For multivariable differentiation, it does not require continuity when when taking partial derivatives. Similarly, when they apply the mean value theorem in $\mathbb R^n$, one can have a function $f:\mathbb R^n \mapsto \mathbb R$ such that its first order partial derivatives need not be continuous.

Can someone explain to me why can one apply this when a function has discontinuity?

Actual Lemma from text book: (shown in $\mathbb R^n$, not generalized)

Let $U$ be an open subset of the plane $\mathbb R^2$ that contains the point $(x_0,y_0)$ and suppose that the function $f:U \mapsto R$ has second-order derivatives. Then there are points $(x_1,y_1)$ and $(x_2,y_2)$ in $U$ at which

$$\frac{\partial ^{2}f}{\partial x \partial y}(x_{1},y_{1}) = \frac{\partial ^{2}f}{\partial y \partial x}(x_{2},y_{2})$$

Any help would be greatly appreciated.

  • $\begingroup$ Which text book states that? $\endgroup$ – Thomas Mar 27 '13 at 9:19
  • $\begingroup$ Advanced calculus Second edition by Patrick M Fitzpatrick $\endgroup$ – Syeung Apr 1 '13 at 17:11

After having a look at Lemma 13.11 in your textbook, I think you are confusing the application of the mean value theorem in one dimension with the one in higher dimensions.

I've never seen that lemma before, so I will try to explain the proof here, if only to have a reference without having to look for the exact page in the book again. Examine the second-order differences $$f(x_0 + r, y_0 + r) - f(x_0 +r, y_0) - f(x_0, y_0 +r) + f(x_0, y_0) \quad (1)$$ for some small positive $r$. You could have different displacements $r_1$ and $r_2$ for the two coordinates, positive or negative, but with small absolute value each, but choosing $r_1 = r_2$ is notationally more convenient (but try the proof with different displacements, it makes for an easy exercise). In any case, you want to get second-order derivatives from that. The mistake that I made when looking at this first was to apply the mean value theorem for $f(x_0+r,y_0+r) - f(x_0+r,y_0)$ in the second variable and likewise for the other difference. But this will lead you to an impasse (think about why). The trick is to consider the functions $$x \mapsto f(x,y_0+r) - f(x,y_0) \quad \text{and}\quad y \mapsto f(x_0+r,y) - f(x_0,y)$$ and apply the mean value theorem to them. This works because $f$ is supposed to have partial derivatives up to order 2 in all directions, so it is continuously differentiable (partially in both variables, if you like that equivalent statement better), which is more than we need to apply the MVT. We get that the differences (1) is equal to both$$r\left(\frac{\partial f}{\partial x}(x_1,y_0 +r) - \frac{\partial f}{\partial x}(x_1,y_0)\right) \quad \text{and} \quad r\left(\frac{\partial f}{\partial y}(x_0+r,y_2) - \frac{\partial f}{\partial y}(x_0,y_2)\right), \quad (2)$$ for some $x_1 \in [x_0,x_0+r]$ and $y_2 \in [y_0,y_0+r]$. But the function $y \mapsto \frac{\partial f}{\partial x}(x_1,y) - \frac{\partial f}{\partial x}(x_1,y)$ is continuous on $[y_0,y_0+r]$ and differentiable on the interior, which is all we need to apply the mean value theorem again. In particular, the derivative need not be continuous! A similar function does the job for the second difference in (2). So (1) is equal to $$r^2 \frac{\partial^2 f}{\partial y \partial x}(x_1,y_1) \quad \text{and} \quad r^2 \frac{\partial^2 f}{\partial x \partial y}(x_2,y_2)$$ for some $y_1,x_2$ in the right intervals. Dividing by $r^2$ we get the result.

Added: See also “The Mixed Partial Derivatives and the Double Derivative” by Donald H. Trahan, The American Mathematical Monthly , Vol. 76, No. 1 (Jan., 1969), pp. 76-77 (JSTOR link)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.