exponential of second derivative, not positive?

Question

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.

Answer 1

I think the answer (with a hint from Adrian), is that for $h(x,y)=e^{\partial_x\partial_y}f(x)g(y)$ to be positive, the eigenvalues of $\partial_x\partial_y$ should be real. Now, $\partial_x\partial_y$ appears to be self-adjoint, and hence have real eigenvalues, but this is at best only true if we define the domain of the operator appropriately. It should be defined on functions with appropriate boundary conditions, and my counter-examples to $e^{\partial_x\partial_y}f(x)g(y)$ being positive did not satisfy those boundary conditions.

In particular, if $\partial_x\partial_y$ is defined on differentiable functions over the reals, and we choose the standard inner product, then it's adjoint is defined via the inner-product as $\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)+\psi^*(x,y) \phi(x,y)\Big]^{\infty}_{-\infty}-\int dx \partial_x\psi^*(x,y) \phi(x,y)\Big]^{y\rightarrow\infty}_{y\rightarrow\infty}-\int dy \partial_y \psi^*(x,y) \phi(x,y)\Big]^{x\rightarrow\infty}_{x\rightarrow\infty} $ and so the domain of the adjoint must be restricted to functions $\psi^*$ which are zero and have zero first derivatives at infinity (or whatever the boundary is) so that the boundary terms vanish.

This means that the domain of definition of $\partial_x\partial_y$ is larger than the domain of it's adjoint, so it is not self-adjoint but only symmetric. Perhaps we can find a self-adjoint extension.

At any rate, the key point is that at best, $\partial_x\partial_y$ only has real eigenvalues on functions which satisfy appropriate boundary conditions. Now, I think that even though $\partial_x\partial_y$ is not self-adjoint, it will have real eigenvalues for functions which are restricted to decay at infinity, because in that case $\int dx dy \psi^*(x,y)\partial_x\partial_y \phi(x,y)=\int dx dy \partial_x\partial_y \psi^*(x,y) \phi(x,y)$ but I'm not completely certain.

Answer 2

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.

Answer 3

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\}&=\sum_{n=0}^{\infty}\left[\frac{1}{n!}\left(\frac{\partial^2}{\partial x \, \partial y}\right)^{\!n}(fg)\right] \\ &=\sum_{n=0}^{\infty}\left[\frac{1}{n!}\frac{\partial^n}{\partial x^n}\frac{\partial^n}{\partial y^n}(fg)\right] \\ &=\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] \\ &=\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.