$$f(x,y)=\left | x-y \right |$$
Hello, Do we have to check for quasiconcavity using leel curves? Or is there any other way?
I'm finding it very difficult to plot or imagine the level curves Any help will be appreciated
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"I'm finding it very difficult to plot" ?? The graph is simply a V-shape whose base is the line $x=y,z=0$. Convexity should follow trivially. – Did Jun 21 '17 at 07:31
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No, the level curves are straight lines parallel to each other – user405925 Jun 21 '17 at 07:45
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Indeed they are, hence, as I said, "The graph is simply a V-shape whose base..." and the answer to your question follows directly. (Why "No"?) – Did Jun 21 '17 at 07:48
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OH! I just imagined it in a sort of 3D graph. Since the level curves are spreading out it is convex? – user405925 Jun 21 '17 at 07:52
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Thankyou, my bad. – user405925 Jun 21 '17 at 07:52
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Hint: You can see this function is the composition of the linear function $g(x,y)=x-y$ and the convex function $h(s)= |s|$. So it is Convex,let alone quasiconvex !
Red shoes
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If we show it is convex, it will imply its q-convexity.
Convexity condition: $$tf(x_1,y_1)+(1-t)f(x_2,y_2)\ge f(tx_1+(1-t)x_2,ty_1+(1-t)y_2) \Rightarrow$$
$$t|x_1-y_1|+(1-t)|x_2-y_2|\ge |tx_1+(1-t)x_2-ty_1-(1-t)y_2|=|t(x_1-y_1)+(1-t)(x_2-y_2)|.$$
This is true because of the triangle inequality: $$|x|+|y|\ge |x+y|.$$
farruhota
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