Takesaki theorem 1.8: Is functional well-defined? Consider the following fragment from Takesaki's book "Theory of operator algebra I":

Why is $f_x$ well-defined? It seems to depend on the choice of decomposition $\omega = \sum_{n=1}^\infty \alpha_n \omega_{\xi_n, \eta_n}$ (with $\{\alpha_n\}\in \ell^1$ and $\{\xi_n\}$ and $\{\eta_n\}$ orthonormal sequences). I think such a decomposition need not be unique in general.
Edit: $\mathscr{L}(\mathfrak{H})$ denotes bounded operators on the Hilbert space $\mathfrak{H}$ and $\mathscr{L}\mathscr{C}(\mathfrak{H})$ denotes compact operators on $\mathfrak{H}$.
 A: There is no uniqueness in the sense you say, but that's not what's needed. If
$$
\omega = \sum_{n=1}^\infty \alpha_n \omega_{\xi_n, \eta_n},\qquad \omega' = \sum_{n=1}^\infty \alpha_n' \omega_{\xi_n', \eta_n'}
$$
and $\omega=\omega'$, this means that for any $x\in\mathscr C(\mathfrak H)$ you have
$$
\sum_{n=1}^\infty \alpha_n \langle x\xi_n, \eta_n\rangle=\sum_{n=1}^\infty \alpha_n' \langle x\xi_n', \eta_n'\rangle,
$$
which, as compacts are sot-dense in $\mathscr L(\mathfrak H)$, is exactly what you need.
A: Claim: $\omega = \sum_n \alpha_n \omega_{\xi_n, \eta_n}$, viewed as a functional on $\mathscr{L}(\mathfrak{H})$, is weakly continuous on bounded subsets.
Proof: Suppose that $x_\lambda \to 0$ in the weak topology and $M:= \sup_\lambda \|x_\lambda\|<\infty$. Then the estimate
\begin{align*}\left|\sum_n \alpha_n \omega_{\xi_n, \eta_n}(x_\lambda)\right| &\le \sum_{n=1}^N \alpha_n |\langle x_\lambda\xi_n, \eta_n\rangle| + M\sum_{n=N+1}^\infty \alpha_n\end{align*}
together with the fact that $\sum_n \alpha_n < \infty$ ensures that $\omega(x_\lambda) \to 0$. This proves the claim. $\quad \square$
Claim: If $\omega = \sum_n \alpha_n \omega_{\xi_n, \eta_n}= \sum_n \alpha_n' \omega_{\xi_n', \eta_n'}$ as functionals on $\mathscr{L}\mathscr{C}(\mathfrak{H})$, then also as functionals on $\mathscr{L}(\mathfrak{H})$.
Proof: By the previous claim, it suffices to show that every element in $\mathscr{L}(\mathfrak{H})$ is a weak limit of a bounded net of elements in $\mathscr{LC}(\mathfrak{H})$. In fact, every such element is even a strong limit of such a bounded net because we can choose a net of finite-rank projections $\{p_\lambda\}$ such that $p_\lambda \to 1$ in the strong topology (such a net can be defined explicitly by considering an orthonormal basis for the Hilbert space) and then $xp_\lambda \to x$ strongly as well. $\quad \square$
