Differentiability as a topological property of vector spaces?
sg...@hotmail.co.uk
Given two normed vector spaces U and V and a function f: U --> V we
say that f is differentiable at x in U and has a linear map f' as a
derivative if
lim_{|h| --> 0} {|f(x+h) - f(x) - f'(h)|/|h|} = 0
(I am guessing that uniqueness of f' is easy to prove, but I use the
indefinite article above since I don't actually care, at least for the
purpose of this post).
It seems as though the existence of metrics on U and V is necessary
for the definition given above, however I *think* I can show that if
we choose a new metric on either U or V which is topologically
equivalent to the old one then the question of whether f' is a
derivative of f is unaffected. The proof involved showing that the
usual chain rule (from IR^n) for the derivative of g o f remains true
provided that f' and g' are both continuous; then using the fact that
the identity on U or V with topologically equivalent metrics is
differentiable with the identity as a continuous derivative.
My question therefore has two parts:
1) is what I have written above true, and if so
2) is there a definition of the derivative of a function from one
topological vector space to another, which coincides with the usual
one when the topologies are metrizable?
-Rotwang
David C. Ullrich
>Hi. I have a possibly dumb question.
>
>Given two normed vector spaces U and V and a function f: U --> V we
>say that f is differentiable at x in U and has a linear map f' as a
>derivative if
>
>lim_{|h| --> 0} {|f(x+h) - f(x) - f'(h)|/|h|} = 0
>
>(I am guessing that uniqueness of f' is easy to prove, but I use the
>indefinite article above since I don't actually care, at least for the
>purpose of this post).
>
>It seems as though the existence of metrics on U and V is necessary
>for the definition given above, however I *think* I can show that if
>we choose a new metric on either U or V which is topologically
>equivalent to the old one then the question of whether f' is a
>derivative of f is unaffected. The proof involved showing that the
>usual chain rule (from IR^n) for the derivative of g o f remains true
>provided that f' and g' are both continuous; then using the fact that
>the identity on U or V with topologically equivalent metrics is
>differentiable with the identity as a continuous derivative.
>
>My question therefore has two parts:
>
>1) is what I have written above true,
Well, it seems possible from what you've written that
you're under the impression that a metrizable TVS is
the same as a TVS where the topology is given by
a norm. That's not so.
If you're really talking about normed TVSs above then
everything above is certainly true, just because if
two norms induce the same topology then they're
equivalent (each is <= a constant times the other).
If you're really asking about just metrizable TVSs
(where presumably the |h| in the definition is
replaced by d(h,0)) then I'm not sure whether
it's all true, I doubt it.
>and if so
>2) is there a definition of the derivative of a function from one
>topological vector space to another, which coincides with the usual
>one when the topologies are metrizable?
>
>-Rotwang
************************
David C. Ullrich
sg...@hotmail.co.uk
> >Given two normed vector spaces U and V and a function f: U --> V we
> >say that f is differentiable at x in U and has a linear map f' as a
> >derivative if
>
> >lim_{|h| --> 0} {|f(x+h) - f(x) - f'(h)|/|h|} = 0
> >
> >It seems as though the existence of metrics on U and V is necessary
> >for the definition given above, however I *think* I can show that if
> >we choose a new metric on either U or V which is topologically
> >equivalent to the old one then the question of whether f' is a
> >derivative of f is unaffected.
I should have written "continuous derivative" here.
> >The proof involved showing that the
> >usual chain rule (from IR^n) for the derivative of g o f remains true
> >provided that f' and g' are both continuous; then using the fact that
> >the identity on U or V with topologically equivalent metrics is
> >differentiable with the identity as a continuous derivative.
>
> >My question therefore has two parts:
>
> >1) is what I have written above true,
David Ullrich wrote:
> Well, it seems possible from what you've written that
> you're under the impression that a metrizable TVS is
> the same as a TVS where the topology is given by
> a norm.
Indeed, that much occured to me after I posted.
> That's not so.
Could you point me in the direction of a counterexample?
> If you're really talking about normed TVSs above then
> everything above is certainly true, just because if
> two norms induce the same topology then they're
> equivalent (each is <= a constant times the other).
