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Show continuity of partial derivatives of piece-wise function

Mathematics Asked by learning_linalg on December 3, 2021

Consider the $f:mathbb{R}^2 mapsto mathbb{R}$ defined by $$f(x,y) = frac{xy(x^2-y^2)}{x^2+y^2}$$ for $(x,y) neq (0,0)$ and $f(0,0)=0$. Show that f $in C^1(mathbb{R}^2), f not in C^2(mathbb{R}^2)$.

To show $f in C^1(mathbb{R}^2)$, I calculated that $nabla f = <frac{yx^4+4x^2y^3-y^5}{(x^2+y^2)^2},frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}>$, which is obviously continuous for $(x,y) neq (0,0)$, but at $(x,y) = (0,0)$, I don’t think this function is of much use. I used the limit definition of partials to show the partials exists at $(0,0)$, but I’m lost as to how to show they are continuous here.

For the second derivative case, I am even more lost as to prove that the partials are not continuous at $(0,0)$, since I don’t have a function representation of the partials at $(0,0)$ to prove continuity with.

One Answer

Firstly calculate $f_x(0,0)$ by definition: $$f_x(0,0)=lim_limits{x to 0}frac{f(x,0)-f(0,0)}{x} = 0$$ Let's proof, that $f_x$ is continuous in $(0,0)$: $$|f_x(x,y)-f_x(0,0)| = frac{|x^4y+4x^2y^3-y^5|}{(x^2+y^2)^2} leqslant frac{6(x^2+y^2)^{5/2}}{(x^2+y^2)^2} = 6(x^2+y^2)^{1/2}$$

Answered by zkutch on December 3, 2021

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