Square Summable coefficients in a Hilbert Space, when the basis is NOT orthonormal/orthogonal.

65 Views Asked by At

If $\{f_j\}_{j = 1}^{\infty}$ is a bounded sequence of vectors in a Hilbert space $\mathcal{H}$, and $g \in \mathcal{H}$ is such that

$$g = \sum_{j = 1}^{\infty}c_jf_j,$$

then must we have that

$$\sum_{j = 1}^{\infty}|c_j|^2 < \infty?$$

I would like the answer to be yes, and this seems like a basic exercise in analysis, but I have unfortunately been unable to solve it. One of the applications that I am hoping to use this for is the following.

Given a Hilbert space $\mathcal{H}$, let us say that $\{f_j\}_{j = 1}^{\infty} \subseteq \mathcal{H}$ is approximately orthogonal if

$$\sum_{j < k}|\langle f_j, f_k\rangle|^2 < \infty.$$

Conjecture (contingent upon the initial question): If $\{f_j\}_{j = 1}^{\infty} \subseteq \mathcal{H}$ is approximately orthogonal, and $g \in \mathcal{H}$ is arbitrary, then

$$\sum_{j = 1}^{\infty}|\langle g,f_j\rangle|^2 < \infty.$$

1

There are 1 best solutions below

1
On BEST ANSWER

Could this be a counterexample?

Choose $\mathcal{H}=l^2$, $c_j=j^{-1/2}$ for all $j$. Set $f_1=e_1$ and (for $j>1$) $f_j=e_{j}-\frac{c_{j-1}}{c_j}e_{j-1}$ where $e_j$ is the standard basis in $l^2$. Thus $$\sum_j c_jf_j=0\in l^2.$$ Of course $\sum|c_j|^2=\infty.$