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.)

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


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