Theorem 1 (von Neumann)

Let $u$ be a closed symmetric operator on a complex Hilbert space E and $(E_+,E_-)$ its deficiency pair.

a) For every boundary condition $b$ for $u$, the partial operator $u_b$ is a closed symmetric extension of $u$;

b) The mapping $b\mapsto u_b$ is a bijection of the set of boundary conditions for $u$ onto the set of closed symmetric extensions of $u$;

c) For every boundary condition $b$ for $u$, one has

Im($u_b+i$) $=$ Im($u+i$)$\oplus$ Im($b$)

Im($u_b-i$) $=$ Im($u-i$)$\oplus$ Ker($b$)$^{\circ}$;

d) For every boundary condition $b$ for $u$, the deficiency pair of $u_b$ is (Ker($b$), Ker($b^*$)).

Let $b$ be a boundary condition for $u$ and $I\subset E_+$ its initial subspace. The restriction of $u^*$ to $\Gamma_{b|I}\subset E_+\oplus E_-$ is the continuous linear mapping defined by $x+b(x)\mapsto ix-ib(x)$ for $x\in I$. The graph of $u_b$ is the direct sum of the graph of $u$ and the graph of this linear mapping; it is therefore closed. The closed operator $u_b$ is an extension of $u$ such that $u_b\subset u^*$.

Let $\gamma_1$ and $\eta_1$ be elements of I. Consider the elements

of the graph $\Gamma_{b|I}$. We calculate

since $b$ is isometric on I. It follows that one has

for all $\gamma$ and $\eta$ in $\Gamma_{b|I}$.

Let $x$ and $y$ be elements of dom($u$)$,\gamma$ and $\eta$ elements of $\Gamma_{b|I}$. Then

where we have used the fact that $u$ is symmetric and that $\gamma$ and $\eta$ belong to the domain of $u^*$, as well as formula (12). The operator $u_b$ is therefore a closed symmetric extension of $u$. Assertion a) is proved.

The mapping $b\mapsto u_b$ is injective. In fact, a partially isometric mapping of $E_+$ into $E_-$ is uniquely determined by its initial subspace and by its restriction to the latter. Now the domain of $u_b$ determines the initial subspace I of $b$ and the graph of the restriction of $b$ to I.

Let us prove that the mapping $b\mapsto u_b$ is surjective. Let $w$ be a closed symmetric extension of $u$. One has $w\subset w^*\subset u^*$. The domain of $w$ is therefore a subspace of dom($u^*$) containing dom($u$), and $w$ is the restriction of $u^*$ to this subspace. Let G be the intersection of the domain of $w$ and of $E_+\oplus E_-$. It is a closed subspace of $E_{u^*}$ (lemma 7) and one has dom($w$) $=$ dom($u$)$\oplus G$ by proposition 26.

Let $x\in E_+$ and $y\in E_-$ be such that $x+y\in G$. Since $w$ is symmetric, and is the restriction of $u^*$ to dom($w$), we have

This equality is equivalent to

that is, to $\|x\|^2=\|y\|^2$. In particular, the canonical projection $p_+: G\rightarrow E_+$ is injective and, if I denotes its image, the space G is the graph of an isometric mapping $b_0$ of I into $E_-$. For $x$ in I and $y$ in $E_-$ such that $x+y\in G$, we have

Thus, considering G as a closed subspace of the Hilbert space $E_{u^*}$, the mapping $x\mapsto \frac{1}{2}p_+(x)$ of G into I is an isometry. In particular, I is closed in $E_+$ (lemma 8 of I, p. 107). There then exists a unique boundary condition $b$ for $u$ whose initial space is I and which coincides with $b_0$ on I. We have dom($w$) $=$ dom($u_b$), whence $w=u_b$. This establishes assertion b).

Let us prove assertions c) and d). Let $b$ be a boundary condition for $u$ and I its initial subspace. For $x_0\in$ dom($u$) and $y\in I$, we have

We have Im($u-i$) $= E^{\circ}_-$ (prop. 25). Since the image of $b$ is contained in $E_-$, this formula shows that Im($u_b-i$) $=$ Im($u-i$)$\oplus$ Im($b$). By loc. cit., we thus have Ker($u^*_b+i$) $= E_-\cap$ Im($b$)$^{\circ}=$ Ker($b^*$).

Analogously, one verifies that Im($u_b+i$) $=$ Im($u+i$)$\oplus I$ and hence (loc. cit.) that Ker($u^*_b-i$) $= I^{\circ}\cap E_+=$ Ker($b$).