description/proof of that for Hilbert space and its normed covectors (dual) space, there is canonical bijective complex-conjugate-linear isometry from Hilbert space onto normed covectors space
Topics
About: Hilbert space
The table of contents of this article
Starting Context
- The reader knows a definition of Hilbert space.
- The reader knows a definition of normed covectors (dual) space of normed vectors space.
- The reader knows a definition of bijection.
- The reader knows a definition of complex-conjugate-linear map.
- The reader knows a definition of 'normed vectors space' isometry.
- The reader knows a definition of orthogonal complement of subset of vectors space with inner product.
- The reader admits the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space.
- The reader admits the proposition that the kernel of any linear map between any vectors spaces is a vectors subspace of the domain.
- The reader admits the proposition that any linear map between any vectors metric spaces induced by any norms is continuous if and only if it is bounded.
- The reader admits the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed.
- The reader admits the proposition that for any Hilbert space and its any closed vectors subspace, the space is the vectors space as direct sum of the subspace and its orthogonal complement.
Target Context
- The reader will have a description and a proof of the proposition that for any Hilbert space and its normed covectors (dual) space, there is the canonical bijective complex-conjugate-linear isometry from the Hilbert space onto the normed covectors 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
\(V\): \(\in \{\text{ the } F \text{ Hilbert spaces }\}\)
\(V^*\): \(= \text{ the normed covectors space of } V\)
\(f\): \(: V \to V^*, v \mapsto \langle \bullet, v \rangle\)
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Statements:
\(f \in \{\text{ the bijective complex-conjugate-linear isometries }\}\)
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2: Note
The part of this proposition that each element of \(V^*\) is \(\langle \bullet, v \rangle\) for the unique \(v \in V\) is prevalently called "the Riesz representation theorem".
But this proposition is more than that: the Riesz representation theorem itself does not say that for each \(v \in V\), \(\langle \bullet, v \rangle \in V^*\), or that \(f\) is injective, or that \(f\) is complex-conjugate-linear, or that \(f\) is a 'normed vectors space' isometry.
3: Proof
Whole Strategy: Step 1: see that \(\langle \bullet, v \rangle \in V^*\); Step 2: see that \(f\) is bijective; Step 3: see that \(f\) is complex-conjugate-linear; Step 4: see that \(f\) is a 'normed vectors space' isometry.
Step 1:
Let us see that \(g: V \to F, v' \mapsto \langle v', v \rangle \in V^*\).
For each \(v_1, v_2 \in V\) and \(r_1, r_2 \in F\), \(g (r_1 v_1 + r_2 v_2) = \langle r_1 v_1 + r_2 v_2, v \rangle = r_1 \langle v_1, v \rangle + r_2 \langle v_2, v \rangle = r_1 g (v_1) + r_2 g (v_2)\), so, \(g\) is linear.
\(sup_{v' \in V \setminus \{0\}} \vert g (v') \vert / \Vert v' \Vert = \vert \langle v', v \rangle \vert / \sqrt{\langle v', v' \rangle} \le \sqrt{\langle v', v' \rangle} \sqrt{\langle v, v \rangle} / \sqrt{\langle v', v' \rangle}\), by the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space, \(= \sqrt{\langle v, v \rangle} \lt \infty\), so, \(g\) is bounded.
So, \(g \in V^*\).
So, \(f\) is well-defined.
Step 2:
Let us see that \(f\) is injective.
Let \(v_1, v_2 \in V\) be any such that \(v_1 \neq v_2\).
Let us suppose that \(f (v_1) = f (v_2)\).
Then, \(\langle \bullet, v_1 \rangle = \langle \bullet, v_2 \rangle\), so, \(\langle v_1 - v_2, v_1 \rangle = \langle v_1 - v_2, v_2 \rangle\), so, \(\langle v_1 - v_2, v_1 \rangle - \langle v_1 - v_2, v_2 \rangle = 0\), but the left hand side was \(\langle v_1 - v_2, v_1 - v_2 \rangle\), so, \(\langle v_1 - v_2, v_1 - v_2 \rangle = 0\), which would imply that \(v_1 - v_2 = 0\), so, \(v_1 = v_2\), a contradiction.
So, \(f (v_1) \neq f (v_2)\).
So, \(f\) is injective.
Let us see that \(f\) is surjective.
Let \(g \in V^*\) be any.
