Let $M\subseteq B(H)$ be a von Neumann algebra. Let $\omega \in M_*$ be a normal functional. In a paper I am reading$^{(\dagger)}$, it is said that it is possible to find a normal functional $\overline{\omega}\in B(H)_*$ such that $$\overline{\omega}\vert_M = \omega, \quad \|\omega\| = \|\overline{\omega}\|.$$
I can see why an extension exists that satisfies the conditions separately (by applying the Hahn-Banach theorem separately for the norm-topology and the $\sigma$-weak topology), but not why we can have both conditions at the same time.
Why is this true? What I can see is true that for any $\epsilon > 0$, you can find a normal extension $\overline{\omega}\in B(H)_*$ with $$\|\overline{\omega}\| < \epsilon + \|\omega\|$$ which is close, but not good enough.
$(\dagger)$ Well, in the paper, they say that you can extend a contractive normal functional $\omega$ on $M$ to a contractive normal functional on $B(H)$. But if $\|\omega\|= 1$, this means that we must have $1=\|\omega\| \le \|\overline{\omega}\|\le 1$ so $\|\overline{\omega}\| = 1$ and then a simple rescale argument shows that this is equivalent with the statement in my question.
Maybe someone will prove me wrong (and I'll be happy about it) but I don't think there is an elementary argument for this.
What one can prove is that any normal functional on a von Neumann algebra $M\subset B(H)$ is of the form $$\tag1 \omega(x)=\operatorname{Tr}(Vx), $$ where $V$ is a trace-class operator, and $\|\omega\|=\|V\|_1$. The extension to $B(H)$ is then automatic.
I'm not aware of a place where this is proven straightforwardly, but here is a possible path: one can prove that normal functionals are the span of the normal states (this is for instance, in Theorem III.4.2 in Takesaki I). And the proof that normal states are of the form $(1)$ can be found for example as Theorem 7.1.12 in Kadison-Ringrose