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Let $$ f(x,y)=\min(|x-c|,|y-c|), $$ where $x,y,c \in[0,1]$. I wish to show that $f$ is quasiconvex.

I wasn't able to prove the claim for this case without tedious case analysis. Any ideas?

Thanks!

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2 Answers 2

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It is quasi-convex.

Proof: for simplicity assume $c=0$ otherwise you can shift $f$ horizontally.

So $f(x,y)=\min (|x|,|y|)$, then each $\alpha-$level set of $f$ i.e., $$\{(x,y) \in R^2 ~ | \quad \min(|x|,|y|) \leq \alpha \}$$ is solid square, and so it is convex subset of $R^2$, therefore $f$ is quasi-convex.

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If $g_1(x)=x^2$ and $g_2(y)=(1-y)^2$, then $f(x,y)$ is not convex along the line $x=y$: $f(0,0)=f(1,1)=0$, $f(\frac12,\frac12)=\frac14$. The same holds for your concrete example if you consider a line that passes through $x=c$ and $y=c$ at different "times".

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  • $\begingroup$ Thanks for your reply. What about quasiconvexity? I edited the question accordingly. $\endgroup$ Jun 2, 2017 at 19:06
  • $\begingroup$ That's right, but you could find easier counterexamples like $g_1 =x$ and $g_2 = 0$ $\endgroup$
    – Red shoes
    Jun 2, 2017 at 20:34

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