News to me I'm afraid. Though now that you mention it it is pretty
obvious.
> If you're really asking about just metrizable TVSs
> (where presumably the |h| in the definition is
> replaced by d(h,0)) then I'm not sure whether
> it's all true, I doubt it.
Thanks.
-Rotwang
Lee Rudolph
>I wrote:
>
>> >Given two normed vector spaces U and V and a function f: U --> V we
>> >say that f is differentiable at x in U and has a linear map f' as a
>> >derivative if
>>
>> >lim_{|h| --> 0} {|f(x+h) - f(x) - f'(h)|/|h|} = 0
>> >
>> >It seems as though the existence of metrics on U and V is necessary
>> >for the definition given above, however I *think* I can show that if
>> >we choose a new metric on either U or V which is topologically
>> >equivalent to the old one then the question of whether f' is a
>> >derivative of f is unaffected.
>
>I should have written "continuous derivative" here.
Still ambiguous. Do you mean that you want f'(x) to be a continuous
linear map from U to V for your given x, or do you mean that you want
the map x |-> f'(x) to be a continuous map from U into an appropriately
topologized space of linear maps from U to V? I guess the former, but
am not sure.
...
>> If you're really asking about just metrizable TVSs
>> (where presumably the |h| in the definition is
>> replaced by d(h,0)) then I'm not sure whether
>> it's all true, I doubt it.
Kriegl and Michor's book, available free online at
http://www.ams.org/online_bks/surv53/, is full of information
about calculus in really hairy TVSs -- much more general (as
I recall; I read the book in hardback, and don't have the kind
of connection it would take to download it now) than Frechet
spaces and other such metrizable, non-normable TVSs. Of course
a lot of the Banach-space theory falls apart, but a lot can be
salvaged, too.
Lee Rudolph
sg...@hotmail.co.uk
> >I should have written "continuous derivative" here.
>
> Still ambiguous. Do you mean that you want f'(x) to be a continuous
> linear map from U to V for your given x, or do you mean that you want
> the map x |-> f'(x) to be a continuous map from U into an appropriately
> topologized space of linear maps from U to V?
The first one, since I needed to use the boundedness of f' and g' in
my attempt to prove the chain rule. I think that, assuming that f is
differentiable at x, continuity (as in boundedness) of f'(x) is
equivalent to continuity of f at x.
> Kriegl and Michor's book, available free online athttp://www.ams.org/online_bks/surv53/, is full of information
> about calculus in really hairy TVSs -- much more general (as
> I recall; I read the book in hardback, and don't have the kind
> of connection it would take to download it now) than Frechet
> spaces and other such metrizable, non-normable TVSs. Of course
> a lot of the Banach-space theory falls apart, but a lot can be
> salvaged, too.
Thanks, I will check it out.
-Rotwang
David C. Ullrich
A very common example: Say r_n is a semi-norm on X.
(So r_n is a norm except that r_n(x) = 0 does not
imply x = 0.) Suppose that the r_n separate points
(so r_n(x) = 0 for _all_ n does imply x = 0.) Define
d(x,y) = sum 2^{-n} r_n(x-y)/(1+r_n(x-y)
Then d is a metric on X making X into a TVS which is
(typically) not normable.
For example, say X is the space of all continuous functions
on R and r_n(f) is the sup of |f(t)| for |t| <= n. Then
we get a metric on C(R) such that f_n -> 0 in this
metric if and only if f_n -> 0 uniformly on compact
sets (the standard topology on C(R).) This topology
is not given by a norm.
>> If you're really talking about normed TVSs above then
>> everything above is certainly true, just because if
>> two norms induce the same topology then they're
>> equivalent (each is <= a constant times the other).
>
>News to me I'm afraid. Though now that you mention it it is pretty
>obvious.
>
>> If you're really asking about just metrizable TVSs
>> (where presumably the |h| in the definition is
>> replaced by d(h,0)) then I'm not sure whether
>> it's all true, I doubt it.
>
>Thanks.
>
>-Rotwang
************************
David C. Ullrich