Proving Continuity of \frac{\partial ^2}{\partial x \partial y}\int_a^x M(t,y)dt

[nonequilibrium]

Given function M from R^2 to R with image M(x,y) and given that $$\frac{\partial M}{\partial y}$$ and $$\frac{\partial M}{\partial x}$$ exist and are continuous, i.e. M is a C^1 function.

Is it true that $$\frac{\partial ^2}{\partial x \partial y}\int_a^x M(t,y)dt = \frac{\partial M(x,y)}{\partial y}$$? If so, how/why? (please note this is not some kind of homework, it's out of my own interest, because it was used as evident in a certain proof although I can't seem to prove it) What you basically want to prove is that the left hand side is continuous (this is not necessary, but it would be sufficient for the equality).

So in simpler terms, under given conditions, how does one know that (if?) $$\frac{\partial ^2}{\partial x \partial y}\int_a^x M(t,y)dt$$ (exists and) is continuous?

 

[jvicens]

I would like to clarify that the given statement is not necessarily true. The continuity of the first partial derivatives of M does not guarantee the continuity of the second partial derivative of the integral in question. This can be seen through counterexamples.

For example, consider the function M(x,y) = xy. The first partial derivatives exist and are continuous, but the second partial derivative of the integral \int_a^x M(t,y)dt with respect to x is 0, while the partial derivative of M(x,y) with respect to y is simply x.

In general, the existence and continuity of the first partial derivatives of M do not necessarily imply the existence and continuity of higher-order partial derivatives of the integral. This is because the integral is a composite function of the variable x and the function M, and the composition of functions can lead to discontinuities in higher-order derivatives.

However, if M is a C^2 function (i.e. it has continuous second partial derivatives), then the statement is true. This can be proven using the Leibniz integral rule, which states that for a function f(x,y) and its integral F(x,y) = \int_a^x f(t,y)dt, the second partial derivative of F with respect to x and y is given by \frac{\partial^2 F}{\partial x \partial y} = f(x,y). This rule holds for C^2 functions, and thus, in this case, the second partial derivative of the integral is indeed equal to the partial derivative of M(x,y) with respect to y.

In conclusion, the statement may be true under certain conditions, but it is not always true. The continuity of the first partial derivatives of M does not guarantee the continuity of the second partial derivative of the integral. However, if M is a C^2 function, then the statement holds true.