How can we compute the adjoint of the inclusion between two Hilbert spaces?

Question

Let

• $\mathbb K\in\left\{\mathbb C,\mathbb R\right\}$
• $(U,\langle\;\cdot\;,\;\cdot\;\rangle_U)$ and $(V,\langle\;\cdot\;,\;\cdot\;\rangle_V)$ be $\mathbb K$-Hilbert spaces such that $U\subseteq V$ and that the inclusion $\iota:(U,\langle\;\cdot\;,\;\cdot\;\rangle_U)\to(V,\langle\;\cdot\;,\;\cdot\;\rangle_V)$ is Hilbert-Schmidt
• $C:=\iota^\ast$ denote the adjoint of $\iota$ and $$\langle u,v\rangle_0:=\langle C^{-\frac 12}u,C^{-\frac 12}v\rangle_U\;\;\;\text{for }u,v\in U$$

We can show that $$U=C^{\frac 12}V\;\;\;\text{and}\;\;\;\langle\;\cdot\;,\;\cdot\;\rangle_0=\langle\;\cdot\;,\;\cdot\;\rangle_U\;.$$

How can we compute what the adjoint $C$ of $\iota$ actually is?

Answer 1

The fact that $\iota$ is HS means that if $\{e_n\}$ is an orthonormal basis of $U$, then $$\tag{1} \sum_n\langle e_n,e_n\rangle_V=\sum_n\|\iota(e_n)\|_V^2<\infty. $$

After identifying $U$ and $V$ with their respective duals, we have $C:V\to U$ given by $$\tag{2} \langle Cv,u\rangle_U=\langle v,\iota (u)\rangle_V=\langle v,u\rangle_V. $$ In particular, if $\{f_m\}$ is an orthonormal basis of $V$, then $$ \langle Cf_m,e_n\rangle_U=\langle f_m,e_n\rangle_V. $$ Since each $e_n\in V$, we can write $e_n=\sum_m\langle e_n,f_m\rangle_V\,f_m$. Then $$ Cf_m=\sum_n\langle Cf_m,e_n\rangle_U\,e_n=\sum_n\langle f_m,e_n\rangle_V\,e_n, $$ and it follows by continuity that
$$\tag{3} Cv=\sum_n\langle v,e_n\rangle_V\,e_n. $$ Note that the sum converges because the sequence $(\langle v,e_n\rangle_V)$ is in $\ell^2$, by Cauchy-Schwarz and $(1)$.

When we think of $C$ as a map $V\to V$ (since $U\subset V$), it is immediate from $(3)$ that $C$ is positive: $$ \langle Cv,v\rangle_V=\sum_n\langle v,e_n\rangle_V\,\langle e_n,v\rangle_V=\sum_n|\langle v,e_n\rangle_V|^2\geq0. $$So it makes sense to talk about $C^{1/2}$. Now, if the inclusion $U\subset V$ is proper and $U$ is $V$-closed, then if $v\in V\cap U^\perp$ (orthogonal taken in $V$), then by $(2)$ we have $$ \langle Cv,u\rangle_U=\langle v,u\rangle_V=0 $$ for all $u\in U$, so $Cv=0$. So, in this case $C$ is not invertible and so it makes no sense to talk about $C^{-1/2}$.

Answer 2

You can not get the expression for the adjoint of the inclusion I: U-> V without having a relation between norms of U and V. Try with a specific example, e.g. inclusion of L^2 into H^(-1). In that case you should easily get the explicit expression for I^*