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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?

Thanks for your time.

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

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  • $\begingroup$ You should give your definition of positive operator. $\endgroup$
    – MaoWao
    Commented Jul 18, 2021 at 16:33
  • $\begingroup$ @MaoWao edited it accordingly. $\endgroup$ Commented Jul 18, 2021 at 17:01
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    $\begingroup$ 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 $0$. $\endgroup$
    – MaoWao
    Commented Jul 18, 2021 at 17:49
  • $\begingroup$ @MaoWao if the inequality is strict then can we say that? $\endgroup$ Commented Jul 19, 2021 at 5:21
  • $\begingroup$ Then it is true. Edit the question and I'll write an answer. $\endgroup$
    – MaoWao
    Commented Jul 19, 2021 at 6:41

2 Answers 2

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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.

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  • $\begingroup$ 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). $\endgroup$ Commented Aug 29, 2021 at 18:16
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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).
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