[proofplan] We fix a complex number and then introduce the shifted operator $A_\lambda = A - \lambda I_H$ on the domain of $A$. If the range of this shifted operator is not dense, the [Hilbert space](/page/Hilbert%20Space) orthogonal-complement argument gives a nonzero vector orthogonal to the range. That orthogonality is exactly the adjoint-domain condition, so self-adjointness turns it into an eigenvalue equation for $A$. For real shifts this immediately destroys injectivity, while for non-real shifts it contradicts the reality of quadratic forms of self-adjoint operators. [/proofplan] [step:Extract a nonzero vector orthogonal to the non-dense range] Fix $\lambda \in \mathbb{C}$. Define the shifted operator \begin{align*} A_\lambda: \mathcal{D}(A) \subset H &\to H \end{align*} by \begin{align*} A_\lambda u = (A - \lambda I_H)u \end{align*} for each $u \in \mathcal{D}(A)$, where $I_H: H \to H$ is the identity operator. Assume that $\operatorname{Range}(A_\lambda)$ is not dense in $H$. Define the closed subspace $M \subset H$ by \begin{align*} M = \overline{\operatorname{Range}(A_\lambda)}. \end{align*} Since $M \neq H$, choose $x_0 \in H \setminus M$. By the Hilbert [projection theorem](/theorems/1985) applied to the nonempty closed convex set $M$, there exists $m_0 \in M$ such that $|x_0 - m_0| = \operatorname{dist}(x_0, M)$. Define $v = x_0 - m_0 \in H$. Since $x_0 \notin M$ and $m_0 \in M$, we have $v \neq 0$. The variational inequality for the minimizer $m_0$ gives \begin{align*} (v, w)_H = 0 \end{align*} for every $w \in M = \overline{\operatorname{Range}(A_\lambda)}$. For each $u \in \mathcal{D}(A)$, the vector $(A - \lambda I_H)u$ belongs to $\operatorname{Range}(A_\lambda) \subset \overline{\operatorname{Range}(A_\lambda)}$. The conjugate-symmetry of the Hilbert space [inner product](/page/Inner%20Product) therefore gives \begin{align*} ((A - \lambda I_H)u, v)_H = \overline{(v, (A - \lambda I_H)u)_H} = 0. \end{align*} [/step] [step:Convert orthogonality to an adjoint eigenvalue equation] For every $u \in \mathcal{D}(A)$, the orthogonality relation gives \begin{align*} (Au, v)_H = (\lambda u, v)_H. \end{align*} Because the Hilbert space inner product is linear in the first variable and conjugate-linear in the second variable, \begin{align*} (\lambda u, v)_H = \lambda (u, v)_H = (u, \bar{\lambda}v)_H. \end{align*} Thus \begin{align*} (Au, v)_H = (u, \bar{\lambda}v)_H \end{align*} for every $u \in \mathcal{D}(A)$. By the definition of the adjoint operator, this proves that $v \in \mathcal{D}(A^*)$ and \begin{align*} A^*v = \bar{\lambda}v. \end{align*} Since $A$ is [self-adjoint](/page/Self-Adjoint%20Operators), $\mathcal{D}(A^*) = \mathcal{D}(A)$ and $A^* = A$. Therefore $v \in \mathcal{D}(A)$ and \begin{align*} Av = \bar{\lambda}v. \end{align*} [guided] The point of the orthogonal vector $v$ is that it converts a density failure into an adjoint equation. Since $v$ is orthogonal to the range of $A-\lambda I_H$, we have \begin{align*} ((A-\lambda I_H)u, v)_H = 0 \end{align*} for every $u \in \mathcal{D}(A)$, using conjugate-symmetry of the inner product to reverse the order from $(v,(A-\lambda I_H)u)_H=0$. Expanding the shifted operator gives \begin{align*} (Au, v)_H - (\lambda u, v)_H = 0. \end{align*} Hence \begin{align*} (Au, v)_H = (\lambda u, v)_H. \end{align*} Now we rewrite the right-hand side in the form required by the definition of the adjoint. The inner product on $H$ is linear in its first argument and conjugate-linear in its second argument, so \begin{align*} (\lambda u, v)_H = \lambda (u, v)_H = (u, \bar{\lambda}v)_H. \end{align*} Therefore \begin{align*} (Au, v)_H = (u, \bar{\lambda}v)_H \end{align*} for all $u \in \mathcal{D}(A)$. This is precisely the adjoint operator domain condition: there exists a vector $w \in H$, namely $w = \bar{\lambda}v$, such that \begin{align*} (Au, v)_H = (u, w)_H \end{align*} for every $u \in \mathcal{D}(A)$. Hence $v \in \mathcal{D}(A^*)$ and $A^*v = \bar{\lambda}v$. Since $A$ is [self-adjoint](/page/Self-Adjoint%20Operators), $A^* = A$ with the same domain, so $v \in \mathcal{D}(A)$ and \begin{align*} Av = \bar{\lambda}v. \end{align*} [/guided] [/step] [step:Rule out residual points on the real axis] Assume first that $\lambda \in \mathbb{R}$. Then $\bar{\lambda} = \lambda$, and the equation $Av = \bar{\lambda}v$ becomes \begin{align*} (A - \lambda I_H)v = 0. \end{align*} Since $v \neq 0$, the operator $A-\lambda I_H$ is not injective. Thus no real $\lambda$ with non-dense range can belong to the residual spectrum. [/step] [step:Rule out non-real residual points using self-adjoint symmetry] Assume now that $\lambda \in \mathbb{C} \setminus \mathbb{R}$. Since $v \in \mathcal{D}(A)$ and $Av = \bar{\lambda}v$, we have \begin{align*} (Av, v)_H = (\bar{\lambda}v, v)_H = \bar{\lambda}(v, v)_H. \end{align*} On the other hand, [self-adjointness](/page/Self-Adjoint%20Operators) implies symmetry on $\mathcal{D}(A)$, so \begin{align*} (Av, v)_H = (v, Av)_H. \end{align*} The conjugate-symmetry of the inner product gives \begin{align*} (v, Av)_H = \overline{(Av, v)_H}. \end{align*} Therefore $(Av,v)_H$ is real. Since $v \neq 0$, we have $(v,v)_H > 0$, and the equality \begin{align*} (Av, v)_H = \bar{\lambda}(v, v)_H \end{align*} forces $\bar{\lambda} \in \mathbb{R}$, contradicting $\lambda \notin \mathbb{R}$. Hence no non-real $\lambda$ can have non-dense range. [/step] [step:Conclude that the residual spectrum is empty] We have shown that whenever $\operatorname{Range}(A-\lambda I_H)$ is not dense in $H$, either $\lambda \in \mathbb{R}$ and $A-\lambda I_H$ is not injective, or $\lambda \notin \mathbb{R}$ and a contradiction occurs. Thus there is no $\lambda \in \mathbb{C}$ for which $A-\lambda I_H$ is injective and has non-dense range. By the definition of the residual spectrum, \begin{align*} \sigma_r(A) = \varnothing. \end{align*} [/step]