# Showing an element is in the resolvent

Let $T:D(T)\to\mathcal{H}$ be a self-adjoint operator on a Hilbert space $\mathcal{H}$ and $\lambda\in\mathbb{R}\backslash\{0\}$. I am trying to show that the resolvent operator $R_\lambda:\mathcal{H}\to\mathcal{H}$ is a bijection.

I have shown injectivity but am stuck on how to show that $i\lambda$ is contained within the resolvent set, so that:

$$Range(T-i\lambda)=\mathcal{H}$$

Any help would be greatly appreciated.

For $\lambda$ real and non-zero, and for $f\in\mathcal{D}(T)$, $$|((T-i\lambda I)f,f)|=|(Tf,f)-i\lambda(f,f)| \ge |\lambda|(f,f) \\ |\lambda| \|f\|^2 \le \|(T-i\lambda I)f\|\|f\| \\ |\lambda| \|f\| \le \|(T-i\lambda I)f\|$$ Knowing that $T$ is closed, this bound forces the range of $T-i\lambda I$ to be closed for $\lambda\in\mathbb{R}\setminus\{0\}$. Hence, $$\mathcal{R}(T-i\lambda I)=\mathcal{N}((T-i\lambda I)^*)^{\perp}=\mathcal{N}(T+i\lambda I)^{\perp}=\{0\}^{\perp}=\mathcal{H}.$$
• I don't understand how that bound shows the range is closed? Also, what does $\mathcal{N}$ represent in this context? – David Apr 26 '17 at 18:37
• @David : $\mathcal{N}$ is null space. If $(T-i\lambda I)f_n$ converges to $g$ in the closure of the range, then the bound shows that $f_n$ is a Cauchy sequence and, hence, converges to some $f$. Then by the closedness of $T$, it follows that $f\in\mathcal{D}(T)$ and $(T-i\lambda I)f=g$. So the range is closed. – DisintegratingByParts Apr 26 '17 at 18:41