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Yes: Just take $u(x,y):=v(x)$.

Indeed, one can use Green's formula to show this, as is done in Christian Remling's answer.

More generally, the result holds for any convex function $v$, without requiring it to be in $C^2$ -- of course, if $u$ is not required to be in $C^2$ and if the derivative $v'(x)$ is understood as (say) the right derivative of $v$ at $x$ for $x\in[-1,1)$ (and the left derivative for $x=1$).

Indeed, with $u(x,y):=v(x)$ for all $x$ and $y$, the inequality in question is invariant with respect to taking positive/conical mixtures of functions $v$. That is, if the inequality in question holds for each suitably integrable function $v_b$ (in place of $v$) with $b$ in a set $B$, then the inequality in question holds for the positive mixture (used in place of $v$) of the form $\int_B\mu(db)\,v_b$ of the $v_b$'s, where $\mu$ is any nonnegative measure over $B$ such that the integral $\int_B\mu(db)\,v_b$ exists in a suitable sense.

It remains to note that the convex function $v$ is a positive mixture of affine functions and the functions of the form $v_a$ with $a\in(-1,1)$, where $v_a(x):=\max(0,x-a)$, and for any one of these latter functions the inequality in question is straightforward to check.

In fact, the constant factor $\frac12$ in the upper bound $\frac12\,(v'(1)-v'(-1))$ can be replaced by the better factor $\frac1\pi$ -- because for affine functions in place of $v$ one can replace $\frac12$ by $0$, and for $v=v_a$ with $a\in(-1,1)$ can replace $\frac12$ by $\frac{\sqrt{1-a^2}}\pi$. So, the worst case here is that of $v_a$ with $a=0$.

Detail: The mentioned mixture representation of the convex function $v$ is as follows: for $x\in(-1,1)$, $$\begin{aligned} &v(x) \\ &=v(-1+)+\int_{[-1,x)}dz\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,\Big(v'(-1)+\int_{[-1,z]}\,dv'(a)\Big) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}dz\,1(z<x)\,\int_{[-1,z]}\,dv'(a) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}\,dv'(a)\int_{[-1,1)}dz\,1(a\le z<x)\, \\ &=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,dv'(a)\,v_a(x) \\ &=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a(x) \\ \end{aligned}$$ where $\mu_{v'}$ is the (nonnegative) Lebesgue--Stieltjes measure corresponding to the nondecreasing function $v'$. So, $$v=C+A\,\text{id}+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a, $$ where $C:=C_v:=v(-1+)+v'(-1)$ and $A:=A_v:=v'(-1)$ are real numbers.

Yes: Just take $u(x,y):=v(x)$, which will be assumed in what follows.

Indeed, one can use Green's formula to show this, as is done in Christian Remling's answer.

More generally, the result holds for any convex function $v$, without requiring it to be in $C^2$ -- of course, if $u$ is not required to be in $C^2$ and if the derivative $v'(x)$ is understood as (say) the right derivative of $v$ at $x$ for $x\in[-1,1)$ (and the left derivative for $x=1$).

Indeed, with $u(x,y):=v(x)$ for all $x$ and $y$, the inequality in question is invariant with respect to taking positive/conical mixtures of functions $v$. That is, if the inequality in question holds for each suitably integrable function $v_b$ (in place of $v$) with $b$ in a set $B$, then the inequality in question holds for the positive mixture (used in place of $v$) of the form $\int_B\mu(db)\,v_b$ of the $v_b$'s, where $\mu$ is any nonnegative measure over $B$ such that the integral $\int_B\mu(db)\,v_b$ exists in a suitable sense.

It remains to note that the convex function $v$ is a positive mixture of affine functions and the functions of the form $v_a$ with $a\in(-1,1)$, where $v_a(x):=\max(0,x-a)$, and for any one of these latter functions the inequality in question is straightforward to check.

In fact, the constant factor $\frac12$ in the upper bound $\frac12\,(v'(1)-v'(-1))$ can be replaced by the better factor $\frac1\pi$ -- because for affine functions in place of $v$ one can replace $\frac12$ by $0$, and for $v=v_a$ with $a\in(-1,1)$ can replace $\frac12$ by $\frac{\sqrt{1-a^2}}\pi$. So, the worst case here is that of $v_a$ with $a=0$.

Detail: The mentioned mixture representation of the convex function $v$ is as follows: for $x\in(-1,1)$, $$\begin{aligned} &v(x) \\ &=v(-1+)+\int_{[-1,x)}dz\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,\Big(v'(-1)+\int_{[-1,z]}\,dv'(a)\Big) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}dz\,1(z<x)\,\int_{[-1,z]}\,dv'(a) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}\,dv'(a)\int_{[-1,1)}dz\,1(a\le z<x)\, \\ &=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,dv'(a)\,v_a(x) \\ &=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a(x) \\ \end{aligned}$$ where $\mu_{v'}$ is the (nonnegative) Lebesgue--Stieltjes measure corresponding to the nondecreasing function $v'$. So, $$v=C+A\,\text{id}+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a, $$ where $C:=C_v:=v(-1+)+v'(-1)$ and $A:=A_v:=v'(-1)$ are real numbers, and $\text{id}$ is the identity function.

