# The space of all positive operators is open in the space of all self-adjoint operators but not in the space of all bounded operators.

How do I show that the space of all positive operators is open in the space of all self-adjoint operators but not in the space of all bounded operators?

This is claimed by our instructor while proving that the square root function is differentiable. But I can't get the point. Could anyone provide me some way to prove it?

EDIT $$:$$ An operator $$A \in \mathcal B(\mathcal H)$$ is said to be positive if for all $$x \in \mathcal H$$ we have $$\left \langle x, Ax \right \rangle \gt 0.$$

• You should give your definition of positive operator. Commented Jul 18, 2021 at 16:33
• @MaoWao edited it accordingly. Commented Jul 18, 2021 at 17:01
• With this definition this statement is not true. The operator $0$ is positive, but $-\epsilon I$ is a non-positive self-adjoint operator with distance $\epsilon$ to $0$. Commented Jul 18, 2021 at 17:49
• @MaoWao if the inequality is strict then can we say that? Commented Jul 19, 2021 at 5:21
• Then it is true. Edit the question and I'll write an answer. Commented Jul 19, 2021 at 6:41

Fix $$A\in B(H)$$ with $$\langle Ax,x\rangle>0$$ for all $$x\in H$$. Since the function $$x\mapsto \langle Ax,x\rangle$$ is convex and strongly continuous, it is weakly semicontinuous.* Thus it takes its minimum on the weakly compact set $$\{x\in H:\|x\|\leq 1\}$$. Therefore $$\langle Ax,x\rangle\geq \lambda\|x\|^2$$ for some $$\lambda>0$$.

If $$B\in B(H)$$ is self-adjoint with $$\|B\|<\lambda$$, then $$\langle Bx,x\rangle$$ is real for all $$x\in H$$ and $$|\langle Bx,x\rangle|\leq \|B\|\|x\|^2<\lambda\|x\|^2$$. Hence $$\langle (A+B)x,x\rangle\geq \lambda\|x\|^2+\langle Bx,x\rangle>0.$$ Thus $$\{C\in B(H):C\,\text{self-adjoint},\,\|A-C\|<\lambda\}$$ is contained in the cone of all strictly positive operators.

To see that this set is not open in $$B(H)$$, simply note that if $$B$$ is not self-adjoint, then $$A+\epsilon B$$ is not positive for all $$\epsilon>0$$.

*The argument goes as follows: A map $$f\colon H\to \mathbb R$$ is weakly lower semicontinuous if and only if the sublevel sets $$\{x\in H: f(x)\leq \alpha\}$$ are closed for every $$\alpha\in \mathbb R$$. But since $$f$$ is convex, its sublevel sets are convex, and by the Hahn-Banach theorem every strongly closed convex set is weakly closed. Thus every strongly lower semicontinuous (in particular every strongly continuous) function on $$H$$ is weakly lower semicontinuous.

• As neither the question nor your proof features any assumptions on the underlying Hilbert space, I wrote an answer of my own to clarify that the statement as OP wrote it only holds for complex finite-dimensional Hilbert spaces (and how to fix this). Commented Aug 29, 2021 at 18:16

Just to clarify some things, the correct statement reads as follows:

Given a finite-dimensional complex Hilbert space $$\mathcal H$$, the set of all strictly positive operators $$A\in\mathcal B(\mathcal H)$$ (i.e. operators which satisfy $$\langle x,Ax\rangle>0$$ for all $$x\in\mathcal H\setminus\{0\}$$) is open in the space of all self-adjoint operators, but not in all of $$\mathcal B(\mathcal H)$$.

The proof given by MaoWao works, although the topological argument in the beginning can be simplified by saying that all eigenvalues of $$A$$ are $$>0$$ due to strict positivity, and because we are in finite dimensions there has to be a smallest one (denoted by $$\lambda$$).

Now I want to point out two important things:

• if $$\mathcal H$$ is not complex then this statement is wrong, because for real Hilbert spaces $$\langle x,Ax\rangle>0$$ does not imply that $$A$$ is self-adjoint (so the set of strictly positive operators cannot be open in the self-adjoint operators as it is not even a subset). The standard example for this is $$\begin{pmatrix}c&-1\\1&c\end{pmatrix}$$ on the real Hilbert space $$\mathbb R^2$$ for any $$c>0$$. The problem is that in general the definition of strict positivity features self-adjointness, and only if $$\mathcal H$$ is complex then self-adjointness is automatically implied by the former. With this general definition of strict positivity, however, the statement in question continues to hold.
• if $$\mathcal H$$ is of infinite dimension then this statement is wrong. For this consider the unique bounded linear operator defined via $$Ae_n:=2^{-n}e_n$$ for some orthonormal system $$\{e_n:n\in\mathbb N\}$$ in $$\mathcal H$$ (and $$Ax:=0$$ on the orthogonal complement). Then for any $$\varepsilon>0$$ there exists $$N\in\mathbb N$$ such that $$2^{-N}<\varepsilon$$ so $$A-2^{-N}|e_{N+1}\rangle\langle e_{N+1}|$$ is not strictly positive as $$\big\langle e_{N+1},\big(A-2^{-N}|e_{N+1}\rangle\langle e_{N+1}|\big)e_{N+1}\big\rangle=\langle e_{N+1},Ae_{N+1}\rangle-2^{-N}=2^{-N-1}-2^{-N}<0 \,,$$ but $$\|A-(A-2^{-N}|e_{N+1}\rangle\langle e_{N+1}|)\|=2^{-N}\||e_{N}\rangle\langle e_{N}|\|=2^{-N}<\varepsilon\,.$$ In other words because the eigenvalues of $$A$$ become arbitrarily small, every ball will feature self-adjoint operators which are not strictly positive anymore. Hence the problem is that, unlike in the finite-dimensional case, one cannot guarantee that a strictly positive operator has a smallest eigenvalue. However, the statement continues to hold if one defines strict positivity as $$\langle x,Ax\rangle>\lambda\|x\|^2$$ for some $$\lambda>0$$ and all $$x\in\mathcal H\setminus\{0\}$$ (which in finite dimensions is equivalent to the original definition).