Let $H$ be a Hilbert space with scalar product $\langle \cdot,\cdot\rangle$ and corresponding norm $\Vert \cdot \Vert$. Let $(f_k)_{k\in\mathbb N}$ be a sequence in $H$. Show TFA.
i) There exists some constant $B>0$ such that $\sum_{k\in \mathbb N}\vert \langle f,f_k\rangle\vert^2\le B\Vert f\Vert^2$ for all $f\in H$.
ii) The operator $T:l_2(\mathbb N)\to H$ given by $T((c_k)_{k\in\mathbb N}):=\sum_{k\in\mathbb N} c_kf_k$ is well-defined and bounded from $l_2(\mathbb N)$ into $H$ and $\Vert T\Vert\le \sqrt B$.
I have shown $i)\to ii)$ with Riesz Representation Theorem. But I do not see how to handle the other direction. I started with $$\sum_{k\in \mathbb N}\vert \langle f,f_k\rangle\vert^2\le \sum_k \Vert f\Vert^2\Vert f_k\Vert^2,$$ therefore we have to show that $\sum_k \Vert f_k\Vert^2\le \Vert T\Vert^2$.
How can one continue here? Any help is welcome!
Best Answer
It is given that $\|T(c_k)\| \leq \sqrt B \sqrt {\sum |c_k|^{2}}$ for all $(c_k) \in l_2$. This gives $|\sum c_k \langle f_k, f \rangle|\leq \|f\| \sqrt B \sqrt {\sum |c_k|^{2}}$. Fix $N$ and take $c_k$ to be the complex conjugate of $\langle f_k, f \rangle|$ for $k \leq N$ and $0$ for $k >N$. After division by $\sqrt {\sum |c_k|^{2}}$ this gives $\sum\limits_{k=1}^{N} |\langle f_k, f \rangle|^{2} \leq \|f\|^{2}B$. Let $N \to \infty$ to finish the proof.