Extending a convex function to a higher dimensional domain

Question

Let $D\subset{\mathbb R}^2$ be the unit disk, and $I=(-1,1)$.

Let $v\in C^2(\bar I)$ be a convex function.

Does there exist a convex function $u\in C^2(\bar D)$, such that on the one hand $u(\cdot,0)=v$, and on the other hand $$\frac1{2\pi}\int_{\partial D}\frac{\partial u}{\partial \nu}d\ell\le \frac{v'(1)-v'(-1)}2\quad ?$$ In other words, the average value of the outer normal derivative of $u$ is less than or equal to that of $v$.

The answer is positive if $v$ is even: just take $u(x)=v(\|x\|)$.

Edit. Christian's answer tells us that the inequality above is true, but not sharp: one has $$\frac1{2\pi}\int_{\partial D}\frac{\partial u}{\partial \nu}d\ell\le c(v'(1)-v'(-1))$$ with $c=\frac1\pi$. Whence the additional question:

What is the smallest constant $c$ with the property above (existence of a convex extension such that the inequality occurs) ?

Answer 1

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 $a=0$.

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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.

Answer 2

$u(x,y)=v(x)$ works. By Green's formula, we can rewrite the LHS of your inequality as $$ \frac{1}{2\pi} \int\!\!\!\int \Delta u\, dA = \frac{1}{2\pi}\int_{-1}^1 dy\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} dx\,v''(x) \le \frac{1}{\pi}(v'(1)-v'(-1)). $$