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Is showing that $f_{xy}$ is not equal to $f_{yx}$ at a given point sufficient/good to show that a function is discontinuous there?

Can you also suggest a clear method to find without graphing a method to tell whether a function is continuous or not for a multivariable function?

Further, but unrelated, does $f_{xy}=f_{yx}$ imply in all cases? I mean is $f_{xxyz}=f_{xyxz}=f_{zyxx}$...?

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2 Answers

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The answer to your first question is no. There are continuous, and even continuously differentiable, functions $f(x,y)$ satisfying $f_{xy}(0,0)\neq f_{yx}(0,0).$ In this situation, Clairaut's theorem ($C^2\Rightarrow f_{xy}=f_{yx}$) implies that $f_{xy}$ and $f_{yx}$ cannot both be continuous.

For the last question, notice that Clairaut's theorem applies to $C^2$ functions. If $f$ is $C^4,$ then your equalities will hold.

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I think there are actually 3 questions up there the OP wants answered. I think you only answered the first and third,unless I'm mistaken. –  Mathemagician1234 Oct 18 '12 at 23:10
@Mathemagician1234, yes that's true. I was hoping somebody else might chime in, as I don't have any concrete general suggestion for the second question. It seems that you too have only answered the first part of the question! –  Andrew Oct 18 '12 at 23:22
My bad. Went back and tried to fix it,but my TeX,as usual,is awful. Someone please fix it! –  Mathemagician1234 Oct 18 '12 at 23:25
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First question: It's very important when studying a delicate theorem of calculus to understand exactly what it says. Let's look at the exact statement of the equality is mixed partial derivatives theorem (which is also attributed to Herman Schwartz,so really should be called the Clairaut-Schwartz Theorem). The theorem states that if $f \colon \mathbb{R}^n \to \mathbb{R}$

has [[continuous function|continuous]] second [[partial derivatives]] at any given point in $\mathbb{R}^n $, say, $(a_1, \dots, a_n),$ then for $1 \leq i,j \leq n,$

$$\frac{\partial^2 f}{\partial x_i\, \partial x_j}(a_1, \dots, a_n) = \frac{\partial^2 f}{\partial x_j\, \partial x_i}(a_1, \dots, a_n).\,\!$$

In words, the partial derivations of this function are [[commutative operation|commutative]] at that point.

The statement clearly assumes the second order partial derivatives of each variable of f are continuous at the point chosen, it says absolutely nothing about the continuity of the function itself at that point-nothing about the lower order partials or the linear derivative function either, for that matter! The classic counterexample is the following (although there are more sophisticated ones):

$$f(x,y) = \begin{cases} \frac{xy(x^2 - y^2)}{x^2+y^2} & \mbox{ for } (x, y) \ne (0, 0)\\ 0 & \mbox{ for } (x, y) = (0, 0). \end{cases} $$

Prove to yourself this function does not have equal mixed second order partial derivatives-but it is nevertheless continuous at $(0,0)$ in the plane!

Second question: Well, clearly, if a multivariable function is differentiable in the strong sense, i.e. there exists a linear map that satisfies the limit that defines it for vector valued functions at a point in n-dimensional space, then the function is continuous there. If f is not differentiable there, then it becomes a very tricky game of calculation to find out if the function is continuous there since the limits for each variable holding the others constant must be calculated and they must be equal to the value of he map at that point regardless of which variables are held constant.This is equivalent to finding the value of the function on paths on the n-dimensional Euclidean space that specifies the domain where (n-1) variables are constant and the limit of f along the path "converges" to to the limit point. For example, in the counterexample above, we can randomly choose any line in the plane that intersect the surface defined by f. It can be proven using an epsilon-delta argument that for any such line, f is continuous at (0.0) regardless of "line of approach". Otherwise, we need to use a trial and error method to find if the limit does not exist. If it doesn't, we should be able to pick 2 lines of approach where we get 2 different one-sided limits of f at (0,0). And be careful here because remember: A function being defined at a point does not guarantee it's continuous there!

Here's a simple example of what I mean:

$$f(x,y) = (x^2 - y^2)$$

It's clear f(0,0) exists and is equal to 0. But this isn't sufficient for continuity at (0,0). Consider taking the limit of f at (0,0) along the line x=0. Then lim f(x,y) ----> (- y^2) as we approach (0,0) along x = 0.But if we approach along y =0, Then lim f(x,y) ----> (x^2) and clearly these limits for f don't match along these 2 paths in the plane. So the limit doesn't exist for his f as we approach (0,0) despite the function being defined at (0,0) and being equal to 0!

The third question is obviously true by the statement of the general theorem.

Hope that answered your question!

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Hi, I added dollar signs ${}$ to put your math in "math mode". I wasn't sure about the hyperlinks, you might have to try those again! –  Andrew Oct 18 '12 at 23:34
Thank You. I realized, my biggest misconception in this case was cleared by "The statement clearly assumes the second order partial derivatives of each variable of f are continuous at the point chosen". I had a misconception that its the original function that is assumed to be continuous. –  007resu Oct 19 '12 at 8:52
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