Suppose you are given fixed $k$ and a (real- or complex-valued) function $f\in C^k(B)$, where $B$ is the closed unit disk in $\mathbb{C}$, which satisfies for all $a\in[-1,1]$, $\|f\|_{\infty,1,a}:=\sup\limits_{\lvert b\rvert\le\sqrt{1-a^2}}\lvert \partial_1^kf(a+bi)\rvert\le1$, and $\|f\|_{\infty,2,b}\le1$ defined similarly. That is, $f$'s $k$th derivative along each vertical and horizontal slice is bounded. (I'd be willing to add a stronger condition about directional derivatives.) Let $S:=[-1,1]+i[-1,1]\supset B$.
- Must there exist a function $g\in C^k(S)$ which agrees with $f$ on $B$, and such that $\|g\|_{\infty,1,a},\|g\|_{\infty,2,b}\le c$ for all $a,b\in[-1,1]$, for some constant $c$ independent of $f$ and $k$?
- If so, is there a way to compute $g$ given $f$? (Some sort of algorithm allowing arbitrarily good eventual approximation would work great.)
- If not, would it change matters to consider instead $h\in C^k((1+\varepsilon)S)$ for small $\varepsilon>0$? (This change allows for an intermediate open set $U$ with $B\subset U\subset(1+\varepsilon)S$, à la this answer.)
(I've come across several similar questions asked here, but mainly about $C^\infty$ functions without reference to bounded derivatives or (approximate) computability.)
I suspect that the answer is "yes" by the following construction, probably requiring analogous conditions on $k$th directional derivatives. Consider the corner $1+i$. Let $g(1+i):=0$. For any point $z$ on $S^1$ in the first quadrant, let $g(tz+(1-t)(1+i)):=tf(z)$. Do the same for all $\pm1\pm i$. Then, the $k$th directional derivatives along the axes at points near the corners are related to $k$th directional derivatives on $S^1$, which we assume to be controlled. However, this may be overlooking some devil in the details.
Any thoughts are appreciated.