We were provided the following example in the context of Schwarz's theorem:

Let be $f:\mathbb{R}^2 \to \mathbb{R}$, given by: $$ f{x \choose y}=\left\{\begin{array}{ll} \frac {xy^3}{x^2+y^2}, & {x \choose y}\neq {0 \choose 0} \\ 0, & {x \choose y}={0 \choose 0}\end{array}\right. . $$

If we calculate the second order partial derivatives for each ${x \choose y} \neq {0 \choose 0}$ we get: $$ D_{12}f{x \choose y} = D_{21}f{x \choose y}. $$

However, at point ${0 \choose 0}$ we have: $$ D_{12}f{0 \choose 0} =1 \neq 0 = D_{21}f{0 \choose 0}. $$

Our lecture notes say that we now can deduce only from Schwarz's theorem, that both second order derivatives $D_{12}f{x \choose y}$ and $D_{21}f{x \choose y}$ cannot be continuous at point ${0 \choose 0} $.

I don't understand how we can deduce this fact only from Schwarz's theorem?

Couldn't it be possible that only one of the two second order derivatives is continuous?


1 Answer 1


You are correct, you can only deduce from Schwarz's Theorem that at least one of the partial deivatives is discontinuous.

However, you can deduce that both must be discontinuous from the stronger version of Scwarz's Theorem proved by Peano.


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