definition of adjoint map of map from dense subset of vectors space with inner product with induced topology into same vectors space
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of inner product on real or complex vectors space.
- The reader knows a definition of norm induced by inner product on real or complex vectors space.
- The reader knows a definition of topology induced by metric.
- The reader knows a definition of dense subset of topological space.
- The reader knows a definition of map.
Target Context
- The reader will have a definition of adjoint map of map from dense subset of vectors space with inner product with induced topology into same vectors space.
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:
\( F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), 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 norm induced by the inner product
\( S\): \(\in \{\text{ the dense subsets of } V\}\)
\( f\): \(: S \to V\)
\( S^*\): \(= \{v \in V \vert \exists v' \in V (\forall s \in S (\langle v, f (s) \rangle = \langle v', s \rangle))\}\)
\(*f^*\): \(: S^* \to V\)
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Conditions:
\(\forall v \in S^* (\langle v, f (s) \rangle = \langle f^* (v), s \rangle)\)
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2: Note
Let us see that \(f^*\) is well-defined.
As \(S^*\) is defined so, there is at least 1 \(v' \in V\) such that \(\langle v, f (s) \rangle = \langle v', s \rangle\).
The issue is whether such a \(v'\) for each \(v \in S^*\) is unique.
Let \(v'' \in V\) be any other one such that \(\forall s \in S (\langle v, f (s) \rangle = \langle v'', s \rangle)\).
\(\langle v, f (s) \rangle = \langle v', s \rangle = \langle v'', s \rangle\), so, \(\langle v', s \rangle - \langle v'', s \rangle = \langle v' - v'', s \rangle = 0\).
As \(S\) is dense, for each \(n \in \mathbb{N} \setminus \{0\}\), there is a \(s_n \in S\) such that \(\langle v' - v'' - s_n, v' - v'' - s_n \rangle \lt 1 / n\), which means that the sequence, \(s_1, s_2, ...\), converges to \(v' - v''\), so, \(lim_{n \to \infty} s_n = v' - v''\).
\(\langle v' - v'', s_n \rangle = 0\), so, \(lim_{n \to \infty} \langle v' - v'', s_n \rangle = 0\).
By the proposition that for any real or complex vectors space with the topology induced by the metric induced by the norm induced by any inner product, the inner product with any 1 argument fixed is a continuous map and the proposition that for any continuous map and any net with directed index set that converges to any point on the domain, the image of the net converges to the image of the point and if the codomain is Hausdorff, the convergence of the image of the net is the image of the point, \(lim_{n \to \infty} \langle v' - v'', s_n \rangle = \langle v' - v'', lim_{n \to \infty} s_n \rangle = \langle v' - v'', v' - v'' \rangle = 0\).
So, \(v' - v'' = 0\), so, \(v' = v''\).
So, \(f^*\) is well-defined.
This concept is valid only for \(f\) that is from any dense subset of \(V\) into \(V\), because if it was into another space, \(V'\), \(\langle v, f (s) \rangle\) would not make sense because \(v\) and \(f (s)\) would be in some different vectors spaces and what vectors space the inner product was defined on?; and the domain needs to be a dense subset, because the proof of the uniqueness of \(v'\) requires it to be dense.
Usually, \(V\) is supposed to be a Hilbert space, but the proof of the well-defined-ness of the definition did not use the completeness.
