Square summable coefficients in infinite linear combination.

69 Views Asked by At

I understand the reasoning behind all this paragraph, what I would like to know is if the condition $\sum_{k=1}^\infty|\beta_k|^2<\infty $ is a necessary condition for all infinite linear combination in a Hilbert Space, or if it is used only for the sake of the proof.

1

There are 1 best solutions below

1
On BEST ANSWER

It's definitely a necessary condition. If $x=\sum_{k=1}^\infty \beta_k x_k\in H$ then, since $<\cdot,\cdot>: H\times H\to\mathbb C$, we have that $$\mathbb C \ni <x,x>=\left<\sum_{k=1}^\infty \beta_k x_k,\sum_{k=1}^\infty \beta_k x_k\right>=\sum_{k=1}^\infty\sum_{j=1}^\infty\bar\beta_k\beta_j<x_k,x_j>=\sum_{k=1}^\infty|\beta_k|^2$$ thus an infinite linear combination of $x_j$ can only be an element of $H$ if the series of its coefficients is square summable.