Yes: Just take $u(x,y):=v(x)$.

Indeed, one can use Green's formula to show this, as is done in Christian Remling's answer.

More generally, the result holds for any convex function $v$, without requiring it to be in $C^2$ -- of course, if $u$ is not required to be in $C^2$ and if the derivative $v'(x)$ is understood as (say) the right derivative of $v$ at $x$ for $x\in[-1,1)$ (and the left derivative for $x=1$).

Indeed, with $u(x,y):=v(x)$ for all $x$ and $y$, the inequality in question is invariant with respect to taking positive/conical mixtures of functions $v$. That is, if the inequality in question holds for each suitably integrable function $v_b$ (in place of $v$) with $b$ in a set $B$, then the inequality in question holds for the positive mixture (used in place of $v$) of the form $\int_B\mu(db)\,v_b$ of the $v_b$'s, where $\mu$ is any nonnegative measure over $B$ such that the integral $\int_B\mu(db)\,v_b$ exists in a suitable sense.

It remains to note that the convex function $v$ is a positive mixture of affine functions and the functions of the form $v_a$ with $a\in[-1,1)$, where $v_a(x):=\max(0,x-a)$, and for any one of these latter functions the inequality in question is straightforward to check. In fact, the constant factor $2$ in the denominator on the right-hand side of the inequality in the OP can be replaced by the better factor $\pi$.

Detail: The mentioned mixture representation of the convex function $v$ is as follows: for $x\in(-1,1)$, $$\begin{aligned} &v(x) \\ &=v(-1+)+\int_{[-1,x)}dz\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,\Big(v'(-1)+\int_{[-1,z]}\,dv'(a)\Big) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}dz\,1(z<x)\,\int_{[-1,z]}\,dv'(a) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}\,dv'(a)\int_{[-1,1)}dz\,1(a\le z<x)\, \\ &=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,dv'(a)\,v_a(x) \\ &=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a(x) \\ \end{aligned}$$ where $\mu_{v'}$ is the (nonnegative) Lebesgue--Stieltjes measure corresponding to the nondecreasing function $v'$. So, $$v=C+A\,\text{id}+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a, $$ where $C:=C_v:=v(-1+)+v'(-1)$ and $A:=A_v:=v'(-1)$ are real numbers.

Yes: Just take $u(x,y):=v(x)$.

Indeed, one can use Green's formula to show this, as is done in Christian Remling's answer.

More generally, the result holds for any convex function $v$, without requiring it to be in $C^2$ -- of course, if $u$ is not required to be in $C^2$ and if the derivative $v'(x)$ is understood as (say) the right derivative of $v$ at $x$ for $x\in[-1,1)$ (and the left derivative for $x=1$).

Indeed, with $u(x,y):=v(x)$ for all $x$ and $y$, the inequality in question is invariant with respect to taking positive/conical mixtures of functions $v$. That is, if the inequality in question holds for each suitably integrable function $v_b$ (in place of $v$) with $b$ in a set $B$, then the inequality in question holds for the positive mixture (used in place of $v$) of the form $\int_B\mu(db)\,v_b$ of the $v_b$'s, where $\mu$ is any nonnegative measure over $B$ such that the integral $\int_B\mu(db)\,v_b$ exists in a suitable sense.

It remains to note that the convex function $v$ is a positive mixture of affine functions and the functions of the form $v_a$ with $a\in(-1,1)$, where $v_a(x):=\max(0,x-a)$, and for any one of these latter functions the inequality in question is straightforward to check. 

In fact, the constant factor $\frac12$ in the upper bound $\frac12\,(v'(1)-v'(-1))$ can be replaced by the better factor $\frac1\pi$ -- because for affine functions in place of $v$ one can replace $\frac12$ by $0$, and for $v=v_a$ with $a\in(-1,1)$ can replace $\frac12$ by $\frac{\sqrt{1-a^2}}\pi$. So, the worst case here is that of $v_a$ with $a=0$.

Detail: The mentioned mixture representation of the convex function $v$ is as follows: for $x\in(-1,1)$, $$\begin{aligned} &v(x) \\ &=v(-1+)+\int_{[-1,x)}dz\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,\Big(v'(-1)+\int_{[-1,z]}\,dv'(a)\Big) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}dz\,1(z<x)\,\int_{[-1,z]}\,dv'(a) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}\,dv'(a)\int_{[-1,1)}dz\,1(a\le z<x)\, \\ &=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,dv'(a)\,v_a(x) \\ &=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a(x) \\ \end{aligned}$$ where $\mu_{v'}$ is the (nonnegative) Lebesgue--Stieltjes measure corresponding to the nondecreasing function $v'$. So, $$v=C+A\,\text{id}+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a, $$ where $C:=C_v:=v(-1+)+v'(-1)$ and $A:=A_v:=v'(-1)$ are real numbers.

