Htykuut

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, $vertvert A_n - A vertvert_{op} rightarrow 0$. 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 $vert vert A_n - A vertvert_{op} rightarrow 0$ 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 $A_nto A$ is operator norm, then $A_n vto A v$ for all $vin H$ (this you already used) and thus also $langle A_n v,vrangleto langle A v,vrangle$. 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 $Ain B(H)^+$ there exists $alpha>0$ such that $Aleq alpha,mathrm{id}$. The canoncial candidate for any bound on $A$ is $lVert ArVert$, and indeed it works:
$$
langle Av,vrangleleq lVert AvrVert lVert vrVert leq lVert ArVert lVert vrVert^2=lVert ArVert langlemathrm{id},v,vrangle.
$$
Hence $Aleq lVert ArVert,mathrm{id}$.