AFA The adjoint map

Section 5.3 The adjoint map

The concept of adjoint map generalizes the matrix transposition or, equivalently, the pull-back of a 1-form.

Definition 5.3.1.

Let $X, Y$ be Banach or Frechet spaces and let $L:X\to Y$ be a linear map with a dense domain $D(J)$. We call adjoint of $L$ the map $L^*:Y^*\to X^*$ defined by

\begin{equation*} L^*\eta(x) = \eta(L(x)) \end{equation*}

Proposition 5.3.2.

In finite dimension, $J$ is injective if and only if $J^*$ is surjective. Indeed, set

\begin{equation*} \ker(J^*)^\perp = \{y\in Y\;|\;\eta(y)=0\text{ for all }\eta\in \ker(J^*)\} \subset Y \end{equation*}

and

\begin{equation*} \ker(J)^\perp = \{\xi\in X^*\;|\;\xi(x)=0\text{ for all }x\in\ker J\}\subset X^*. \end{equation*}

Then it is easy to verify that

\begin{equation*} J(X) = \ker(J^*)^\perp \end{equation*}

and

\begin{equation*} J^*(Y^*) = \ker(J)^\perp. \end{equation*}

Now, if $J$ is surjective, then $J(X)=Y$ and so every covector in $\ker(J^*)$ is zero on each vector of $Y\text{,}$ namely $\ker(J^*)=\{0_Y\}\text{,}$ i.e. $J^*$ is injective. Similarly, if $J$ is injective, then $\ker(J)=\{0_X\}$ and so $J^*(Y^*) = \ker(J)^\perp=X^*\text{,}$ i.e. $J^*$ is surjective. In particular, this shows that $J$ is an isomorphism between $X$ and $Y$ if and only if $J^*$ is an isomorphism between $X^*$ and $Y^*\text{.}$

In infinite dimension, things are more subtle, for instance the image $J(X)$ can be dense in $Y\text{.}$

Proposition 5.3.3.

Let $X, Y$ be Banach spaces and let $J\in B(X,Y)\text{.}$ Then:

$\ker(J^*)^\perp = \overline{J(X)}\text{;}$
$\ker(J)^\perp = \overline{J^*(Y^*)}\text{.}$

Proof.

(1) Assume first that $y\in\ker(J^*)^\perp\text{,}$ namely that $\eta(y) = 0$ for every $\eta\in\ker J^*\text{.}$

The following is a fundamental result by S. Banach and shows that, when the image is dense, things cannot go as in the finite-dimensional case. We do not include the proof because it is long. A complete proof appear in the textbook by Yosida.

Theorem 5.3.4. Closed Range Theorem.

Let $X, Y$ be Banach spaces and let $J$ be a closed operator with dense domain. The the following are equivalent:

$J(X)$ is closed in $Y\text{;}$
$J^*(Y^*)$ is closed in $X^*\text{;}$
$\ker(J^*)^\perp = J(X)\text{;}$
$\ker(J)^\perp = J^*(Y^*)\text{.}$

Using the theorem above, it is easy to prove the following infinite-dimensional extension of the properties of the transposed of a matrix we saw above.

Proposition 5.3.5.

Let $X, Y$ be Banach spaces and let $J\in B(X,Y)\text{.}$ Then:

$J^*$ is continuous.
$J$ is surjective if and only if $J^*$ is injective and $J^*(Y^*)\subset X^*$ is closed;
$J$ is injective if and only if $J^*$ is surjective and $J^*(Y^*)\subset X^*$ is closed;

Proof.

(1) Since

\begin{equation*} \|J^*_\eta\|_{X^*} = \sup_{\|x\|_X=1}|\eta(J(x))|\leq\|\eta\|_{Y^*}\cdot\|J\|_{op}, \end{equation*}

we have that

\begin{equation*} \|J^*\|_{op} = \sup_{\|eta\|_{Y^*}=1}\|J^*_\eta\|_{X^*}\leq\|J\|_{op}, \end{equation*}

so $J^*$ is bounded -- in fact, as usual one can use the Hahn-Banach theorem to show that $\|J^*\|_{op}=\|J\|_{op}\text{.}$

(2) Assume first that $J(X)=Y\text{.}$ Then, by (2) and (3) in Theorem 5.3.4, $\ker(J^*)=\{0_{Y^*}\}\text{,}$ i.e. $J^*$ is injective, and $J^*(Y^*)$ is closed in $X^*\text{.}$ Assume now that $J^*$ is injective, so that $\ker J^*=\{0_{Y^*}\}\text{,}$ and $J^*(Y^*)$ is closed. Then, by (3) in Theorem 5.3.4, $J(Y) = \ker(J^*)^\perp = X\text{.}$ (3) Similar to (2).

Proposition 5.3.6.

Let $X, Y$ be Banach spaces and let $J:X\to Y$ be a continous embedding such that $\overline{J(X)}\neq Y\text{.}$ Then $J^*$ is not injective.

Proof.

By the Hahn-Banach theorem, there is a non-zero $\eta\in Y^*$ such that $\ker\eta\supset\overline{J(X)}\text{.}$ Hence, $J^*\eta(x) = \eta(J(x))=0$ for every $x\in X\text{,}$ namely $J^*_\eta=0_{X^*}\text{,}$ so $J^*$ is not injective.

Proposition 5.3.7.

Let $X, Y$ be Banach spaces and let $J:X\to Y$ be a continous dense embedding. Then the adjoint map $J^*:Y^*\to X^*$ is a continuous dense embedding as well. Moreover, $J^*$ is surjective if and only if $J$ is.

Proof.

$J^*$ is injective by the same argument used in Proposition 5.3.5. By [cross-reference to target(s) "t-kerperp" missing or not unique], the injectivity of $J$ implies that the closure of $J^*(Y^*)$ is the whole space $X^*\text{.}$ Finally, if $J^*(Y^*)=X^*$ then, by [cross-reference to target(s) "transpose" missing or not unique], also $J$ is surjective.

We will meet several examples of these properties in next section.