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:


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, $$ \mathcal{R}(T-i\lambda I)=\mathcal{N}((T-i\lambda I)^*)^{\perp}=\mathcal{N}(T+i\lambda I)^{\perp}=\{0\}^{\perp}=\mathcal{H}. $$

  • $\begingroup$ I don't understand how that bound shows the range is closed? Also, what does $\mathcal{N}$ represent in this context? $\endgroup$ – David Apr 26 '17 at 18:37
  • $\begingroup$ @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. $\endgroup$ – DisintegratingByParts Apr 26 '17 at 18:41
  • $\begingroup$ Of course! Thank you. $\endgroup$ – David Apr 26 '17 at 18:47
  • $\begingroup$ @David : You're welcome. Another way to look at this: the inverse is a bounded operator. And because it is a bounded operator that is closed, it follows that the domain of the bounded inverse is a closed subspace, which is what the sequential argument shows. $\endgroup$ – DisintegratingByParts Apr 26 '17 at 18:52

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