# Does the continuity of the second order mixed partial derivative imply the continuity of the partial derivative of $f$

Theorem. $$f(x,y):\mathbb{R}^2 \to \mathbb{R}$$ is a function, and $$f$$ has partial derivatives $$f_x$$, $$f_y$$, $$f_{xy}$$, $$f_{yx}$$. Moreover, $$f_{xy}$$ and $$f_{yx}$$ are continuous. Then $$f_{xy}=f_{yx}$$.

I am trying to understanding the change of sequence of higher order derivatives. In my thinking the questions comes to me: If $$f_{xy}$$ and $$f_{yx}$$ are continuous, then $$f_x$$ and $$f_y$$ exists, but are $$f_x$$ and $$f_y$$ continuous?

This questions claims that $$f_x$$ and $$f_y$$ are continuous but I can not see why. I know if $$g$$ is two variable function and $$g_x, g_y$$ are continuous, then $$g$$ is differentiable and so continuous. But in this question we do not know the continuous of $$f_{xx}$$ ($$f_{yy}$$), so we cannot get the continuous of $$f_x$$ ($$f_y$$) by using $$f_{yx}$$'s ($$f_{xy}$$'s) continuity.

Thank you in advance.

• Please add enough info that we do not need to follow a link. Commented May 27 at 16:41
• @Steen82: Ok. It is my fault. And now I think I am getting it right. Commented May 27 at 16:49

## 1 Answer

The claim is false. Let $$g\colon\Bbb R\to\Bbb R$$ be any differentiable function whose derivative is not continuous (say, at $$0$$) — plenty of standard examples of this phenomenon have appeared numerous times on this site. Now set $$f(x,y)=g(x)+g(y)$$. Then $$f_{xy}=0=f_{yx}$$ everywhere, and yet $$f_x(x,y) = g'(x)$$ and $$f_y(x,y)=g'(y)$$, so $$f_x$$ and $$f_y$$ are not continuous at the origin.

• great answer and what I suspected, but real life prevented me from being sure. Commented May 27 at 19:53
• Thanks. Set $f(x,y)=g(x)+g(y)$ is genius. Commented May 28 at 1:48