exponential of second derivative, not positive? I'm interested in whether the following function $h(x,y)=e^{\partial_x\partial_y}f(x)g(y)$, is positive in the sense that $h(x,y)\geq 0$, whenever $f(x)\geq 0$ and $g(y)\geq 0$ (positive and real). I would have thought that it was positive because the derivatives are in the exponential, but it appears not to be. Any idea why it isn't positive? Maybe because ${\partial_x\partial_y}$ is not a normal operator? But I think it is even self-adjoint (or at least symmetric), so should have real eigenvalues.

To see that it is not positive one can expand the exponential $h(x,y)=f(x)g(y)+f'(x)g'(y)+\frac{1}{2}f''(x)g''(y)+...$ and take (for example) $f(x)=x^2$ and $g(y)=y^2$. Then the series terminates at second order and is $(xy+2)^2-2$, which is negative at $x=-\sqrt{2}$, $y=\sqrt{2}$ etc.

To see that $\partial_x\partial_y$ is symmetric we have $\int dx dy \psi^*(x,y)\partial_x\partial_y \phi(x,y)=\int dx dy \partial_x\partial_y \psi^*(x,y) \phi(x,y)$ after integrating by parts twice, plus a boundary term. So for the boundary term to vanish we should have that $\psi^*(x,y)$ is zero at infinity, so maybe  $e^{\partial_x\partial_y}f(x)g(y)$ is positive on functions which decay to zero  at infinity? This is not the case with polynomials which are the examples I know that give negative values.
We will assume Clairaut's Theorem holds, such that $\partial_x\partial_y=\partial_y\partial_x.$ Using the generalized Leibnitz Rule for the higher-order derivatives of a product, we have
\begin{align*}
e^{\partial_{x}\partial_{y}}\{fg\}&amp;=\sum_{n=0}^{\infty}\left[\frac{1}{n!}\left(\frac{\partial^2}{\partial x \, \partial y}\right)^{\!n}(fg)\right] \\
&amp;=\sum_{n=0}^{\infty}\left[\frac{1}{n!}\frac{\partial^n}{\partial x^n}\frac{\partial^n}{\partial y^n}(fg)\right] \\
&amp;=\sum_{n=0}^{\infty}\left[\frac{1}{n!}\frac{\partial^n}{\partial x^n}\left(\sum_{i=0}^{n}\binom{n}{i}\left\{\frac{\partial^{n-i}}{\partial y^{n-i}}\,f\right\}\left\{\frac{\partial^i}{\partial y^i}\,g\right\}\right)\right] \\
&amp;=\sum_{n=0}^{\infty}\left[\frac{1}{n!}\sum_{j=0}^{n}\sum_{i=0}^{n}\binom{n}{i}\binom{n}{j}\left\{\frac{\partial^{n-j}}{\partial x^{n-j}}\frac{\partial^{n-i}}{\partial y^{n-i}}\,f\right\}\left\{\frac{\partial^j}{\partial x^j}\frac{\partial^i}{\partial y^i}\,g\right\}\right].
\end{align*}
Because you're dealing with the derivatives of functions that could be decreasing, even if they are positive, there is no guarantee that the result here will be positive. 

An operator being normal just means it's diagonalizable (infinite-dimensional space: spectral theorem holds). There's still no guarantee that its spectrum is real, which is what you'd need here, I think. Unitary operators are normal, and their eigenvalues are definitely not guaranteed to be real. If you're exponentiating an Hermitian operator, then I think you'd have yourself a positive operator coming out.
I presume you mean $e^{(\partial_x f(x))(\partial_y g(y))}= e^{f_x(x)g_y(y)}$.  For given x and y, assuming they are real numbers, whether the partial derivatives are positive or not, the exponential is always positive.