As we know, $ f(x,y)={\frac {x^{3}y-xy^{3}}{x^{2}+y^{2}}}$ , if $(x,y) \ne (0,0)$ and $f(0,0)=0$.All of its second-order partial derivatives exist on $ \mathbb R^2$. But $\frac{\partial ^{2}}{\partial x \partial y}f(0,0)=1 $ and $\frac{\partial ^{2}}{\partial y \partial x}f(0,0)=-1.$ the set $\{f_{xy}(x,y)\ne f_{yx}(x,y)\vert (x,y)\in \mathbb R^2 \}=(0,0)$ has a zero Lebesgue measure.

Let a function $f$ defined on an open set $O$ in $\mathbb R^n,$ $f\colon O \to\mathbb R$, all of its partial derivatives of order $k$ $-$that is, all the derivatives $\partial_{i_{1}}\partial_{i_{2}}\cdots\ \partial_{i_{k}}f-$ exist in $O$.

For any fixed integer $s(\leq k),$ whether we have $\partial_{i_{1}}\partial_{i_{2}}\cdots\ \partial_{i_{s}}f$=$\partial_{j_{1}}\partial_{j_{2}}\cdots\ \partial_{j_{s}}f$ holds almost-everywhere on $O$,whenever the sequence$\{j_1,\cdots,j_{s}\}$ is a reordering of the sequence$\{i_1,\cdots,i_{s}\}$ ?


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