# Continuity of projection

1. Dec 25, 2007

### jostpuur

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.

2. Dec 25, 2007

### jostpuur

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: Dec 25, 2007
3. Dec 25, 2007

### mathwonk

just take a non closed subspace and a complement and project on the complement.