Proving U ⊕ V is dense in ℓ₂^∞ using orthogonal projections

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
noospace
Messages
69
Reaction score
0

Homework Statement



Let [itex]e_i = (0,0,\ldots, 1, 0 , \ldots)[/itex] be the basis vectors of the Hilbert space [itex]\ell_2^\infty[/itex].

Let U and V be the closed vector subspaces generated by [itex]\{ e_{2k-1}|k \geq 1 \}[/itex] and [itex]\{ e_{2k-1} + (1/k)e_{2k} | k \geq 1 \}][/itex].

Show [itex]U \oplus V[/itex] dense in [itex]\ell_2^\infty[/itex]

I am looking for hints that anyone can offer.

The Attempt at a Solution



My main problem seems to be finding the most direct method of proof.

First I tried proving that [itex]\overline{U\oplus V} = \ell_2^\infty[/itex] using the set theoretic procedure of show each side of the equation is a subset or equal to the opposite side. I tried to use the fact that every element of the LHS has a sequence in U + V converging towards it. This got me nowhere.

I next tried showing that if x is any point in the hilbert space, there should be a point in the direct sum within a distance epsilon > 0 of it. I managed to construct an expression for x in terms of the basis vectors, but this expression is precisely x so is useless.

Currently I'm thinking about projecting x onto the subspaces and then using the sum of the projections somehow. It seems like this won't work however, since there no dependence on epsilon.

Any help would be greatly appreciated.

Thanks,.
 
Physics news on Phys.org
If (xn) is in l2, then given [itex]\epsilon[/itex] > 0, there exists N such that

[tex]\sum _{n > 2N}|x_n|^2 < \epsilon[/tex]

So at this point it suffices to show that for any N, (x1, x2, ..., x2N-1, x2N, 0, 0, ...) is in [itex]U \oplus V[/itex].