Distance of convex combination of pairs of points in $\mathbb{R}^n$

Question

Given 4 points $w,x,y,z \in \mathbb{R}^n$ define for $t\in [0,1]$ $f(t)=d(wt + (1-t)x, yt + (1-t)z)$. Is this function convex? I have found a proof by differentiating twice and calculating a lot but it's not very "nice". Does somebody know any more proofs? Maybe using some geometry or clever argument (but still rigorous)?

Answer 1

We have \begin{align*} f(t) &= \|tw+(1-t)x-ty-(1-t)z\| \\ &= \|(x-z)+t(w-y-x+z)\| \\ &= \|u+tv\| \end{align*} where I've written $u=x-z$ and $v=w-y-x+z$ for simplicity. The convexity of this function comes directly from the triangle inequality: \begin{align*} \|u+((1-\lambda)t_1+\lambda t_2)v\| &= \|(1-\lambda)(u+t_1 v)+\lambda (u+t_2 v)\| \\ &\le (1-\lambda)\|u+t_1 v\| + \lambda\|u+t_2 v\| \end{align*} (One nice thing about this proof is that it works for any norm on $\mathbb{R}^n$, not just the usual Euclidean one.)