- #1

nonequilibrium

- 1,439

- 2

Let [tex]M[/tex] be a function from R^2 to R with image M(x,y).

Given is that [tex]M[/tex] has continuous partial derivatives [tex]\frac{\partial M(x,y)}{\partial y}[/tex] & [tex]\frac{\partial M(x,y)}{\partial x}[/tex].

**Question:**

Does

[tex]\frac{\partial^2}{\partial x \partial y} \int M(x,y) \mathrm d x[/tex]

*exist*and is it

*continuous*?

It is used in my DE course, but I don't find it self-evident. The professor's argument was that since M was already differentiable with respect to x and y, and integrating makes it

*more*continuous, it will definitely be differentiable. Okay I find it somewhat self-evident (with this reasoning) that

[tex]\frac{\partial}{\partial y}\int M(x,y) \mathrm d x[/tex]

exists, but how does one convince himself of this last expression being differentiable with respect to x? And even if it is differentiable with respect to x, the derivative might have an essential discontinuity (of course I'm not saying it can in this case: I believe my professor; but I don't believe in his hand-waving reasoning and I'm looking for a more rigorous/insightful argument)