Let $f(x,y) = 2x+y-x^2$ be a two variable function.
(a) Show that $f$ is concave.
(b) Show that $g = -e^{-f(x,y)}$ is concave.
I have proved part (a) using Hessian matrix.
We have $f_x = 2-2x$ and $f_y = 1$.
Then we can compute: $f_{xx} = -2, f_{xy} = 0, f_{yy} = 0$.
Hence the Hessian matrix is the diagonal matrix with diagonal entries $-2,0$.
Thus it is negative semi-definite.
And $f$ is concave.
But I am stuck with (b). Need some help.
I have tried to find the Hessian matrix, but the entries are coming very complicated. Is there any other method?
Best Answer
Theorem :
If $\;\varphi:A\to\Bbb R\;$ and $\;\psi:\Bbb R\to\Bbb R\;$ are two convex functions (where $A$ is a convex set of $\,\Bbb R^n$) and the function $\;\psi\;$ is monotonically non-decreasing$\,,\;$ then also the function $\;\psi\!\circ\!\varphi:A\to\Bbb R\;$ is convex.
Proof :
We will prove that for any $\,x,y\in A\,$ and for any $\,t\in[0,1]\,,\,$ it results that $\;(\psi\!\circ\!\varphi)\big((1-t)x+ty\big)\leqslant (1-t)(\psi\!\circ\!\varphi)(x)+t(\psi\!\circ\!\varphi)(y)\,.$
For any $\,x,y\in A\,$ and for any $\,t\in[0,1]\,,\,$ we get that \begin{align} (\psi\!\circ\!\varphi)\big((1-t)x+ty\big)&=\psi\left[\varphi\big((1-t)x+ty\big)\right]\\[3pt] &\leqslant\psi\big((1-t)\varphi(x)+t\varphi(y)\big)&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\color{brown}{\text{( }\varphi\text{ is convex and }\psi\text{ is non-decreasing )}}\\[3pt] &\leqslant(1-t)\psi(\varphi(x))+t\psi(\varphi(y))\quad\color{brown}{\text{( }\psi\text{ is convex )}}\\[3pt] &=(1-t)(\psi\!\circ\!\varphi)(x)+t(\psi\!\circ\!\varphi)(y)\,. \end{align}
Since the function $\;f:\Bbb R^2\to\Bbb R\;$ is concave, it follows that the function $\;\varphi=-f\;$ is convex.
Moreover, the function $\;\psi(x)=e^x:\Bbb R\to\Bbb R\;$ is convex and non-decreasing ( in fact is increasing ), so we can apply the previous Theorem and get that
the function $\;\psi\!\circ\!\varphi=e^{-f}:\Bbb R^2\to\Bbb R\;$ is convex ,
consequently ,
the function $\,g=-e^{-f}\,$ is concave.