Clairaut's Theorem
Let $f$ be a function of two variables,let$(x_{0},y_{0})$ be a point and let $U$ be an open disk with center $(x_{0},y_{0})$.Assume that $f$ is defined on $U$ and its partial derivatives $f_{x},f_{y}$,$f_{xy}$ and $f_{yx }$ exist on $U$.Assume,further,that $f_{xy}$ and $f_{yx}$ are continuous at $(x_{0},y_{0})$.Then $f_{xy}(x_{0},y_{0})$=$f_{yx}(x_{0},y_{0})$.
My question arise from the above theorem.We know $\frac{\partial^{2} f}{\partial x \partial y}(x_{0},y_{0})$ exists $\Rightarrow\frac{\partial f}{\partial x }(x_{0},y_{0})$ exists;$\frac{\partial^{2} f}{\partial y \partial x}(x_{0},y_{0})$ exists $\Rightarrow\frac{\partial f}{\partial y }(x_{0},y_{0})$ exists.So I think the condition: $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y }$ exist everywhere on $U$ must be eliminated from the theorem,because $\frac{\partial^{2} f}{\partial x \partial y};\frac{\partial^{2} f}{\partial y \partial x}$exist everywhere on $U$.Why do many textbooks persist with this condition in the theorem?
My question focus on :If $\frac{\partial^{2} f}{\partial x \partial y},\frac{\partial^{2} f}{\partial y \partial x} $exist on $U$,then naturally $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y }$ exist on $U$ .
So this theorem only can be weakened like this:
Let $f$ be a function of two variables,let$(x_{0},y_{0})$ be a point and let $U$ be an open disk with center $(x_{0},y_{0})$.Assume that $f$ is defined on $U$ and its partial derivatives $f_{xy}$ and $f_{yx }$ exist on $U$.Assume,further,that $f_{xy}$ and $f_{yx}$ are continuous at $(x_{0},y_{0})$.Then $f_{xy}(x_{0},y_{0})$=$f_{yx}(x_{0},y_{0})$.
Remark: $\frac{\partial^{2} f}{\partial x \partial y},\frac{\partial^{2} f}{\partial y \partial x} $exist on $U$ $\nRightarrow$ $\frac{\partial f}{\partial x},\frac{\partial f}{\partial y }$ are continuous on $U$.
Technically, you're correct-you can't have the second order mixed partials exist throughout the open set U if the first order partials don't exist. But you'll notice the theorem doesn't just require the first order partials to exist on the open set-it requires them to be continuous throughout U. This turns out to be the important condition and if you drop the statement that the first order partials exist on U, it might give the mistaken notion to the beginner that they don't have to be continuous on U for the mixed order partials to be equal. Which of course would be a big problem,because this is false. You indeed can have a function defined on an open set in $R^2$ where partial derivatives $f_{x}$,$f_{y}$ and $f_{xy}$ and$f_{yx }$ exist and and the second order mixed partials are not necessarily all continuous.In that case, Clairaut's theorem is false. Here's a counterexample.
That answer your question?