Invertibility of a linear combination of self-adjoint operators Assume a bounded linear operator $T:\mathcal{H}\to\mathcal{H},$ where $\mathcal{H}$ is an infinite dimensional separable Hilbert space, is invertible. Let $$A={1\over 2}(T+T^*),\quad B={1\over 2i}(T-T^*)$$ Does there exist   a real constant $r$ such that the operator $A+rB$ is injective ?
The operators $A$ and $B$ are self-adjoint and $T=A+iB.$ The assumptions imply that $\ker A\cap \ker B=\{0\}.$
Assume  that $\ker (A-rB)\neq \{0\}$ for any positive value of $r.$ Let $0\neq x_r\in\ker (A-rB).$ Then for $s\neq r$ we get
$$\langle Ax_r,x_s\rangle =r\langle Bx_r,x_s\rangle,\quad \langle x_r,Ax_s\rangle =s\langle x_r,Bx_s\rangle$$
Therefore $$\langle Ax_r,x_s\rangle=0,\quad s\neq r$$
The conclusion is valid if one of the operators $A$ or $B$ is nonnegative. Indeed, consider the case $A\ge 0.$
Then
$$0=\langle Ax_r,x_s\rangle =\langle A^{1/2}x_r,A^{1/2}x_s\rangle \quad s\neq r$$
The family of nonzero orthogonal vectors $\{A^{1/2}x_r\}_{r>0}$ is uncountable, which leads to a contradiction.
The conclusion is valid also for any  normal invertible operator $T.$ In that case the operators $A$ and $B$ commute. We have $$\ker(A^2-r^2B^2)=\ker [(A+rB)(A-rB)]\supset \ker (A-rB)$$
The operator $A^2$ is nonnegative, hence there is $r\neq 0$ such that $\ker(A^2-r^2B^2)=\{0\}.$ Thus $A-rB$ is injective.
We cannot expect the invertibility of the operator $A+rB.$ I was able (spoiler) to get an example of an invertible operator $T$ such $A+rB$ is not invertible for any real value $r.$

Let $\mathcal{H}=L^2(0,\pi)$ and $(Tf)(x)=e^{ix}f(x).$ Then $(Af)(x)=\cos x\,f(x) $ and $(Bf)(x)=\sin x\, f(x).$ Thus $[(A+rB)f](x)=(\cos x+r\sin x)\,f(x).$ The function $x\mapsto \cos x+r\sin x$ is not bounded away from $0,$ Therefore the operator $A+rB$ is not invertible.

 A: Not a solution, but a summary of partial progress, failed approaches, and sharpness tests.

*

*The claim is known (from the OP) if $[A,B]=0$, if $A\geq0$, or if $B\geq0$.  

*The claim is also known if $0\notin\sigma(A)\cup\sigma(B)$.  
For example, if $0\notin\sigma(B)$, then there exists $m$ such that $\|Bv\|\geq m\|v\|$ for all $v\in\mathcal{H}$.  Thus if $r\geq(1+\epsilon)\frac{\|A\|}{m}$, then for all $v\in\mathcal{H}$ \begin{align*}
\|(A+rB)v\|&\geq r\|Bv\|-\|Av\|\\
&\geq(1+\epsilon)\frac{\|A\|}{m}\cdot m\|v\|-\|A\|\cdot\|v\|\\
&=\epsilon\|A\|\cdot\|v\|
\end{align*}  Thus $A+rB$ is injective.

*Unfortunately, there are circumstances in which $0\in\sigma(A)\cap\sigma(B)$.  For example, the multiplication operator mentioned behind the spoiler in the OP is of this form.

*The claim is false for inseparable Hilbert spaces; the OP's spoiler provides a counter-example.

*The claim is, however, true if $\mathcal{H}$ is finite-dimensional.  
Let $f(r)=\det(A+rB)$, so that $f(r)=0$ iff $A+rB$ is injective.  The map $f(r)\in\mathbb{R}[r]$ and has a holomorphic extension to $\mathbb{C}$; by the previous sentence, $f(i)\neq0$; and so the holomorphic extension $f\neq0$.  But nonzero holomorphic functions have isolated roots, so that $f(r)\neq0$ for generic $r$.  

*Generalizing the previous point to $\ell^2(\mathbb{N})$ faces substantial obstacles.  
First, one loses holomorphism.  In $n$ dimensions, $\det(2)=2^n$.  For a finite value in arbitrary dimensions, one must use the normalized determinant $\sqrt[n]{|\det|}$; but the latter function is only subharmonic.  In general, subharmonic functions can accumulate roots.  For example, $z\mapsto\max(0,\log(|z|))$ on $\mathbb{C}$ is subharmonic, but vanishes on $\mathbb{D}$.
Worse, the most relevant infinite-dimensional determinant I know is the Fuglede-Kadison determinant $$\Delta(M)=e^{\frac{1}{2}\operatorname{tr}(\log(M^*M))}$$ which is well-defined as long as $0\notin\sigma(M)$ and has a continuous extension vanishing iff $0\in\sigma(M)$.  Thus $\Delta(A+rB)$ only detects whether $0\in\sigma(A+rB)$, not whether $A+rB$ is injective.

