Suppose I have a function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $f$ has continuous first partial derivatives AND $f_{xy}$ exists. Let $x_1 \ne x_2, y_1 \ne y_2 \in \mathbb{R}$. Define $$\Delta_xf(y) = f(x_2, y) -f(x_1,y) \\ \Delta_y \Delta_xf = \Delta_xf(y_2) – \Delta_xf(y_1) = f(x_2,y_2) + f(x_1,y_1) – f(x_1,y_2) -f(x_2,y_1)$$ Is it true that there is some mean value theorem type result that states that there exist $\tilde x, \tilde y$ between $x_1$ and $x_2$ and $y_1$ and $y_2$ respectively such that $$\frac{\Delta_y \Delta_x f}{\Delta y \Delta x} = f_{xy}(\tilde x, \tilde y)$$
for $\Delta y = y_2 – y_1, \Delta x = x_2 – x_1$? I've tried to prove this using the 1 dimensional mean value theorem for single derivatives. We know that for every $y \in \mathbb{R}$, there exists an $x(y) \in (x_1, x_2)$ or $(x_2, x_1)$ such that $$\frac{\Delta_x f(y)}{\Delta x} = f_x(x(y),y) \implies \frac{\Delta_y \Delta_x f}{\Delta y \Delta x} = \frac{f_x(x(y_2), y_2) – f_x(x(y_1), y_1)}{\Delta y}$$
If $x(y_1)$ and $x(y_2)$ were necessarily the same, we could just apply the mean value theorem in the univariate sense in the second argument of $f_x$ but this is not the case surely. Therefore I'm thinking that this doesn't hold. Any advice/help would be massively appreciated. Thanks!
Best Answer
Your notation actually makes it more difficult.
Let $g(u)=f(u,y_2)-f(u,y_1)$. Then $$g'(u) = f_x(u,y_2)-f_x(u,y_1) = f_{xy}(u,\tilde y)\Delta y$$ for some $\tilde y$. Thus, for some $\tilde x$, we have $$\Delta_y\Delta_x f = g(x_2)-g(x_1) = g'(\tilde x)\Delta x = f_{xy}(\tilde x,\tilde y)\Delta x\Delta y,$$ as desired.