If $T$ is a positive operator then $I+T$ is invertible Let $T$ be a positive operator on a Hilbert space $H$, prove that $I+T:H\to H$ is invertible and $(I+T)^{-1} \in B(H)$.
Now, If I prove $I+T$ is invertible, the bounded inverse theorem implies the second part. Now while proving that $I+T$ is invertible, I have proved that $I+T$ is one-one. But now I have to prove that $I+T$ is onto. In doing so my idea is to prove that $I+T$ is bounded below, so that $Range(I+T)$ becomes closed and then show that $Range(I+T)^{\perp}=\{\ 0 \}\ $, then by projection theorem we will have $Range(I+T)=H$.
But I couldn't execute this idea. Other ideas will also be appreciated
Thanks in advance!!
 A: Note that 
$$
I+T-\lambda I=T-(\lambda -1)I.
$$
So $\lambda\in\sigma(I+T)\iff \lambda-1\in\sigma(T)$. In other words, 
$$
\sigma(I+T)=\{\lambda+1:\ \lambda\in\sigma(T)\}.
$$
As $T$ is positive, $\sigma(T)\subset[0,\infty)$. Thus $\sigma(I+T)\subset [1,\infty)$. 
It follows that $0\not\in\sigma(I+T)$, so $I+T$ is invertible. 
A: Let $A=I+T$. Recall $\ker A^* = (\operatorname{ran} A)^\perp$. For injective self-adjoint operators, this already implies that the range is dense. But your operator is not just injective but is bounded from below, meaning  there is $c>0$ such that 
$$\|Ax\|\ge c\|x\| \quad \forall x \tag{1}$$
Property (1) implies $\operatorname{ran} A$ is closed.  To summarize: a self-adjoint operator that is bounded from below is invertible. 
The proof of (1) is an application of monotonicity:
$$
\langle x+Tx, x+Tx\rangle  \ge \langle x, x\rangle
$$ 
so $c=1$ works.
A: A linear operator $T$ on a $\mathbb R$-Hilbert space is nonnegative if and only if $-T$ is dissipative. This immediately implies that $\lambda+T$ is injective for all $\lambda>0$; see, for example, Proposition 3.14-i in One-Parameter Semigroups for Linear Evolution Equations by Engel and Nagel.
In particular, $\lambda+T$ is bounded from below (in the operator sense) with constant $\lambda$ and $(\lambda+T)^{-1}$ has operator norm at most $\lambda^{-1}$.
