Let $H$ be a Hilbert space. Define its positive cone by $$\mathcal{B}(H) = \{A \in \mathcal{B}(H) : \langle v, A(v)\rangle \geq 0, \forall v \in H\}.$$
Show that $\mathcal{B}(H)^+$ is closed with respect to the operator norm and it is generated by $id \in \mathcal{B}(H)$.
1) $\mathcal{B}(H)^+$ is closed.
Let $(A_n)_{n \in \mathbb{N}}$ be a sequence of positive elements of $\mathcal{B}(H)$ and $A = \lim_{n \in \mathbb{N}} A_n$, ie, $\vert\vert A_n - A \vert\vert_{op} \rightarrow 0$. This means that $ \sup_{v \in H, ||v|| = 1} |A_n(v) - A(v)| \rightarrow 0$.
As $A_n \in \mathcal{B}(H)$, for all $n \in \mathbb{N}$, we have that for all $v \in H$ and for all $n \in \mathbb{N}$, $\langle v, A_n(v) \rangle \geq 0$. In order to show that $\mathcal{B}(H)^+$ is closed, we need to show that for all $v \in H$, $\langle v, A(v) \rangle \geq 0$.
Suppose that $v \in H$ is such that $||v|| = 1$ (otherwise we can divide and multiply by the norm and do some similar calculation). Then,
$\langle v, A(v) \rangle = \langle v, A(v) - A_n(v) \rangle + \langle v, A_n(v) \rangle \geq 0$, because the first term goes to $0$ and $\vert \vert A_n - A \vert\vert_{op} \rightarrow 0$ and the second term is positive.
Problem: I think this argument is not totally precise. Could someone help me here?
And I am struggling to show that $\mathcal{B}(H)^+$ is generated by $id$. I would appreciate very much some hint :)
Thanks in advance!
1) The argument is not written very clearly, but it's essentially ok. If $A_n\to A$ is operator norm, then $A_n v\to A v$ for all $v\in H$ (this you already used) and thus also $\langle A_n v,v\rangle\to \langle A v,v\rangle$. The latter follows directly from the continuity of the inner product. Since every sequence member is nonnegative, so is the limit.
2) You have to show that for every $A\in B(H)^+$ there exists $\alpha>0$ such that $A\leq \alpha\,\mathrm{id}$. The canoncial candidate for any bound on $A$ is $\lVert A\rVert$, and indeed it works: $$ \langle Av,v\rangle\leq \lVert Av\rVert \lVert v\rVert \leq \lVert A\rVert \lVert v\rVert^2=\lVert A\rVert \langle\mathrm{id}\,v,v\rangle. $$ Hence $A\leq \lVert A\rVert\,\mathrm{id}$.