*Nevertheless, one might investigate the set $R=\{r:A+rB\text{ injective}\}\subseteq\mathbb{C}$, hoping to show that $R\cap\mathbb{R}\neq\emptyset$.  On the one hand, $R$ is closed under complex conjugation: an operator is invertible iff its adjoint is, and $$(A+rB)^*=A+\overline{r}B$$

*On the other hand, for any countable $Q\subseteq\mathbb{C}\setminus\{\pm i\}$, I think there is a choice of $T$ such that $R\cap Q=\emptyset$.  (My original construction of $T$ has a technical flaw, but can almost certainly be repaired; I sketch the argument below behind a spoiler.)
Fix $0<l<u<\infty$ to be determined later and enumerate $Q$ as $\{q_n\}_{n=0}^{\infty}$.  For each $n$, I claim (see spoilered section below) that there are self-adjoint $A_n,B_n\in\mathbb{C}^{2\times2}$ such that $A_n+iB_n$ is invertible with $\|A_n+iB_n\|<\sqrt{u}$ and $\|(A_n+iB_n)^{-1}\|<\frac{1}{\sqrt{l}}$, but $A_n+q_nB_n$ has nontrivial kernel.
Now let $H=\bigoplus_n{\mathbb{C}^2}$ and let $\mathcal{H}$ be the completion of $H$ under the $l^2$ norm.  Let $T$ act block-diagonally on $H$, with each block $A_n+iB_n$; then $T$ extends to $\mathcal{H}$ by continuity with $\|T\|\leq\sqrt{u}$ and $\|T^{-1}\|\leq\frac{1}{\sqrt{l}}$.  The associated operators $A$, $B$, and $T^{-1}$ are similarly block-diagonal: each block of $A$ is $A_n$, etc.  But for any $q_n$ there is a $v\in\mathcal{H}$ such that $$(A+q_nB)v=(A_n+q_nB_n)v=0$$  Thus $q_n\notin R$.

! To construct such $A_n$ and $B_n$, let the two eigenvalues of $A_n+iB_n$ be $\lambda_1$ and $\lambda_2$.  Then we seek $A_n,B_n$ self-adjoint with \begin{gather*}
\det(A_n+q_nB_n)=0 \\
\sqrt{l}<|\lambda_2|\leq|\lambda_1|<\sqrt{u}
\end{gather*}  To address the eigenvalue condition, it suffices to instead work with \begin{alignat*}{3}
&d=&|\det(A_n+iB_n)|^2&=|\lambda_1|^2|\lambda_2|^2\\
&n=&\|A_n+iB_n\|_F^2&=|\lambda_1|^2+|\lambda_2|^2
\end{alignat*} by Vieta's formulas.  In the latter language, we require \begin{gather*}
\det(A_n+q_nB_n)=0 \\
2l<n<2u \\
2l(n-2l)<4d<n^2
\end{gather*}
!
! I originally thought I had an explicit solution to these conditions, because I forgot about the lower bound $l$, which it violates.  Nevertheless, the space of pairs of self-adjoint matrices is $8$-dimensional (over $\mathbb{R}$), and there are three conditions here, polynomial in the matrix entries (two quadratic and one quartic).  In principle there should be a $5$-dimensional moduli space of solutions.
!
! The only tricky condition is the last line, because if $q_n\approx\pm i$, then we should expect $d\approx0$ as well.  But although the last condition forces $d\neq0$, it does allow $d$ to be arbitrarily small — just choose $n\approx 2l$.
!
! Readers skeptical that the moduli space is nonempty are invited to investigate further.



*Even in the previous point, the possibility remains that $R$ is (path-)connected.  Because $\{\pm i\}\subseteq R$, if $R$ is (path-)connected, then it must intersect $\mathbb{R}$.  But there is no clear reason why $R$ should exhibit either property.  

*Likewise for full measure or comeagerity.

*The claim is false if the requirement that $\exists(A+Bi)^{-1}$ is weakened to $\ker(A)\cap\ker(B)=0$.  Indeed, let $\mathcal{H}=\mathbb{C}^3$; $$A=\begin{bmatrix}0&1&0\\1&0&0\\0&0&0\end{bmatrix}\quad\quad\text{ and }\quad\quad B=\begin{bmatrix}0&0&1\\0&0&0\\1&0&0\end{bmatrix}$$  Then $\ker(A)=\mathbb{C}(0,0,1)$ and $\ker(B)=\mathbb{C}(0,1,0)$, but $\det(A+rB)=0$ for any $r$.

*The claim also appears to hold if $A$ and $B$ are allowed to be unbounded.  For example, let $A=x$ and $B=i\frac{d}{dx}$ on $L^2$.  Then solutions to $(A+rB)f=0$ are oscillatory, whereas solutions to $(A+iB)f=0$ grow or decay without bound.  Thus if we attempt to tweak $L^2$ to make $A+rB$ not injective, then we will almost certainly be forced to include function-classes that make $A+iB$ not injective (and thus not invertible).
  
