Correspondence between linear maps of a vector space into itself and linear maps of the dual into itself.

Question

I was wondering about vector spaces and their dual. Specifically, in the context of finite-dimensional vector spaces, I asked myself if it is true that there is a one-to-one correspondence between the set of linear maps of a vector space into itself and the set of linear maps of the dual into itself. I came to a conclusion, but I am not sure it is correct. Here is what I found.

Let $V$ be a finite-dimensional, complex vector space, and let $V^{*}$ be its dual. Denote with $\mathcal{Lin}(V\,;V)$ the set of linear maps from $V$ to $V$, and with $\mathcal{Lin}(V^{*}\,;V^{*})$ the set of linear maps from $V^{*}$ to $V^{*}$.

If $\phi\in\mathcal{Lin}(V\,;V)$, then we can define $\phi^{*}:V^{*}\longrightarrow V^{*}$ by:

$$ \left(\phi^{*}(w)\right)(x):=w\left(\phi(x)\right) $$ for all $w\in V^{*}$ and for all $x\in V$. Then, considering $w_{1},w_{2}\in V^{*}$ and $\alpha,\beta\in\mathbb{C}$:

$$ \left(\phi^{*}\left(\alpha w_{1} + \beta w_{2}\right)\right)(x):=(\alpha w_{1} + \beta w_{2})(\phi(x))=w_{1}\left(\alpha\phi(x)\right) + w_{2}\left(\beta\phi(x)\right)= $$ $$ =w_{1}\left(\phi(\alpha x)\right) + w_{2}\left(\phi(\beta x)\right)=(\phi^{*}(w_{1}))(\alpha x) + (\phi^{*}(w_{2}))(\beta x)=\alpha(\phi^{*}(w_{1}))(x) + \beta(\phi^{*}(w_{2}))(x)= $$ $$ =\left(\alpha(\phi^{*}(w_{1})) + \beta(\phi^{*}(w_{2}))\right)(x) $$ and thus, since $x\in V$ is arbitrary:

$$ \phi^{*}\left(\alpha w_{1} + \beta w_{2}\right)=\alpha(\phi^{*}(w_{1})) + \beta(\phi^{*}(w_{2})) $$ which means that $\phi^{*}\in\mathcal{Lin}(V^{*}\,;V^{*})$. Note that the linearity of $\phi$ is needed, otherwise we can not write $w_{1}(\alpha(\phi(x)))=w_{1}(\phi(\alpha x))$.

Next, if $\psi\in\mathcal{Lin}(V^{*}\,;V^{*})$, we can define $\psi_{*}:V\longrightarrow V$ by:

$$ w(\psi_{*}(x)):=(\psi(w))(x) $$ for all $x\in V$ and for all $w\in V^{*}$. Then, considering $x,y\in V$ and $\alpha,\beta\in\mathbb{C}$, we have:

$$ w(\psi_{*}(\alpha x + \beta y)):=(\psi(w))(\alpha x + \beta y)=(\alpha\psi(w))(x) + (\beta\psi(w))(y)=(\psi(\alpha w))(x) + (\psi(\beta w))(y)= $$ $$ =(\alpha w)(\psi_{*}(x)) + (\beta w)(\psi_{*}(y))=w(\alpha\psi_{*}(x)) + \beta w(\beta\psi_{*}(y))=w(\alpha\psi_{*}(x) + \beta\psi_{*}(y)) $$ and thus, since $w\in V^{*}$ is arbitrary:

$$ \psi_{*}(\alpha x + \beta y)=\alpha\psi_{*}(x) + \beta\psi_{*}(y) $$ which means that $\psi_{*}\in\mathcal{Lin}(V\,;V)$. Note that the linearity of $\psi$ is needed, otherwise we can not write $(\alpha\psi(w))(x)=(\psi(\alpha w))(x)$.

