# Stronger version of equality of mixed partials

It is well-known that for function $$f:\mathbb{R}^2\to\mathbb{R}$$, if both $$\partial_x f$$ and $$\partial_y f$$ exist and are differentiable at $$(x_0,y_0)$$, then $$\partial_x\partial_y f(x_0,y_0)=\partial_y\partial_x f(x_0,y_0)$$ (see Dieudonné's Foundations of Modern Analysis theorem 8.12.2).

Can this theorem be strengthened? For example, if $$\partial_x f$$ is differentiable at $$(x_0,y_0)$$ and $$\partial_y f$$ is continuous at $$(x_0,y_0)$$, does it follow that the mixed partials are equal? Or is there a counterexample? If this is true, then can we weaken the condition further?

(Note: the continuity of $$\partial_x\partial_y f$$ at $$(x_0,y_0)$$ is not assumed here. For example, $$f(x,y)=(x^2+y^2)^3\sin\big(\frac{1}{x^2+y^2}\big)$$ (and $$f(0,0)=0$$) has differentiable partial derivatives, but both are not continuous at the origin.)

• I’d use the tag real-analysis instead of multivariable-calculus.
– Joe
Commented Sep 9, 2019 at 9:39