This may be too easy, but:



> 
> Is there a function $f$ on the first quadrant of $\mathbb R^2$ such
>  that $$ f(x,1)=x,\qquad f(x,0)=0, $$ and $f$ is convex or concave?
> 
> 
> 


Note there is no solution of the form $f(x,y)=x\cdot g(y)$, since (i) $-f$ is convex iff $f$ is concave, and (ii) the Hessian is
$$
H=\left(
\begin{matrix}
0& g'(y) \\
g'(y)&x\cdot g''(y)
\end{matrix}
\right)
$$
which is positive semi-definite only if 
$$
-(g'(y))^2\ge 0
$$
which would mean $g(y)$ is constant.



Convex extensibility of combination of two lines
There is; for example, the following function is concave:
$f(x,0)=0, f(x,y)=x$ for $y>0$.


