Proving the Continuity of Projections in Vector Spaces

  • Context: Graduate 
  • Thread starter Thread starter jostpuur
  • Start date Start date
  • Tags Tags
    Continuity Projection
Click For Summary
SUMMARY

Projections in vector spaces are not always continuous, as demonstrated by the mapping from l1 to l1. The discussion highlights that while projections can map vectors to shorter ones, this does not guarantee continuity. The example provided shows that the mapping can lead to undefined behavior when considering vectors with infinite norms. A valid proof of continuity requires the subspace to be closed, which is contingent upon the properties of the projection operator.

PREREQUISITES
  • Understanding of vector spaces and subspaces
  • Familiarity with the concepts of projections in functional analysis
  • Knowledge of Hilbert spaces and their properties
  • Basic principles of continuity in mathematical mappings
NEXT STEPS
  • Study the properties of projections in Hilbert spaces
  • Learn about the projection theorem and its implications for closed subspaces
  • Investigate the differences between l1 and l spaces
  • Explore examples of continuous and discontinuous mappings in functional analysis
USEFUL FOR

Mathematicians, students of functional analysis, and anyone studying the properties of vector spaces and projections.

jostpuur
Messages
2,112
Reaction score
19
Are projections always continuous? If they are, is there simple way to prove it?

If P:V->V is a projection, I can see that P(V) is a subspace, and restriction of P to this subspace is the identity, and it seems intuitively clear that vectors outside this subspace are always mapped to shorter ones, but I don't know how to prove it.

If V was a Hilbert space, and we knew P(V) is closed, then I could prove this using the projection theorem. However only way to prove that P(V) is closed, that I know, is to use continuity of P.
 
Physics news on Phys.org
Is this a counter example?

P:l^1\to l^1

(1,0,0,0,0,0,0,\ldots)\mapsto (0,2,0,0,0,0,0,\ldots)
(0,1,0,0,0,0,0,\ldots)\mapsto (0,1,0,0,0,0,0,\ldots)
(0,0,1,0,0,0,0,\ldots)\mapsto (0,0,0,3,0,0,0,\ldots)
(0,0,0,1,0,0,0,\ldots)\mapsto (0,0,0,1,0,0,0,\ldots)
(0,0,0,0,1,0,0,\ldots)\mapsto (0,0,0,0,0,4,0,\ldots)
(0,0,0,0,0,1,0,\ldots)\mapsto (0,0,0,0,0,1,0,\ldots)
\cdots

hmh.. no it is not, because the mapping is not well defined, since (1,1,1,...) would be mapped to have infinite norm. But if we choose such vector spaces, where only finite amount of components can have non-zero values, then that could be it.

(It seems I mixed l^1 and l^{\infty}.)
 
Last edited:
just take a non closed subspace and a complement and project on the complement.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K