Spectrum of Self-Adjoint Operators This is an exercise (5-i) from here. It has two parts as follows. For a self-adjoint operator $A$. Show that


*

*$A \geq k I, \ k \in \mathbb R$ if and only if $\lambda \geq k$ for all $\lambda$ belonging to the spectrum of $A$, i.e., $\sigma(A)$.

*If $A \geq I$, then $A^n \geq I$ for every natural number $n \geq 1$.


Update: The following is my attempt.


*For part one, I remember a result saying that $A \geq 0$ if and only
if $\sigma (A) \subset [0, \infty)$. However, I do not know how
to prove this result. How to show this, please? If for the moment
I assume this result, then by assumption, $A-kI \geq 0$ if and only
if $\sigma(A-kI) \subset [0, \infty)$. How to get the desired
result from this, please?

*For part two, I tried to prove by induction. For $n = 2$, we have
$$A^2 - I = (A+I) (A-I), $$ where the first term on the right is sum
of positive operators and hence positive. The second by assumption
is positive. The two are commutative and self-adjoint. Hence, the
product is positive. Then assume that the claim is true for $n=k$,
i.e., $A^k - I \geq 0$. Then it is only left to show that it is true
for $n=k+1$, i.e., $A^{k+1} - I\geq 0$. I can have $$A^{k+1}-I = A^{k+1} - I^{k+1} = (A-I)(A^k + A^{k-1} + \cdots + A + I) .$$ However, this does not seem to work as it stands. How do I proceed from there, please? Thank you!
 A: For the first part, if $k-\lambda > 0$, then
$$
           ((A-\lambda I)x,x) = ((A-kI)x,x)+(k-\lambda)(x,x) \ge (k-\lambda)\|x\|^{2}.
$$
Therefore
$$
                (k-\lambda)\|x\|^{2} \le \|(A-\lambda I)x\|\|x\|,\\
                   (k-\lambda)\|x\| \le \|(A-\lambda I)x\|.
$$
The last inequality can be used to show (a) $\mathcal{N}(A-\lambda I)=\{0\}$ and (b) that the range of $A-\lambda I$ is closed and $A-\lambda I$ has a bounded inverse on that range. Because the range is closed,
$$
         \mathcal{R}(A-\lambda I)=\mathcal{N}((A-\lambda I)^{\star})^{\perp}=\mathcal{N}(A-\lambda I)^{\perp}=\{0\}^{\perp}=H,
$$
where $H$ is the underlying Hilbert space. Therefore $\lambda\in\rho(A)$ for all real $\lambda < k$, which proves $\sigma(A)\subseteq[k,\infty)$.
For the converse, you can use the spectral integral in the reference. If $\sigma(A)\subseteq[k,\infty)$, then
$$
           (Ax,x) = \int_{k}^{M}\lambda d(E(\lambda)x,x) \ge k\int_{k}^{M} d(E(\lambda)x,x)=k\|x\|^{2}.
$$
For the last part you use the spectral mapping theorem and combine it with the first part, or simply write
$$
                  A^{n} = \int_{1}^{M}\lambda^{n}dE(\lambda) \ge \int_{1}^{M}dE(\lambda) = I.
$$
A: For the part two: use spectral mapping theorem from your pdf and the part one.
You know that $A\ge I \iff\forall \lambda\in\sigma(A)\, \lambda\ge 1$ (first part).
Then, by the spectral mapping theorem you know that $\mu\in\sigma(A^n)\iff \exists\lambda\in\sigma(A):\, \mu=\lambda^n$. As all such $\lambda$ are superior to $1$ (hypothesis of the second part and first part), we conclude that $\mu\ge 1$ and therefore $A^n\ge 1$ (again, first part).
As to how to prove the fact that $A\ge 0\Rightarrow\sigma(A)\ge 0$, it depends on some additional hypothesis (dimension of the space, boundedness of $A$). In the finite-dimensioned case take $\lambda\in\sigma(A)$ and the corresponding eigenvector $v$,
then $$k\|v\|^2\le (Av,v) = \lambda \|v\|^2,$$hence $\lambda\ge k$. Similarly, if all eigenvalues are superior to $k$, then use the fact that eigenvectors form an orthonormal basis of the whole space.
