hilbert space

mona123 · Mar 2, 2015

Hi can someone please help me to prove this equivalence: Let A∈B(X) and let {A_n}⊂B(x)

such that there is a constant C with ∥A∥≤C and ∥A_n∥≤C for all n.

Show that A_n→A in B(X) if and only if there is a dense subset Y⊂X such that A_nx→Ax for all x∈Y.

here is my answer can you please correct it:

Let X be a normed vector space. Let A ∈ B (X) and let {A_n} ⊂ B (X) such that exist a

constant C with || A || ≤ C || A_n || ≤ C for all n.

1) assume that A_n → A in B (X) and show that there exist a dense subset Y ⊂ X such that

A_nx → Ax for all x ∈ Y we have

|| A_n - A || = sup_{x∈X \ {0}} || A_n (x) - A (x) ||_{/ || x ||}

So we have for all x∈X || A_n (x) - A (x) || ≤ || A_n - A || → 0 as n tends to ∞

( by assumption since A_n → A into B (X))

This implies that for any x∈X A_n (x) → A (x) hence there exist a subset Y = X dense in

X such that A_n (x) → A (x) for all x ∈ Y

2) assume that there exist a dense subset Y ⊂ X such that A_nx → Ax for all x ∈ Y and

prove that A_n → A into B (X)

either x∈ X

*] As Y is dense in X, there exists a sequence (y_n)∈Y such that y_n → x

So ∀ x ∈ X, ∀ε> 0, ∃N1> 0 telque ∀n≥N1 || x-y_n || ≤ε_{/ 3c}

*] Moreover by hypothesis we have

∀y ∈ Y, ∀ε> 0, ∃N2> 0 such that ∀n≥N2 ||A_n (y) - A (y) || ≤ε_{/ 3}

so ∀ x ∈ X ,∀ε> 0 ∃N = max (N1, N2)> 0 such that

∀n≥N: || A_n x-Ax || = || (A_n x-A_n y_n) + (A_n y_n-Ay_n) + (Ay_n-A x) ||

≤ || A_n (x) - A_n (y_n) || + || A_n (y_n) - A (y_n) || + || A (y_n) -. A (x) ||

≤ || A_n ||. || x-y_n || + || A_n (y_n) - A (y_n) || + || A ||. ||x-y_n ||

≤c.ε_/3c + ε _{/ 3} + c.ε_{/ 3c} = ε

thus we have for all ∀ x ∈ X,∀ε> 0 ∃N> 0 such that ∀n≥N: || A_n x-Ax ||≤ε

which gives A_n → A in B (X). thanks in advance