A separable family of semi-norms on $\mathcal{B(H)}$

Question

Let $\mathcal{H}$ be a Hilbert space and $\mathcal{B(H)}$ be the space of bounded operators on $\mathcal{H}$.

Let $p_x:\mathcal{B(H)}\rightarrow \mathbb{R}^+\cup \{0\}$ such that $p_x(u)=\Vert u(x)\Vert$. This function is trivially a semi-norm on $\mathcal{B(H)}$. I would like to show that $\{p_x\}_{x\in\mathcal{H}}$ is a separable family of semi-norms, i.e. for every $u_1\neq u_2 \in \mathcal{B(H)}$ there there exists $x\in \mathcal{H}$ s.t. $\Vert u_1(x) \Vert \neq \Vert u_2(x) \Vert$.

If the operators $u_1,u_2$ have distinct kernels then the result is easy to show. We just pick a $x$ which is in the Kernel of $u_1$ and not in the kernel of $u_2$. But if they have identical kernels it boils down to proving that if $\Vert u_1(x) \Vert=\Vert u_2(x) \Vert \forall x\in \mathcal{H}$, then $u_1=u_2$. I am not able to show the second step.

I would appreciate if someone can shed some light on this process, also, it is equally welcome if you provide an alternate method to prove this.

Answer 1

If $\lVert{u_1(x)}\rVert = \lVert{u_2(x)}\rVert \; \forall x \in \mathcal{H}$, then $u_1 = u_2$.

This is false for any nontrivial hilbert space $\mathcal{H}$. Just consider $u_1 = \operatorname{id}_{\mathcal{H}}$ and $u_2 = -\operatorname{id}_{\mathcal{H}}$.