Show that an operator with certain properties is an isomorphism

Question

Suppose $H$ is an infinite dimensional separable Hilbert space, with $\langle \cdot,\cdot\rangle$ denoting the duality pairing between the dual $H'$ and $H$ and $(\cdot,\cdot)$ the inner product on $H$. Let $A\,:\,H \to H'$ be a bounded linear functional such that

(a) there exists $\gamma \in H$ such that for all $v \in H$ we have $\langle A\gamma, v\rangle = \mu (\gamma,v)$, where $\mu < 0$;

(b) there exists $c > 0$ such that $\langle Aw,w\rangle \geq c \|w\|^2$ for all $w \in W$, where $W := \{ v \in H\,|\, (v,\gamma) = 0\}$.

It seems untuitively clear that $A$ is invertible, because we have a sole negative eigenvalue $\mu$ and the rest of the spectrum is bounded below by $c$. Is it an isomorphism? For that we would have to be able to find $c_2 > 0$ such that for all $v \in H$ we would have $\|Av\| \geq c_2 \|v\|$.

My futile attempt at establishing such an inequality centered around the decomposition $v = \alpha \gamma + w$, where $w \in W$ and $\alpha$ is some scalar. For instance we then have $$ \langle Av,v \rangle = \alpha^2\mu\|\gamma\|^2 + \langle Aw,w \rangle \geq \alpha^2 \mu \|\gamma\|^2 + c\|w\|^2, $$ but the fact that $\mu < 0$ seems to destroy the usual argument. I have a strong imppresion I am missing something obvious here. I'll gladly accept any comments you might have.

Edit: I probably should have also mentioned that $A$ is assumed self-adjoint in the sense that $\langle Av,w \rangle = \langle Aw,v \rangle$ for any $v,w \in H$.

Answer 1

Let $v = \alpha \gamma + w$. Instead of $v$ we use $ -\alpha \gamma + w$ as test function: $$ \langle Av, -\alpha \gamma + w \rangle = \alpha\mu (\gamma,-\alpha \gamma + w) + \langle A(-\alpha \gamma + w),w\rangle =- \alpha^2 \mu \|\gamma\|^2 + \langle Aw,w\rangle \\ \ge \alpha^2 |\mu| \cdot\|\gamma\|^2 + c \|w\|^2. $$ Now $\|{\pm}\alpha\gamma + w\|^2 = \alpha^2 \|\gamma\|^2 + \|w\|^2$, which implies $$ \langle Av, -\alpha \gamma + w \rangle \ge \min(|\mu|,c) \|v\|\cdot \|-\alpha \gamma + w\|$$ and $$ \|Av\|_{H'} \ge \min(|\mu|,c) \|v\|. $$

Answer 2

Here is a different way to look at this. If you identify $H$ with $H'$ and decompose $H$ as $H=\mathrm{lin}\{\gamma\}\oplus\{\gamma\}^\perp$, then $A$ decomposes as $$ A=\begin{pmatrix}B&C\\D&E\end{pmatrix}. $$ By property (a), $B=-\mu$ and $D=0$. Self-adjointness implies $C=0$ and $E=E^\ast$. Finally, $E$ is invertible by property (b). Thus $A$ is invertible as diagonal operator matrix with invertible diaogonal entries with inverse $$ A^{-1}=\begin{pmatrix}\mu^{-1}&0\\0&E^{-1}\end{pmatrix}. $$