# Equivalence of norms

1. Aug 24, 2009

### quasar987

Let H be a separable Hilbert space and let {e_k} be a Hilbert basis (aka total orthonormal sequence) for H. Then

$$|||u|||_1:=\sum_{k=1}^{+\infty}\frac{1}{2^k}|(e_k,u)|$$

is a norm. If {f_k} is another Hilbert basis, we get another norm by setting

$$|||u|||_2:=\sum_{k=1}^{+\infty}\frac{1}{2^k}|(f_k,u)|$$

How to show that these two norm are equivalent???

2. Aug 24, 2009

### morphism

Try using the fact that

$$f_k = \sum_l (f_k,e_l) e_l.$$

3. Aug 24, 2009

### quasar987

I did:

$$|||u|||_2 =\sum_k\frac{1}{2^k}|(f_k,u)|=\sum_k\frac{1}{2^k}\left|\left(\sum_l(f_k,e_l)e_l,u\right)\right| = \sum_k\frac{1}{2^k}\left|\sum_l(f_k,e_l)(e_l,u)\right|$$
$$\leq\sum_k\frac{1}{2^k}\sum_l|(f_k,e_l)||(e_l,u)|$$

and then I'm stuck...