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Consider the Chebyshev distance in two dimensions: $$ C[x,y] := \max\left(\text{abs}(x-x_0),\text{abs}(y-y_0)\right) $$

Is $C[x,y]$ a convex function of $(x,y)$? Now I think, say $\frac{dC[x,y]}{dx}$ is not smooth so I don't think we can use the condition $f^{''}>0$ to prove the convexity of the Chebyshev distance.

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up vote 2 down vote accepted

If $f$ and $g$ are two convex functions, so is $h:=\max\{f,g\}$. Indeed, if $u,v$ are in the domain of definition and $t\in [0,1]$ then $$h(tu+(1-t)v)=\max\{f(tu+(1-t)v),g(tu+(1-t)v)\}\\\leq \max\{tf(u)+(1-t)f(v),tg(u)+(1-t)g(v)\}.$$ Since $$tf(u)+(1-t)f(v)\leq t\max\{f(u),g(u)\}+(1-t)\max\{f(v),g(v)\},$$ and the same is true replacing $f$ by $g$ in the RHS, we get the result.

To conclude this problem, use the fact that the map $x\mapsto |x-x_0|$ is convex using triangular inequality.

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