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Consider the following functional equation: $$TV(k)=\max[W(k),\beta V(f(k))]$$ where $\beta\in (0,1)$, $W(k)$ is continuous, increasing, bounded, and strictly concave function defined on $[0,\bar{k}]$, and $f(k)$ is a continuous, increasing, and bounded function defined on $[0,\bar{k}]$. Does $T$ map strictly concave functions into strictly concave functions? If so, how can I prove it? If not, is there a counter example?

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No. The maximum of two concave functions is usually not concave, so this is pretty hopeless.

Here is an explicit counterexample: Let $\bar{k}=1$, $W(k)=\sqrt{k}$, $f(k)=k$, $\beta=0.999$. Let $V(k)=k+1/2\sqrt{k}$. Draw a picture.

Michael Greinecker
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  • Hi. Please check out this graph: https://www.desmos.com/calculator/abkp1m5s6t. I'm still very confused, because it seems that $Tv(k)=W(k)$ when $k<0.25$ and $Tv(k)=\beta v(k)$ when $k>0.25$. And therefore, $Tv(k)$ is not strictly concave. Am I correct? –  Nov 16 '23 at 02:48
  • Yes. Though I'm not sure the point is exactly at 0.25. – Michael Greinecker Nov 16 '23 at 07:06