# Adjoint of a linear mapping mapping into a product of Banach Spaces

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.

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 $$Tx=(T_1x,T_2x)$$.
• Thanks, this makes sense. You were right I was not appreciating the linearity on $Y_{1} \times Y_{2}$. Jun 29, 2022 at 1:20