# Thread: Inequality in Separable Hilbert space.

1. ## Inequality in Separable Hilbert space.

Prove if H is a separable Hilbert space and $\{y_\alpha\}_{\alpha\epsilon A}$ is a set of pairwise orthogonal vectors in H such that for any $\alpha\epsilon A, 0<\|y_\alpha\|\leq b$ for some constant b, then $\sum_{\alpha\epsilon A}||^2\leq b^2\|x\|^2$ for any $x \epsilon H.$

What I have:
$\sum_{\alpha\epsilon A}||^2\leq\sum_{\alpha\epsilon A}(\|x\|\|y_\alpha\|)^2$ by Schwarz Inequality
$=\sum_{\alpha\epsilon A}\|x\|^2\|y_\alpha\|^2$
$\leq\sum_{\alpha\epsilon A}\|x\|^2b^2$
$=\|x\|^2b^2$

I just want to make sure that I can drop the sum. I think I can do this since the $y_\alpha$ are pairwise orthogonal, but I am not sure.

2. You can't just drop the sum. You'd have to multiply $\|x\|^{2}\,b^{2}$ by the cardinality of $A$, which I don't think you want to do. It's no different, mathematically, from the sum

$\displaystyle{\sum_{k=1}^{n}1=n}.$

Your proof doesn't appear to use separability. I'm not sure off-hand how that can work, at the moment, but I have a feeling that's important. I do know that there's a theorem that says any orthornomal set in a separable Hilbert space is countable. Your set is not at the moment orthonormal, but it is pairwise orthogonal. You could simply run through your set and normalize each one to get an orthonormal set. What you don't know is if your set $\{y_{\alpha}\}_{\alpha\in A}$ is a basis. If it were, you could write $x$ as a linear combination of elements of the set, and then I think your proof would be rather straight-forward. As it is, you'll have to do something else. Assuming you've orthonormalized your set, you do know the distance between any two elements in the set (think unit vectors).

So, while I don't know the proof solution, I can answer your question. Hope this helps.

3. I don't think it matters whether the space is separable or not. Only countably many of the inner products $\langle x,y_\alpha\rangle$ can be nonzero, so we may as well assume that the index set A is countable.

As Ackbeet suggests, you can normalise the set by letting $z_\alpha = y_\alpha/\|y_\alpha\|$. Then $\{z_\alpha\}$ is an orthonormal set, and it follows from Bessel's inequality that $\sum_{\alpha\in A}|\langle x,z_\alpha\rangle|^2 \leqslant\|x\|^2$. Thus $\displaystyle\sum_{\alpha\in A}\frac{|\langle x,y_\alpha\rangle|^2 }{\|y_\alpha\|^2}\leqslant\|x\|^2$. But $1/b^2\leqslant1/\|y_\alpha\|^2$, and that gives you the inequality that you want.