When \(g = 0\), \(g = \langle \bullet, 0 \rangle\).
Let us suppose that \(g \neq 0\) hereafter.
Let \(V^` := g^{-1} (0)\).
\(V^`\) is a vectors subspace of \(V\), by the proposition that the kernel of any linear map between any vectors spaces is a vectors subspace of the domain.
\(V^`\) is a closed vectors subspace of \(V\), because \(\{0\}\) is a closed subset of \(F\), by the proposition that any linear map between any vectors metric spaces induced by any norms is continuous if and only if it is bounded and the proposition that any topological spaces map is continuous if and only if the preimage of any closed subset of the codomain is closed.
Let \({V^`}^{\perp}\) be the orthogonal complement of \(V^`\).
\(V\) is a vectors space as direct sum of \(V^`\) and \({V^`}^{\perp}\), by the proposition that for any Hilbert space and its any closed vectors subspace, the space is the vectors space as direct sum of the subspace and its orthogonal complement.
As \(g \neq 0\), \(\{0\} \subset {V^`}^{\perp}\), because if \({V^`}^{\perp} = \{0\}\), for each \(v \in V\), \(v = v^` + 0\) where \(v^` \in V^`\), because \(V\) was a vectors space as direct sum of \(V^`\) and \({V^`}^{\perp}\), so, \(v = v^` \in V^`\), so, \(V^` = V\), which would mean that \(g = 0\), a contradiction.
So, there is a \(w \in {V^`}^{\perp}\) such that \(w \neq 0\), but let us take \(\Vert w \Vert = 1\), which is possible because \({V^`}^{\perp}\) is a vectors subspace: take \(w / \Vert w \Vert \in {V^`}^{\perp}\) instead.
For each \(v' \in V\), \(g (v') w - g (w) v' \in V^`\), because \(g (g (v') w - g (w) v') = g (v') g (w) - g (w) g (v') = 0\).
So, \(0 = \langle g (v') w - g (w) v', w \rangle = \langle g (v') w, w \rangle - \langle g (w) v', w \rangle = g (v') \langle w, w \rangle - g (w) \langle v', w \rangle = g (v') - \langle v', \overline{g (w)} w \rangle\), so, \(g (v') = \langle v', \overline{g (w)} w \rangle\).
That means that \(g (v') = \langle v', v \rangle\) where \(v := \overline{g (w)} w\), which means that \(g = \langle \bullet, v \rangle\).
So, \(f\) is surjective.
Step 3:
Let us see that \(f\) is complex-conjugate-linear.
Let \(v_1, v_2 \in V\) and \(r_1, r_2 \in F\) be any.
\(f (r_1 v_1 + r_2 v_2) = \langle \bullet, r_1 v_1 + r_2 v_2 \rangle = \overline{r_1} \langle \bullet, v_1 \rangle + \overline{r_2} \langle \bullet, v_2 \rangle = \overline{r_1} f (v_1) + \overline{r_2} f (v_2)\).
So, \(f\) is complex-conjugate-linear.
Step 4:
Let us see that \(f\) is a 'normed vectors space' isometry.
For each \(v \in V\), \(\Vert f (v) \Vert = sup_{v' \in V \setminus \{0\}} \vert f (v) (v') \vert / \Vert v' \Vert = sup_{v' \in V \setminus \{0\}} \vert \langle v', v \rangle \vert / \Vert v' \Vert \le sup_{v' \in V \setminus \{0\}} \sqrt{\vert \langle v', v' \rangle \vert} \sqrt{\vert \langle v, v \rangle \vert} / \sqrt{\vert \langle v', v' \rangle \vert}\), by the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space, \(= sup_{v' \in V \setminus \{0\}} \sqrt{\vert \langle v, v \rangle \vert} = \sqrt{\vert \langle v, v \rangle \vert} = \Vert v \Vert\), and \(\vert f (v) (v') \vert / \Vert v' \Vert = \Vert v \Vert\) is in fact realized by \(v' = v\), because \(\vert f (v) (v) \vert / \Vert v \Vert = \vert \langle v, v \rangle \vert / \Vert v \Vert = \Vert v \Vert^2 / \Vert v \Vert = \Vert v \Vert\), so, \(\Vert f (v) \Vert = \Vert v \Vert\).
So, \(f\) is a 'normed vectors space' isometry.
