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 $u(x)=v(\|x\|)$.

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) ?

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 $$\frac1{2\pi}\int_{\partial D}\frac{\partial u}{\partial \nu}d\ell\le \frac{v'(b)-v'(a)}2\quad ?$$ 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 $u(x)=v(\|x\|)$.

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 $u(x)=v(\|x\|)$.