Adjoint of linear mapping into $\mathbb{R}$. I am self studying adjoint operators on Banach spaces and want to check my thinking regarding the case when a linear mapping maps into $\mathbb{R}$.
In the general case we let $T:X \rightarrow Y$ be a linear mapping with $X$ and $Y$ Banach spaces. Then the adjoint of $T$ is itself a continuous linear mapping, $T^{*}:Y^{*} \rightarrow X^{*}$ defined by
\begin{equation}
   T^{*}(y^{*})(x) = y^{*} \circ (Tx). 
\end{equation}
Now look at the case where $Y = \mathbb{R}$. Two points come to mind.
First, $T:X \rightarrow \mathbb{R}$, i.e. we already have $T \in X^{*}$.
Second, by definition
\begin{equation}
   T^{*}: \mathbb{R}^{*} \rightarrow X^{*}.
\end{equation}
However, since the dual space to $\mathbb{R}$ is itself $\mathbb{R}$, we in fact have
\begin{equation}
   T^{*}:\mathbb{R} \rightarrow X^{*}
\end{equation}
defined by
\begin{equation}
   T^{*}(c)(x) = cT(x)
\end{equation}
where $c \in \mathbb{R}$ and $x \in X$. Since $T$ is a linear mapping, this could be re-written as $T^{*}(c)(x) = T(cx)$. However, $X$ is a Banach space and is hence linear.  Therefore, $x \in X$ implies $cx \in X$ for any $c \in \mathbb{R}$. My question is, given the seeming redundancy of $c$ and the fact $T \in X^{*}$, can we say the adjoint of $T:X \rightarrow \mathbb{R}$ is
\begin{equation}
\begin{aligned}
&T^{*}:\mathbb{R} \rightarrow X^{*} \\
&T^{*}(x) = Tx,
\end{aligned}
\end{equation}
or is the $c$ a necessary part of the definition, even given its seeming redundancy?
 A: As pointed out in the comments, your thinking is correct until the very last line. The issue boils down to understanding the domains of functions. For fun, let's walk through your thinking again to make sure we understand every step :)
Following your convention, we let $X,Y$ denote Banach spaces with $X^*,Y^*$ denoting their respective continuous dual spaces.
Your definition of the adjoint of a linear map $T : X \to Y$ is a common and great starting point. I'd just like to note that the adjoint $T^*$ is continuous given that $T$ itself is continuous. If $T$ is not continuous, then the adjoint need not be continuous. Also, in case you are interested, there are generalizations of the concept of adjoints to when $X,Y$ are simply normed vector spaces. One may also replace the continuous duals $X^*,Y^*$ in the definition with the algebraic duals $X^\sharp,Y^\sharp$. This Wikipedia link gives a nice summary. (Note the distinction between continuous and algebraic dual spaces is often an important one to make)
Now, onto your question. let $T : X \to \mathbb{R}$ be an arbitrary linear map. If we also assume $T$ is continuous (or equivalently bounded, see this Wikipedia link), then yes $T \in X^*$ by definition. Hence, by definition, $T^* : \mathbb{R}^* \to X^*$ is the continuous linear map where $T^*(y^*) = y^* \circ T$ for any $y^* \in \mathbb{R}^*$. Importantly, $y^* \in \mathbb{R}^*$ is a continuous linear functional on $\mathbb{R}$, and not a scalar. So, $T$ maps a linear functional on $\mathbb{R}$ to a linear functional on $X$. Yet, as you say, one can isomorphically identify $\mathbb{R}^*$ with $\mathbb{R}$. This is because every continuous linear functional on $\mathbb{R}$ is simply multiplication by a real number while, conversely, multiplication by a real number is always a continuous linear functional on $\mathbb{R}$. Try to prove that this is an isomorphism yourself as an exercise :) Denoting this isomorphism by $\Phi : \mathbb{R} \to \mathbb{R}^*$, we have that $T^* \circ \Phi : \mathbb{R} \to X^*$ is a continuous linear map. Since we may view $T^* \circ \Phi$ as "essentially the same" operator as $T^*$, we may abuse notation and write $T^* : \mathbb{R} \to X$ instead of $T^* \circ \Phi$. This is what the majority of people do the majority of the time. Now, as you acknowledge in the comments, we may conclude that $T^*(c) = cT$ for any real number $c$, meaning $T^*$ maps the number $c$ to the continuous linear operator $cT$. So, we may indeed write $T^*(c)(x) = cT(x) = T(cx)$ for any real number $c$ and any $x \in X$. But, this does not mean $T^* = T$, as again $T^*$ maps real numbers to elements of $X^*$ while $T$ is itself an element of $X^*$. You have acknowledged this in the comments, which is great!
The point of this answer is to walk through some of the finer details that may or may not have been overlooked. It is always easy to get confused when dealing with functions between function spaces, and so it is a good exercise to slow down and take note of the domains, codomains, and identifications that are at play. Best of luck in your studies!
