3
$\begingroup$

I am self studying adjoint operators on Banach spaces. The adjoint of a linear mapping $L:X \rightarrow Y$, where $X,Y$ are Banach spaces, is a unique mapping \begin{equation} L^*:Y^{*} \rightarrow X^* \end{equation} defined by \begin{equation} L^*y^* = y^* \circ L. \end{equation} The above definition can be found on Page 98 of Functional Analysis by Rudin. My question concerns the case where the mapping goes to a product of Banach spaces. For example, let $T:X \rightarrow Y_1 \times Y_2$ be a linear mapping, with $X,Y_{1},Y_2$ Banach spaces. Following the above definition, the adjoint of $T$ would be \begin{equation} T^*:(Y_{1} \times Y_{2})^* \rightarrow X^*. \end{equation} Since the dual of a product space is a product space of the duals, the above is equivalent to \begin{equation} T^*:Y_{1}^* \times Y_{2}^* \rightarrow X^*. \end{equation} My question regards how to explicitly define $T^*$. The adjoint is itself a linear operator so my intuition says that \begin{equation} T^*(y_{1}^*,y_{2}^*) = y_{1}^* \circ T + y_{2}^* \circ T \end{equation} but I have no proof of this. Is this correct? Any proof or reference would be welcome.

$\endgroup$

1 Answer 1

1
$\begingroup$

Your question reduces to understand in what sense is $Y_1^*\times Y_2^*$ the dual of $Y_1\times Y_2$. After all, an ordered pair of linear functionals is not a linear functional.

If $f:Y_1\times Y_2\to\mathbb C$ is linear, we can write $f(y_1,y_2)=f_1(y_1)+f_2(y_2)$, where $f_j\in Y_j^*$ is given by $$ f_1(y)=f(y_1,0),\qquad\qquad f_2(y)=f(0,y_2). $$ This induces a natural isomorphism $(Y_1\times Y_2)^*\to Y_1^*\times Y_2^*$, where the norm on $Y_1^*\times Y_2^*$ will depend on the norm you gave to $Y_1\times Y_2$: $$ \|(f,g)\|=\sup\{|f(y)+g(z)|:\ \|(y,z)\|=1\}. $$ With this point of view, $$ T^*(f,g)(x)=(f,g)(Tx)=f(T_1x)+g(T_2x). $$ That is, $$ T^*(f,g)=f\circ T_1+g\circ T_2, $$ where $T_1:X\to Y_1$ and $T_2:X\to Y_2$ are given by $Tx=(T_1x,T_2x)$.

$\endgroup$
1
  • $\begingroup$ Thanks, this makes sense. You were right I was not appreciating the linearity on $Y_{1} \times Y_{2}$. $\endgroup$
    – user273331
    Commented Jun 29, 2022 at 1:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .