description/proof of that for vectors space with inner product with induced topology, map from dense subset into space, and map from dense subset that contains 1st subset into space whose adjoint has space as domain, adjoint of sum of \(2\) maps is sum of adjoints of \(2\) maps
Topics
About: vectors space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any vectors space with any inner product with the induced topology, any map from any dense subset into the space, and any map from any dense subset that contains the 1st subset into the space whose adjoint has the space as the domain, the adjoint of the sum of the \(2\) maps is the sum of the adjoints of the \(2\) maps.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\mathbb{R}, \mathbb{H}\}\), with the canonical field structure
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), with any inner product, with the topology induced by the metric induced by the inner product
\(S_1\): \(\in \{\text{ the dense subsets of } V\}\)
\(S_2\): \(\in \{\text{ the dense subsets of } V\}\) such that \(S_1 \subseteq S_2\)
\(f_1\): \(: S_1 \to V\)
\(f_2\): \(: S_2 \to V\)
\({f_1}^*\): \(= \text{ the adjoint of } f_1\), \(: {S_1}^* \to V\)
\({f_2}^*\): \(= \text{ the adjoint of } f_2\), \(: V \to V\)
\(f_1 + f_2\): \(: S_1 \to V\)
\((f_1 + f_2)^*\): \(= \text{ the adjoint of } f_1 + f_2\)
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Statements:
\((f_1 + f_2)^*: {S_1}^* \to V = {f_1}^* + {f_2}^*\)
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2: Note
Typically, \(S_2 = V\), but be aware that that does not guarantee that \({S_2}^* = V\).
An example of that \(S_2 = V\) and \({S_2}^* = V\) is \(f_2: V \to V, v \mapsto r v\) for any \(r \in F\): \({f_2}^*: V \to V, v \mapsto \overline{r} v\), because for each \(v, v' \in V\), \(\langle v', r v \rangle = \langle \overline{r} v', v \rangle\).
3: Proof
Whole Strategy: Step 1: see that for each \(v' \in {S_1}^*\), for each \(s \in S_1\), \(\langle ({f_1}^* + {f_2}^*) v', s \rangle = \langle v', (f_1 + f_2) s \rangle\); Step 2: see that there is no \(v' \in V \setminus {S_1}^*\) such that there is a \(v'' \in V\) such that for each \(s \in S_1\), \(\langle v'', s \rangle = \langle v', (f_1 + f_2) s \rangle\).
Step 1:
Let \(v' \in {S_1}^*\) be any.
For each \(s \in S_1\), \(\langle v', (f_1 + f_2) s \rangle = \langle v', f_1 s + f_2 s \rangle = \langle v', f_1 s \rangle + \langle v', f_2 s \rangle = \langle {f_1}^* v', s \rangle + \langle {f_2}^* v', s \rangle = \langle {f_1}^* v' + {f_2}^* v', s \rangle = \langle ({f_1}^* + {f_2}^*) v', s \rangle\).
That means that the domain of \((f_1 + f_2)^*\) contains \({S_1}^*\) and over \({S_1}^*\), \((f_1 + f_2)^* = {f_1}^* + {f_2}^*\), but does not mean that the domain of \((f_1 + f_2)^*\) equals \({S_1}^*\), yet.
Step 2:
Let us see that there is no \(v' \in V \setminus {S_1}^*\) such that there is a \(v'' \in V\) such that for each \(s \in S_1\), \(\langle v'', s \rangle = \langle v', (f_1 + f_2) s \rangle\).
Let us suppose that there was such a \(v' \in V \setminus {S_1}^*\).
\(\langle v', (f_1 + f_2) s \rangle = \langle v', f_1 s + f_2 s \rangle = \langle v', f_1 s \rangle + \langle v', f_2 s \rangle = \langle v', f_1 s \rangle + \langle {f_2}^* v', s \rangle\), because \({S_2}^* = V\).
On the other hand, \(\langle v'', s \rangle = \langle v'' - {f_2}^* v' + {f_2}^* v', s \rangle = \langle v'' - {f_2}^* v', s \rangle + \langle {f_2}^* v', s \rangle\).
So, \(\langle v', f_1 s \rangle + \langle {f_2}^* v', s \rangle = \langle v'' - {f_2}^* v', s \rangle + \langle {f_2}^* v', s \rangle\), which would imply that \(\langle v', f_1 s \rangle = \langle v'' - {f_2}^* v', s \rangle\), which would mean that \(v' \in {S_1}^*\), a contradiction against \(v' \in V \setminus {S_1}^*\).
So, there is no such a \(v' \in V \setminus {S_1}^*\).
That means that the domain of \((f_1 + f_2)^*\) equals \({S_1}^*\).
