Re: must specify free variable for consistent operator on L^2(R)?

On Tue, 12 Jun 2007 07:52:53 -0000, Dan Greenhoe <dgreenhoe@xxxxxxxxx>
wrote:

Suppose I want to define a "dilation" operator D on L^2(R) such that
D x(t) = x(2t) ... (t is the free variable).

What about D x(u(t))?

The problem here is not whether it's possible to define a consistent
whatever, the problem is just that the notation is ambiguous, or
at least may not mean what you intend.

If you _want_ to say x(2u(t)) that's exactly the same
thing as (Dx) (u(t)). If you want to specify x(u(2t))
that's D(f o u)(t) (where the o is that little circle
denoting composition.)

Is it possible to define a *consistent* dilation operator D that
"dilates" with respect to the argument u(t) rather than the free
variable t?

That is, is it possible to define D in a consistent manner such that
D x(u(t)) = x(2u(t)) ... (rather than D x(u(t)) = x(u(2t)) )

In this case, D seems inconsistent because
x(t) = (3t+1)^2 ==> D x(t) = (6t+1)^2
y(t) = t^2 ==> D y(3t+1) = (6t+2)^2

which is inconsistent because
D (3t+1)^2 = Dx(t) = (6t+1)^2
not=
(6t+2)^2 = D y(3t+1) = D (3t+1)^2

So then my question is, when defining an operator on L^2(R), is it
always necessary to also specify the free variable (e.g. "t")? Because
without defining exactly what the free variable is, won't this lead to
inconsistent definitions of operators on L^(R)?

For example, when defining a "dilation" operator D on L^2(R), must I
also specify exactly what the free variable is?

Again, there's no problem with the way you defined the dilation,
the problem, if any, is just that it's not clear what function
you mean to apply the dilation _to_ when you write D x(u(t)).

(If I saw the notation exactly as you wrote it I'd probably
assume that what you intended was (Dx)(u(t)); if you mean the
other you need some other notation.)

Many thanks in advance,
Dan Greenhoe


************************

David C. Ullrich
.

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