Finally, let $\phi\in\mathcal{Lin}(V\,;V)$ and set $\psi\equiv\phi^{*}$, then:

$$ (\psi(w))(x)=(\phi^{*}(w))(x):=w(\phi(x)) $$ hence:

$$ w(\psi_{*}(x)):=(\psi(w))(x)=(\phi^{*}(w))(x):=w(\phi(x)) $$ which means:

$$ w(\psi_{*}(x) - \phi(x))=0 $$ and, since $w\in V^{*}$ and $x\in V$ are arbitrary:

$$ \phi=\psi_{*}=(\phi^{*})_{*} $$ Therefore, we can conclude that every $\phi\in\mathcal{Lin}(V\,;V)$ induces a $\psi\in\mathcal{Lin}(V^{*}\,;V^{*})$ given by $\psi=\phi^{*}$, and every $\psi\in\mathcal{Lin}(V^{*}\,;V^{*})$ induces a $\phi\in\mathcal{Lin}(V\,;V)$ given by $\phi=\psi_{*}$. Moreover, $\phi=(\phi^{*})_{*}$ and $\psi=(\psi_{*})^{*}$.

Having said that, my questions are the following:

1. are there errors I am not able to see?
2. if not, are there some obstruction in extending this result to the case of an infinite-dimensional Banach space $V$?

Thank You

Answer 1

Here are some partial answers to your questions. First (assuming the Axiom of Choice), if $V$ is infinite-dimensional, then its algebraic dual $V^*$ always has larger cardinality than $V$ itself. Thus, the extension of the theorem fails for general infinite-dimensional vector spaces. If the Axiom of Choice is assumed false, then there are vector spaces that have no non-trivial linear functionals (i.e., $V^* = \{ 0 \}$.) Again, the proposed extension fails.

I'm not sure about the situation for Banach spaces.

To address your first question. Your construction of $\psi_*$ doesn't make reference to the finite-dimensionality of $V$, and would seem to hold for all vector spaces. Given the above remarks, this is problematic. My guess is that your definition of $\psi_*$ isn't well-defined. (What does $\psi_*(x)$ equal? It equals the vector in $V$ with the property that whenever you plug it into a linear functional $w$, it takes on the value $(\psi(w))(x)$. How can you know that such a vector exists and is unique?)

The upshot of all this: If $V$ is finite-dimensional, then the result you're after is true. My guess is that the only way to prove it is to start with a basis for $V$ (which determines a basis for $V^*$) and then define your maps in terms of these bases.

Answer 2

As has been correctly pointed out by @PeterJL, the definition of $\psi_{*}$ in my question is ill-posed, so something else is needed. In accordance with @Giuseppe Negro's comment, I looked for "Banach space adjoint" and I came up with the following answer.

Let $X$ be a Banach space. It is well known that its topological dual $X'$ (i.e., the set of all the linear functionals on $X$ that are continuous with respect to the topology of $X$ and $\mathbb{C}$) is a Banach space itself, nevertheless, $X'$ is, in general, only a subset of the algebraic dual $X^{*}$ of $X$ (namely, the set of all linear functionals on $X$). Moreover, the topological bidual $X''$ of $X$ is in general "greater" than $X$, and we have a canonical injective isometry $B:X\longrightarrow B(X)\subseteq X''$. If $X$ is such that $B(X)=X''$ then it is called reflexive.

Now, let $\phi:X\longrightarrow X$ be a bounded linear map of a Banach space into itself, then we can define the so-called Banach adjoint $\phi^{*}:X'\longrightarrow X'$ as:

$$ \left(\phi^{*}(\omega)\right)(x)=\omega(\phi(x)) $$ for every $\omega\in X'$ and every $x\in X$. In Conway's book: "A course in functional analysis" it is proved that $\phi^{*}$ is linear, it is bounded with $||\phi^{*}||=||\phi||$, and it is such that $(\phi^{*})^{*}|_{B(X)}=\phi$. Therefore, every bounded linear map $\phi:X\longrightarrow X$ induces a bounded linear map $\phi^{*}:X'\longrightarrow X'$. Obviously, since $X'$ is itself a Banach space, the same applies for a bounded linear map $\Phi:X'\longrightarrow X'$, and we have that every such map induces a bounded linear map $\Phi^{*}:X''\longrightarrow X''$.

If $X$ is reflexive, we have that $X$ and $X''$ can be identified by means of the bijective isometry $B$, and thus every bounded linear map $\phi:X\longrightarrow X$ induces a bounded linear map $\phi^{*}:X'\longrightarrow X'$ and viceversa.