# If the second derivatives $f_{xx}$ and $f_{yy}$ exist, does $f_{xy}$ exist?

If the second derivative with respect to to $x$ exists ($f_{xx}$) and the second derivative with respect to $y$ ($f_{yy}$), does it follow that $f_{xy}$ exists?

• Do you want the derivative to exist over the whole domain or just at a point? Oct 5, 2014 at 17:51
• Just one pointt Oct 5, 2014 at 17:52
– Did
Oct 15, 2014 at 12:23

• Yes, if the second derivatives $\:f_{xx}$ and $\:f_{yy}$ are continuous, and
• No (not necessarily), if $\:f_{xx}$ and $\:f_{yy}$ are discontinuous.

For your specific case some clarification is required.

If by saying "second derivatives $\:f_{xx}$ and $\:f_{yy}$ exist at a point $P$" you mean that the corresponding right and left limits exist and are equal, i.e. for a point $P = (x,y)$

$$\lim\limits_{\Delta x \to 0} \dfrac{f_x(x + \Delta x,y) - f_x(x , y)}{\Delta x} = \lim\limits_{\Delta x \to 0} \dfrac{f_x(x,y) - f_x(x - \Delta x, y)}{\Delta x} \tag{1a}$$ and $$\lim\limits_{\Delta y \to 0} \dfrac{f_y(x,y + \Delta y) - f_y(x , y)}{\Delta y} = \lim\limits_{\Delta y \to 0} \dfrac{f_y(x,y) - f_y(x, y - \Delta y)}{\Delta y}, \tag{1b}$$ then YES, $\ f_{xy}$ exists, as conditions $\eqref{1a}$ and $\eqref{1b}$ imply that $f_{xx}$ and $f_{yy}$ are continuous at $P$.

Proving first case is not difficult. One write something like

For any function $f: \mathbb{R}^2 \to \mathbb{R}$ with continuous second partial derivatives at a point $P = (x,y)$ we have $f_{xy} = f_{yx}$ (Schwartz-Clairaut theorem). That means that the order in which you differentiate with respect to different variables does not matter. The existence and continuity of $f_{yy}$ implies that you can differentiate function $f$ at least once w.r.t. $y$ and preserve continuity. Similarly, we conclude that we can differentiate $f$ at least once w.r.t. $x$ and preserve continuity. These two operations are independent, so the result of applying both of them to $f$ will still be continuous, i.e. the corresponding right and left limits exist and are equal. That means that $f_{xy}$ exists and is continuous at the point $P$.

As for the second case, we can provide a counterexample. For instance, assume $P = (0,0)$ and consider function

$$f(x) = \begin{cases} \dfrac{xy(x^2-y^2)}{x^2+y^2} & \text{for } \ (x,y) \neq P,\\ (0,0) & \text{for } \ (x,y) = P. \end{cases}$$

Clearly $f$ is continuous at $P=(0,0)$, but its second partial derivatives are not.