Let $X$ be a Banach space over the field $\mathbb{R}$, and denote by $\mathcal{L}(X)$ the space of continuous linear operators acting on $X$. The spectrum $\sigma(T)$ of an operator $T\in\mathcal{L}(X)$ is defined as the set of all $\lambda\in\mathbb{C}$ for which the complexification shift $\lambda-T_\mathbb{C}$ is not invertible in $\mathcal{L}(X_\mathbb{C})$. (See here for the definition of a complexification.) However, I am interested in invertibility in the real setting.
The real spectrum of $T$, denoted $\sigma_\mathbb{R}(T)$, is the set of all $a\in\mathbb{R}$ such that $a-T$ does not have an inverse in $\mathcal{L}(X)$. It is the complement of the real resolvent, defined as \begin{equation}\rho_\mathbb{R}(T):=\left\{a\in\mathbb{R}:\text{ there is }\;S\in\mathcal{L}(X)\;\text{ such that }\;S(a-T)=(a-T)S=I_X\right\},\end{equation} where $I_X\in\mathcal{L}(X)$ is the identity operator ($I_Xx=x\;\forall x\in X$).
Question 1. I suspect (although I haven't verified it yet) that $\sigma_\mathbb{R}(T)=\mathbb{R}\cap\sigma(T)$. Is there a reference for this?
Question 2. Suppose $r\in\partial\sigma_\mathbb{R}(T)$ and $(a_n)_{n=1}^\infty\subseteq\rho_\mathbb{R}(T)$ with $a_n\to r$. Is it true that $\|(a_n-T)^{-1}\|\to\infty$?
I would also be interested in any good literature on the real spectrum, if it is available.
Thank you!
I cannot contribute references. But, regarding question 1, note that for $\lambda\in\mathbb R$, you have that $T-\lambda I$ is invertible if and only if $T_{\mathbb C}-\lambda I$ is invertible. Then $$ \sigma_\mathbb{R}(T)=\mathbb{R}\cap\sigma(T). $$
To show the above invertibility, let us write $T$ and $T_{\mathbb C}$. If $ST=TS=I$, then $$ S_{\mathbb C}T_{\mathbb C}(x+iy)=S_{\mathbb C}(Tx+iTy)=STx+iSTy=x+iy. $$ In a similar way we obtain $T_{\mathbb C}S_{\mathbb C}=I$.
If $S'T_{\mathbb C}=T_{\mathbb C}S'=I$, then define $Sx$ to be the "real part" of $S'(x+i0)$. Then $$ STx=S'(Tx+i0)=S'(Tx+iT0)=S'T_{\mathbb C}(x+i0)=x+i0=x, $$ and $$ TSx=TSx+iT0=T_{\mathbb C}(Sx+i0)=T_{\mathbb C}S'(x+i0)=x+i0=x $$
Regarding question 2, suppose $\|(a_n-T)^{-1}\|\leq c$ for all $n$. Then since $r-a_n\to0$, for $n$ big enough we have $$ \|(a_n-T)^{-1}\|\leq\frac1{|r-a_n|}. $$ But then, for $n$ big enough $$\tag{1} \|(r-T)-(a_n-T)\|=|r-a_n|\leq\|(a_n-T)^{-1}\|^{-1}. $$ Since $a_n-T$ is invertible, this would imply that $r-T$ is invertible, a contradiction. So $$ \|(a_n-T)^{-1}\|\to\infty. $$