A sequence $\{f_{n}\}_{n\in I}$ is a frame for a separable Hilbert space $H$ if there exists $0<A\leq B<\infty$ such that $$ A\|f\|^{2} \leq \sum_{n\in I}|\langle f,f_{n}\rangle|^{2}\leq B\|f\|^{2} $$ for all $f\in H$. Some books define a frame for just "Hilbert space" and not mentioning the ...

A sequence $\{f_{n}\}_{n\in I}$ is a frame for a separable Hilbert space $H$ if there exists $0<A\leq B<\infty$ such that $$ A\|f\|^{2} \leq \sum_{n\in I}|\langle f,f_{n}\rangle|^{2}\leq B\|f\|^{2}$$ for all $f\in H$. If the sequence only satisfying the upper bound it is called a Bessel sequence...

$\mathcal H$ being a Hilbert space, $\{g_k\}_{k \in N}$ is a Bessel sequence if there exsits $B >0$ such that $\forall f \in \mathcal H$, $\sum_{k\in N} |\langle f,g_k\rangle|^2 \leq B \| f \|^2$. And it is a frame if there exists also $A >0$ such that $A \| f \|^2 \leq \sum_{k\in N} |\langle f,g...

I'm posting this question again because I'm still confused about the answer! A sequence $\{f_{n}\}_{n\in I}$ is called a Bessel sequence in a Hilbert space $H$, if there exists $B>0$ such that $$\sum_{n\in I}|\langle f,f_{n}\rangle|^{2}\leq B\|f\|^{2}$$ for all $f\in H$. Now my questin is: if...

Assume $H$ is a separable Hilbert space and $\{f_k\}_{k=1}^\infty$ is a sequence in $H$ such that : $$\forall f\in H : \sum_{k=1}^\infty \left|\langle f,f_k\rangle\right|^2 < \infty$$ Why $\{f_k\}_{k=1}^\infty$ is then a Bessel sequence ? That means that $$\exists B>0 : \forall f\in H : \sum_...

I've shown that if for a sequence $\{f_{n}\}_{n=1}^{\infty}$ in a Hilbert space $H$ we have $$\sum_{n=1}^{\infty}|\langle f,f_n\rangle|^{2}< \infty$$ for all $f\in H$ (i.e., it is a Bessel sequence in $H$), then the map $T$ from $\mathcal{l}^2$ to $H$ sending $(a_n)$ to $\Sigma_n a_n f_n$ is wel...