When is partial differentiation commutative

Question

Consider a function: $$f:\mathbb R ^2\to\mathbb R. $$ When does $$\dfrac {\partial f(x,t)}{\partial t\partial x}=\dfrac {\partial f(x,t)}{\partial x\partial t}?$$

Thinking about it in terms of the limit definition of derivative, and thinking about it as taking slices of surfaces and measuring the slope on the edge, seems to be giving me the feeling that answer is always.

However I have some memories of there being requirements like piecewise continuous, and smooth.

What is the full set of conditions?

It's true where the partial derivatives are all continuous. (But not an iff statement as far as I'm aware). — ah11950, Mar 25 '14 at 01:44
ah11950: That sounds like an answer (not a comment). Why don't you post it as an answer? — Frames Catherine White, Mar 25 '14 at 01:48
The point is that the discrete version of your formula, that is $$\Delta_t\Delta_x f = \Delta_x \Delta_t f, $$ where $\Delta_t f(x, t)=f(x, t +h)$ for a fixed increment $h$, is always true. However, to infer from this that partial derivatives commute you need a limiting process and so you need at least some continuity. — Giuseppe Negro, Mar 25 '14 at 02:10
A notational quibble: I would write $\partial^2f$ where you wrote $\partial f.$ — David K, Jan 19 '25 at 04:27

score 4 · Accepted Answer · answered Mar 25 '14 at 01:58

4

It's true where the mixed partial derivatives are all continuous. (But not an iff statement as far as I'm aware).

An example of a function that fails to satisfy equality of mixed partial derivatives is

$$f(x,y) = \begin{cases} \frac{xy(x^2-y^2)}{x^2+y^2} \; &\text{if}\; (x,y) \neq 0\\ 0 &\text{if}\; (x,y) = 0\end{cases}$$

At the origin we have $f_{xy}(0,0) = 1 \neq -1 = f_{yx}(0,0)$

answered Mar 25 '14 at 01:58

ah11950

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How did you calculate $f_{xy}, f_{yx}$ so easily? Is there some special technique to this? – PythonSage Mar 31 '20 at 23:56
@PythonSage i think it is the other way around. We note (using the definition of derivative, and relabelling the infinitesimals, let's say $(h,k)$, with $(x,y)$, and that $f(x,y)=0$ for $x=0$ or $y=0$) that
$$f_{xy}(0,0)= \lim_{y\to0}\frac{1}{y} \lim_{x\to 0}\frac{1}{x} f(x,y)= \lim_{y\to0}\frac{1}{y} (-y)=-1 $$

and $$f_{yx}(0,0)= \lim_{x\to 0}\frac{1}{x} \lim_{y\to0}\frac{1}{y} f(x,y) =\lim_{x\to 0}\frac{1}{x}(+x) = +1$$
– Benjamin Wang Jan 19 '25 at 02:45

score 1 · Answer 2 · answered Jan 19 '25 at 03:02

Late to the party: this does have associated keyword "Clairault's theorem", though I hesitate to think that was the first consideration of this issue. :)

Yes, additional hypotheses are needed for pointwise equality of the two mixed partials.

The point I'd want to make is that the two mixed partials are always equal, in fact for tempered distributions, as tempered distributions. The proof is easy: the question of whether $\partial_x\partial_y f$ is equal to $\partial_y\partial_x f$ is equivalent, by Fourier transform, to the question of whether $xy\hat{f}=yx\hat{f}$. Of course, "yes". :)

Once I realized this, I stopped worrying about the fretful pointwise comparison. :)

When is partial differentiation commutative

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