{R}$ is convex, then for almost all $x\in\mathbb{R}^n$ where $f$ is differentiable, we have $$ \lim_{y\to x}\sup_{\sigma_y\in\partial f(y)}\frac{|\sigma_y-Df(x)-D^2f(x)(y-x)|}{|y-x|}=0. $$

If we want to prove differentiability of $\partial f$ at $x$, we cn always assume that $x=0$ by placing the origin at $x$.

If $f$ is twice differentiable say, at $0$ as in Theorem 1, then we have $$ f(x)=f(0)+Df(0)x+\frac{1}{2}x^TD^2f(0)x+R(x)= f(0)+Df(0)x+\langle Ax,x\rangle +R(x), $$ where $A=\frac{1}{2}D^2f(0)$ and $R(x)=0(|x|^2)$. Note that $$ a(r):=\sup_{0<|x|\leq 2r}\frac{|R(x)|}{|x|^2}\to 0 \qquad \text{as $r\to 0^+$.} $$ Moreover, $$ |R(x)|\leq a\Big(\frac{|x|}{2}\Big)\, |x|^2\leq a(|x|)|x|^2. $$ It suffices to prove:

Proposition. If a convex function $f:\mathbb{R}^n\to\mathbb{R}$ is twice differentiable at $0$ as in Theorem 1, then $$ \lim_{x\to 0}\frac{\sigma_x-Df(0)-D^2f(0)x}{|x|}=0, $$ whenever $\sigma_x\in\partial f(x)$.

Proof. For points $x,y\neq 0$, we have $$ f(x)=f(0)+Df(0)x+\langle Ax,x\rangle +R(x), \quad f(y)=f(0)+Df(0)y+\langle Ay,y\rangle +R(y). $$ Since $f(x)+\langle \sigma_x,y-x\rangle\leq f(y)$, we have $$ \langle\sigma_x,y-x\rangle \leq f(y)-f(x)= Df(0)(y-x)+\langle A(x+y),y-x\rangle+R(y)-R(x). $$ We used here the fact that $A$ is symmetric and hence $\langle Ax,y\rangle=\langle Ay,x\rangle$. Let $$ y=x+w, \quad \text{where} \quad w=\sqrt{a(|x|)}\,|x|z,\ |z|=1. $$ Then $$ \langle\sigma_x,w\rangle\leq Df(0)w+ \langle A(2x+w),w\rangle +R(y)-R(x), $$ $$ \langle\sigma_x-Df(0)-2Ax,w\rangle\leq \langle Aw,w\rangle +R(y)-R(x). $$ If $|x|$ is sufficiently small, then $a(|x|)\leq 1$ and hence $|w|\leq |x|$, so $|y|\leq 2|x|$. Therefore, $$ |R(y)|\leq a\Big(\frac{|y|}{2}\Big)\, |y|^2\leq 4a(|x|)|x|^2, \qquad |R(y)-R(x)|\leq 5a(|x|)|x|^2. $$ Taking the supremum over all $z$ with $|z|=1$ we get $$ |\sigma_x-Df(0)-2Ax|\sqrt{a(|x|)}|x|\leq |A|a(|x|)|x|^2+5a(|x|)|x|^2, $$ and hence $$ \frac{|\sigma_x-Df(0)-2Ax|}{|x|}\leq (|A|+5)\sqrt{a(|x|)}\to 0 \quad \text{as $x\to 0$.} $$ Since $2A=D^2f(0)$, the result follows.