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The answer is negative. Example: $$\max\{ |y|,|x|-1\}.$$ This is a convex piecewise linear function which is not a convex combination of your "tripods", and not a limit of such combinations.

If you have an arbitrary system of linear equations and add to it one equation, the dimension of the space of solutions can decrease by at most 1. Do take any $d$ equations of your hyperplanes, order …

This paper is contained in the 1-st volume of Aleksandrov's Selected works. This has an English translation: Alexandrov, A. D. Reshetnyak, Yu. G. (ed.) Selected works. Part 1: Selected scientific pape …

There are no such $C$ at all, when $n>2$. Indeed your conditions imply $h_\phi=\phi$ on $\partial C$. But $h$ is bounded and continuous on $\partial C$, while when $n\geq 2$ there is always a subharmo …

This fact is trivial. Your assumptions imply that $$f(x)=ax^m\prod(1+x/x_k),$$ where $a>0$ and $x_k>0$. All zeros $-x_k$ are negative because you assume that $a_k\geq 0$, so there are no positive zero …

Let me explain first what Anton wrote: we construct piecewise linear function $f$ by on $[0,n]$ by induction in $n$. Suppose that it is already constructed on $[0,n]$. Define the right derivative $D^+ …

Take $g_1(x)=x\log^2x$. Properties 1,2 are evidently satisfied, and computation of the second derivative shows that it is negative for $0< x<1/e$. Now rescale: $g(x)=g_1(x/e)$.

Yes. The boundary even has a locally finite Haudsorff $(m-1)$-measure. No. A convex function of 1 variable has increasing derivative, but this derivative can have a dense set of jumps. In general, …