Yes: Just take $u(x,y):=v(x)$.

Indeed, one can use Green's formula to show this, as is done in Christian Remling's answer.

More generally, the result holds for any convex function $v$, without requiring it to be in $C^2$ -- of course, if $u$ is not required to be in $C^2$ and if the derivative $v'(x)$ is understood as (say) the right derivative of $v$ at $x$ for $x\in[-1,1)$ (and the left derivative for $x=1$).

Indeed, with $u(x,y):=v(x)$ for all $x$ and $y$, the inequality in question is invariant with respect to taking positive/conical mixtures of functions $v$. That is, if the inequality in question holds for each suitably integrable function $v_b$ (in place of $v$) with $b$ in a set $B$, then the inequality in question holds for the positive mixture (used in place of $v$) of the form $\int_B\mu(db)\,v_b$ of the $v_b$'s, where $\mu$ is any nonnegative measure over $B$ such that the integral $\int_B\mu(db)\,v_b$ exists in a suitable sense.

It remains to note that the convex function $v$ is a positive mixture of the constant functions, the identity function $\text{id}$, its opposite $-\text{id}$, and the functions of the form $v_a$ with $a\in[-1,1)$, where $v_a(x):=\max(0,x-a)$, and for any one of these functions the inequality in question (in fact, with the constant factor $\pi$ in the numerator on the right-hand side of the inequality, better than the corresponding constant factor $2$ in the OP) is straightforward to check.

Detail: The mentioned mixture representation of the convex function $v$ is as follows: for $x\in(-1,1)$, $$\begin{aligned} &v(x) \\ &=v(-1+)+\int_{[-1,x)}dz\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,\Big(v'(-1)+\int_{[-1,z]}\,dv'(a)\Big) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}dz\,1(z<x)\,\int_{[-1,z]}\,dv'(a) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}\,dv'(a)\int_{[-1,1)}dz\,1(a\le z<x)\, \\ &=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,dv'(a)\,v_a(x) \\ &=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a(x) \\ \end{aligned}$$ where $\mu_{v'}$ is the (nonnegative) Lebesgue--Stieltjes measure corresponding to the nondecreasing function $v'$. So, $$v=C+A\,\text{id}+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a, $$ where $C:=C_v:=v(-1+)+v'(-1)$ and $A:=A_v:=v'(-1)$ are real numbers.

Yes: Just take $u(x,y):=v(x)$.

Indeed, one can use Green's formula to show this, as is done in Christian Remling's answer.

More generally, the result holds for any convex function $v$, without requiring it to be in $C^2$ -- of course, if $u$ is not required to be in $C^2$ and if the derivative $v'(x)$ is understood as (say) the right derivative of $v$ at $x$ for $x\in[-1,1)$ (and the left derivative for $x=1$).

Indeed, with $u(x,y):=v(x)$ for all $x$ and $y$, the inequality in question is invariant with respect to taking positive/conical mixtures of functions $v$. That is, if the inequality in question holds for each suitably integrable function $v_b$ (in place of $v$) with $b$ in a set $B$, then the inequality in question holds for the positive mixture (used in place of $v$) of the form $\int_B\mu(db)\,v_b$ of the $v_b$'s, where $\mu$ is any nonnegative measure over $B$ such that the integral $\int_B\mu(db)\,v_b$ exists in a suitable sense.

It remains to note that the convex function $v$ is a positive mixture of affine functions and the functions of the form $v_a$ with $a\in[-1,1)$, where $v_a(x):=\max(0,x-a)$, and for any one of these latter functions the inequality in question is straightforward to check. In fact, the constant factor $2$ in the denominator on the right-hand side of the inequality in the OP can be replaced by the better factor $\pi$.

Detail: The mentioned mixture representation of the convex function $v$ is as follows: for $x\in(-1,1)$, $$\begin{aligned} &v(x) \\ &=v(-1+)+\int_{[-1,x)}dz\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,v'(z) \\ &=v(-1+)+\int_{[-1,1)}dz\,1(z<x)\,\Big(v'(-1)+\int_{[-1,z]}\,dv'(a)\Big) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}dz\,1(z<x)\,\int_{[-1,z]}\,dv'(a) \\ &=v(-1+)+v'(-1)(x+1) \\ &\qquad\qquad+\int_{[-1,1)}\,dv'(a)\int_{[-1,1)}dz\,1(a\le z<x)\, \\ &=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,dv'(a)\,v_a(x) \\ &=[v(-1+)+v'(-1)]+v'(-1)x+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a(x) \\ \end{aligned}$$ where $\mu_{v'}$ is the (nonnegative) Lebesgue--Stieltjes measure corresponding to the nondecreasing function $v'$. So, $$v=C+A\,\text{id}+\int_{[-1,1)}\,\mu_{v'}(da)\,v_a, $$ where $C:=C_v:=v(-1+)+v'(-1)$ and $A:=A_v:=v'(-1)$ are real numbers.