Independence of Leibniz rule and locality from other properties of the derivative? - MathOverflow

Independence of Leibniz rule and locality from other properties of the derivative?

2012-10-04T02:39:29Z

The following is meant to be an axiomatization of differential calculus of a single variable. To avoid complications, let's say that $f$, $g$, $f'$, and $g'$ are smooth functions from $\mathbb{R}$ to $\mathbb{R}$ ("smooth" being defined by the usual Cauchy-Weierstrass definition of the derivative, not by these axioms, i.e., I don't want to worry about nondifferentiable points right now). In all of these, assume the obvious quantifiers such as $\forall f \forall g$.

Z. $\exists f : f'\ne 0$

U. $1'=0$

A. $(f+g)'=f'+g'$

C. $(g \circ f)'=(g'\circ f)f'$

P. $(fg)'=f'g+g'f$

L. The value of $f'(x)$ is determined by knowing $f$ in any neighborhood of $x$.

Axioms A and P hold in any differential algebra. C and L mean that we're talking about something more specific than a differential algebra; they're meaningful only because we're talking about a ring of functions.

I could choose to omit U, since it can be proved from the others. I would prefer to keep U and omit P. Is P superfluous, or can anyone find a model in which P fails?

Likewise, is L independent of the others?

What seems to be tricky is to rule out models of the general flavor of $f'(x)=f^{CW}(x-17)$, where $f^{CW}$ is the usual Cauchy-Weierstrass derivative of $f$.

Are there models that are not the same as CW? Is this a nice axiomatization? Could it be improved?

[EDIT] Tom Goodwillie didn't say so explicitly, but his answer, along with one of my comments below his answer, shows that Z, A, and C suffice, so U, P, and L are not needed.

It looks like you can also take P as an axiom and recover the standard derivative, i.e., either P or C can be proved from the other: http://mathoverflow.net/questions/44774/do-these-properties-characterize-differentiation

Answer by Pietro Majer for Independence of Leibniz rule and locality from other properties of the derivative?

2012-10-04T09:20:15Z

Actually L excludes that $f'$ is a translate of the usual derivative $D$ (for instance there are $C^{\infty}$ functions with compact support). Moreover, by C, with $g=f=\mathrm{id}$ we have $x'=(x')^2$, and by the assumed continuity of $x'$, $x'$ is either $1$ or $0$ identically; the former is excluded taking, again in C, $g=\mathrm{id}$ and $f$ as in Z. By P we get $f'=Df$ at least for any polynomial function. So you'd have it for all smooth function, provided you also assume that $f\mapsto f'$ is continuous in the $C^1_{loc}$ sense.

With no other assumption, I'm not sure if one can reach this conclusion. However, the set of $f$ for which $f'=Df$ is quite a wide one; for instance, by C and L, it includes any $f$ that locally solves a functional equation like $p\circ f=q$, for non-zero polymonials $q$ and $p$; or also $f\circ p=q$, with $p$ of odd degree (hence surjective).

Answer by Tom Goodwillie for Independence of Leibniz rule and locality from other properties of the derivative?

2012-10-04T11:11:28Z

If you grant that $c'=0$ when $c$ is a constant, you can argue as follows:

EDIT: Actually $c'=0$ follows from $1'=0$ using the chain rule, since $c=c\circ 1$.

Let $I(x)=x$. Since $I=I\circ I$, the chain rule gives $(I')^2=I'$, so $I'$ is the characteristic function of a set $A$ of real numbers. Let $T_c(x)=x+c$. Then $T_c'=I'+c'=I'$, and the chain rule applied to $T_c=I\circ T_c$ then implies $I'=(I'\circ T_c)I'$. That means that the set $A$ is either all or nothing. But you can't have $I'$ identically $0$ because that would imply $f'=(f\circ I)'=0$ for all $f$. So $I'=1$ as expected.

Now consider linear functions. If $L_m(x)=mx$, then since $L_m(x+a)-L_m(x)$ is constant, $L_m'(x+a)-L_m'(x)$ is zero. Thus $L_m'$ is a constant depending on $m$, say $h(m)$. The map $h$ is an additive homomorphism, and by the chain rule it is also multiplicative. We also know $h(1)=1$. This implies that $h(m)=m$ for all $m$. (A ring map from reals to reals preserves squares, therefore preserves ordering, therefore is continuous ...)

So $f'$ is what it should be when $f$ is polynomial of degree at most one. Now let $S(x)=x^2$. From $S(x+t)=S(x)+2tx+t^2$ we get $S'(x+t)=S'(x)+2t$, therefore $S'(x)=S'(0)+2x$. But $S'(0)=0$ using $S(-x)=S(x)$. So the derivative of squaring is what it should be.

Now the special case of Leibniz that says $(f^2)'=2ff'$ follows by the chain rule. The general case follows by expressing $fg$ in terms of $f^2$, $g^2$, and $(f+g)^2$.

EDIT: This was all about global functions. But it can be extended. Let me spell out what I hope your axioms are: $f'$ is defined when $f$ is a $C^\infty$ real function whose domain is an open set $U\subset \mathbb R$, and $f'$ is another such function with the same domain. The axioms are

(U) $1_U'=0_U$ where $1_U$ is the constant function on $U$.

(A) $(f+g)'=f'+g'$ where $f$ and $g$ (and $f+g$) have the same domain.

(C) $(f\circ g)'=(f'\circ g)g'$, when $f$ has domain $U$ and $g$ has domain $V$ and $g(V)\subset U$, so that $f\circ g$ and $f'\circ g$ have domain $V$.

(Z) For every nonempty $U$ there is some $f$ with domain $U$ such that $f'$ is not identically zero.

The arguments that I gave above can be adapted to show then that:

$c_U'=0$ for any constant function on any $U$.

$I_U'=1$ where $I_U$ with domain $U$ is defined by $I_U(x)=x$. (Here you have to mess around with compositions $I_U\circ (I_V+c_V)$.)

So in the end you get the desired localization property, too: the operator commutes with restriction from $U$ to $V\subset U$ by the chain rule, because restriction is composition with $I_V$.