| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1129.1 | | AITG::DERAMO | like a candle in the wind | Sun Oct 29 1989 12:46 | 11 |
| >> Find an example of a normed vector space X such that the only open
>> convex subsets of X are the empty set and X itself.
So far my only example is the one point space. :-)
It seems that if the space is R^n, you can delete an n-1
dimensional piece containing the origin, and separating
what's left into two open convex pieces. Do the only
nontrivial examples have infinite dimension? Is the
Hilbert Cube an example?
Dan
|
| 1129.2 | | JAKES::XIA | In my beginning is my end. | Sun Oct 29 1989 16:38 | 4 |
| p
Hint: consider L ([0,1]) where 0 < p < 1.
Eugene
|
| 1129.3 | | AITG::DERAMO | ga naar je kamer | Tue Nov 28 1989 22:26 | 11 |
| >> p
>> Hint: consider L ([0,1]) where 0 < p < 1.
p
One of my books states that L [a,b] is a normed linear
p
space for p >= 1 with norm || f || = integral(a,b) |f(x)| dx
What norm are you using for 0 < p < 1?
Dan
|
| 1129.4 | | HPSTEK::XIA | In my beginning is my end. | Wed Nov 29 1989 10:45 | 13 |
| p
> One of my books states that L [a,b] is a normed linear
Well, the same definition works for 0 < p < 1 too.
p
> space for p >= 1 with norm || f || = integral(a,b) |f(x)| dx
> What norm are you using for 0 < p < 1?
p 1/p
It should be ||f|| = (integral(a,b) |f| dx)
Eugene
|
| 1129.5 | | AITG::DERAMO | ga naar je kamer | Wed Nov 29 1989 18:22 | 17 |
| re .4
>> > One of my books states that L [a,b] is a normed linear
>>
>> Well, the same definition works for 0 < p < 1 too.
No it doesn't, the triangle inequality is violated.
p
>> > space for p >= 1 with norm || f || = integral(a,b) |f(x)| dx
>> > What norm are you using for 0 < p < 1?
>>
>> p 1/p
>> It should be ||f|| = (integral(a,b) |f| dx)
Oops ... I meant to say that. :-)
Dan
|
| 1129.6 | | HPSTEK::XIA | In my beginning is my end. | Wed Nov 29 1989 22:16 | 13 |
| re .5,
You are right Dan. I am afraid I screwed up big on this one. The
problem is that when I saw L^p, I immediately thought it was a normed
vector space. Oh well, I apologize to whoever is trying on this
problem. The problem will work if we relax "normed vector space" to
"topological vector space" with the metric defined as:
p
d(f,g) = integral (a,b) |f-g| dx. With this the triangular
inequality works, but the scalar multiplications rule of the
norm space does not work. Sorry about that.
Eugene
|
| 1129.7 | | AITG::DERAMO | Dan D'Eramo, nice person | Mon Feb 26 1990 19:27 | 34 |
| In R^2 if you use the function
f(a,b) = |a1 - b1|^p + |a2 - b2|^p
with a=(0,0) and consider the set of all points b such
that f(a,b) < 1, then you get:
1) At p=2, the set is the interior of a ball of radius 1,
and is convex
2) At p=3, the set includes a ball of radius 1, and is
convex, and lies insides the square centered at the origin
with side 2.
3) As p->oo, the set approaches being the interior of the
square centered at the origin and with side 2, and is convex.
4) At p=1, the set is the interior of the diamond centered
at the origin with edges the lines connecting (+/-1,0) and
(0,+/-1), and is convex.
5) For 0 < p < 1, the set extends out towards (+/-1,0) and
(0,+/-1) but lies within the diamond (square) of part 4, and
is not convex.
Now if f is a distance function for 0 < p < 1 then it gives
a distance function in which the open balls fail to be convex.
So it is possible that open sets made up of them can fail to
be convex if they aren't the entire space. If you are thinking
of some specific function space with d(f,g) = integral (a,b) |f-g|^p dx
as an example then it probably works for a reason analogous to
this.
Dan
|