Theorem 6.2.4.FDE87BE964C7 Let $H_0,H_1,H_2$ be Hilbert spaces, let $T$ from $H_0$ to $H_1$ be a closed densely defined operator, and let $S$ from $H_1$ to $H_2$ be a closed operator with $\operatorname {Ran}T\subset \ker S$. Let $g\in \ker S$ and $0\le M<\infty $, and assume that
\[ |\langle g,f\rangle |^2\le M\bigl (\| T^*f\| ^2+\| Sf\| ^2\bigr ) \qquad (f\in \operatorname{Dom}T^*\cap \operatorname{Dom}S). \]
Then the equation $Tu=g$ has a solution $u\in \operatorname{Dom}T$ with $\| u\| ^2\le M$. Among all solutions there is a unique one of smallest norm, and it is orthogonal to $\ker T$.
Proof.
Since $S$ is closed, $\ker S$ is a closed subspace, and $H_1=\ker S\oplus (\ker S)^\perp $ by proposition 6.2.1. Since $\operatorname {Ran}T\subset \ker S$, lemma 6.2.3 gives $(\ker S)^\perp \subset (\operatorname {Ran}T)^\perp =\ker T^*$. Let $f\in \operatorname{Dom}T^*$, and write $f=f_0+f_1$ with $f_0\in \ker S$ and $f_1\in (\ker S)^\perp $. Then $f_1\in \operatorname{Dom}T^*$ and $T^*f_1=0$, so $f_0\in \operatorname{Dom}T^*\cap \operatorname{Dom}S$, $T^*f_0=T^*f$ and $Sf_0=0$. Moreover $\langle g,f\rangle =\langle g,f_0\rangle $, because $g\in \ker S$. The hypothesis applied to $f_0$ therefore gives
\[ |\langle g,f\rangle |\le \sqrt M\, \| T^*f\| \qquad (f\in \operatorname{Dom}T^*). \]
Consequently the map $\lambda \colon T^*f\mapsto \langle g,f\rangle $ is well defined on $\operatorname {Ran}T^*$: if $T^*f=T^*f'$, then $|\langle g,f-f'\rangle |\le \sqrt M\, \| T^*(f-f')\| =0$. It is conjugate-linear and bounded by $\sqrt M$, hence uniformly continuous, so it extends uniquely to a conjugate-linear functional $\tilde\lambda $ on the closed subspace $W=\overline{\operatorname {Ran}T^*}$, again bounded by $\sqrt M$. If $P_ W$ is the orthogonal projection onto $W$, then $x\mapsto \tilde\lambda (P_ Wx)$ is a conjugate-linear functional on $H_0$ of norm at most $\sqrt M$, because $\| P_ Wx\| \le \| x\| $. By the representation of conjugate-linear functionals in proposition 6.2.1, there is $u\in H_0$ with $\| u\| \le \sqrt M$ and
\[ \langle u,T^*f\rangle =\langle g,f\rangle \qquad (f\in \operatorname{Dom}T^*). \]
Taking complex conjugates, we get $\langle T^*f,u\rangle =\langle f,g\rangle $ for all $f\in \operatorname{Dom}T^*$. Thus the functional $f\mapsto \langle T^*f,u\rangle $ is bounded on $\operatorname{Dom}T^*$, so $u\in \operatorname{Dom}T^{**}$ and $T^{**}u=g$. Since $T^{**}=T$ by lemma 6.2.3, we conclude that $u\in \operatorname{Dom}T$ and $Tu=g$.
The solutions of $Tu=g$ form the affine subspace $u+\ker T$, which is closed because $\ker T$ is closed. If $P$ denotes the orthogonal projection onto $\ker T$, then $\| u+k\| ^2=\| u-Pu\| ^2+\| Pu+k\| ^2$ for $k\in \ker T$. Hence $u-Pu$ is the unique solution of smallest norm; it is orthogonal to $\ker T$, and its norm is at most $\| u\| \le \sqrt M$.
$\square$
Pome
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