Consider the function: $f(x,y)=e^{ax+by^{2}}$
I have to find the values for $b$
such that $f(x,y)$ is convex and concave.
These are my calculations:
$f_{xx}=a^{2}e^{ax+by^{2}}$
$f_{yy}=4b^{2}y^{2}e^{ax+by^{2}}$
$f_{xy}=2abye^{ax+by^{2}}$
So, I found the determinant of the Hessian to be equal to zero. If I remember correctly, this means that the function can be either concave/convex.
For $f(x,y)$
to be concave, $f_{yy}<0$
So i want to find the values of b such that $4b^{2}y^{2}e^{ax+by^{2}}<0$
. This is where I get lost.
Since we have $b^{2}$, it can never be zero? The $e$ term can also never be zero. So does this means that the function is strictly convex?
All help is greatly appreciated.
Best Answer
You forgot a term in $f_{yy}$. With $f_{yy}=\left(4b^2y^2+2b\right)f$, the sign of the Hessian is the sign of $b$. You can also use the trace of the Hessian, which is $\left(a^2+4b^2y^2+2b\right)f$. The determinant and the trace are the product and the sum, respectively, of the eigenvalues of the Hessian.
For $b\gt0$, the product and sum are both positive, so both eigenvalues are positive and $f$ is strictly convex.
For $b=0$ and $a=0$, the product and sum are both $0$, so both eigenvalues are $0$ (not surprisingly, since in this case $f$ is constant). In this case $f$ is both convex and concave, but neither strictly convex nor strictly concave.
For $b=0$ and $a\ne0$, the product is $0$ and the sum is positive, so one eigenvalue is $0$ and the other is positive; $f$ is convex but not strictly convex.
For $b\lt0$, the product is negative, so $f$ is neither convex nor concave.