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Here's the question:

Let $A$ be a positive operator on a (possibly infinite dimensional) Hilbert space. Let $I$ denote the identity operator. Suppose that $A \geq I$, which is to say that $A - I$ is a positive operator. Prove that $A$ is invertible.

I think that this is true, but I haven't been able to find a proof one way or the other. I would like to avoid invoking any heavy machinery (like the spectral theorem) if possible. I would also be interested in a proof that carries over to more general $C^*$ algebras.

Of course, the proof in the case of finite dimensional spaces is fairly obvious, since it suffices to show that the operator has a trivial kernel. Since that does not suffice here, I really have no clue what my next move should be.

Any guidance here would be greatly appreciated.

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    $\begingroup$ Maybe I didn't understand your question, but here is what I have: $$\|Ax\|\|x\|\ge(Ax,x)\ge (Ix,x)=\|x\|^2\Rightarrow \|Ax\|\ge 1\cdot \|x\|$$ So if $Ax=Ay$ then $0\ge \|x-y\|$. $\endgroup$ – Svetoslav Sep 29 '15 at 16:50
  • $\begingroup$ @Svetoslav right. So, certainly $A$ is injective. However, in infinite dimensional spaces, this is not enough. $\endgroup$ – Omnomnomnom Sep 29 '15 at 16:51
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    $\begingroup$ No. For example, take the right shift operator, or an integration operator. $\endgroup$ – Omnomnomnom Sep 29 '15 at 16:55
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    $\begingroup$ @Svetoslav the right-shift operator has no two-sided inverse. $\endgroup$ – Omnomnomnom Sep 29 '15 at 17:00
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    $\begingroup$ Yes, if you want an inverse that's defined on $R(A)$, then that's enough (no, this inverse will not necessarily be bounded). However, what I mean (and what mathematicians generally mean) when they refer to the inverse of a function $f:A \to B$ is a function $g:B \to A$ satisfying $g \circ f = id_A$ and $f \circ g = id_B$. $\endgroup$ – Omnomnomnom Sep 29 '15 at 17:19
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It shouldn't be to hard to show that $A$ is injective, for if $Ax = 0$ then $$\langle x,x \rangle \le \langle Ax,x\rangle =0.$$

It will follow that $A$ is invertible once we show that $A$ is surjective: the range $R(A)$ satisfies $R(A) = H$.

Let $A^*$ denote the adjoint of $A$. Suppose that $y \in N(A^*)$ so that $A^*y = 0$. Then $$0 = \langle y,A^*y \rangle = \langle Ay,y \rangle \ge \langle y,y \rangle$$ so that $N(A^*) = \{0\}$. Consequently $R(A)^\perp = \{0\}$.

You can use the projection theorem to conclude that $R(A) = H$ provided that $R(A)$ is closed. Suppose that $\{y_k\}$ is a sequence in $R(A)$ that converges to a point $y \in H$ and let $x_k \in H$ satisfy $A x_k = y_k$. Then $$ \langle x_k - x_j, x_k - x_j \rangle \le \langle Ax_k - Ax_j, x_k - x_j \rangle \le \|Ax_k - Ax_j\| \|x_k - x_j\|.$$ Since $\{A x_k\}$ is Cauchy in $H$, it follows that $\{x_k\}$ is Cauchy too, hence $x_k \to x$ for some $x$, which by continuity will satisfy $Ax = y$. Thus $R(A)$ is closed.

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  • $\begingroup$ To clarify for future readers: we then have then met the hypothesis of the bounded inverse theorem. $\endgroup$ – Omnomnomnom Sep 29 '15 at 16:57
  • $\begingroup$ You're welcome. The formula $R(A)^\perp = N(A^*)$ comes in handy frequently. $\endgroup$ – Umberto P. Sep 29 '15 at 17:00
  • $\begingroup$ @UmbertoP. Do we assume that $A$ is linear ? $\endgroup$ – Svetoslav Sep 29 '15 at 17:01
  • $\begingroup$ I do not know how adjoint is defined for nonlinear operator. $\endgroup$ – Svetoslav Sep 29 '15 at 17:02
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    $\begingroup$ @Svetoslav: "operator" here means linear operator. $\endgroup$ – hardmath Sep 29 '15 at 17:02
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Since $A$ is positive, its spectrum is contained in $[0,\infty)$. Then the spectrum of $A+I$ is contained in $[1,\infty)$; thus $0$ is not in the spectrum of $A+I$ and $A+I$ is invertible.

To justify that the spectrum translates note that $A+I-\lambda I=A-(\lambda-1)I$. So $A+I-\lambda$ is not invertible precisely when $A-(\lambda -1)I$ is not invertible, i.e. when $\lambda-1$ is in the spectrum of $A$. So $$ \sigma(A+I)=\{\lambda+I:\ \lambda\in\sigma(A)\} $$

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  • $\begingroup$ Care to explain the downvote, please? $\endgroup$ – Martin Argerami Sep 29 '15 at 21:46
  • $\begingroup$ Nice answer. Any quick justification that the spectrum is in $[0,\infty)?$ $\endgroup$ – Omnomnomnom Sep 29 '15 at 22:09
  • $\begingroup$ That depends on your definition of positivity. You asked about an operator in $B(H)$, but also asked about arbitrary C$^*$-algebras. What does "$A$ positive" mean? $\endgroup$ – Martin Argerami Sep 29 '15 at 22:22
  • $\begingroup$ $A = T^*T$ for some $T$ would be nice, but $\langle x,Ax \rangle \geq 0$ for all $x$ will do. $\endgroup$ – Omnomnomnom Sep 29 '15 at 22:34
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    $\begingroup$ The canonical way is to define "positive" as "selfadjoint with non-negative spectrum". If you want to start from $A=T^*T$, I think it is a bit of a struggle to understand the spectrum. If you start with $\langle Ax,x\rangle\geq0$, then probably the easy way is to argue as in Umberto's answer to show that $A-\lambda I$ is invertible whenever $\lambda<0$. $\endgroup$ – Martin Argerami Sep 29 '15 at 22:51

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