description/proof of that normed covectors (dual) space of Hilbert space is Hilbert 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 admits 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.
- The reader admits the proposition that for any normed vectors space, if the norm satisfies the parallelogram law, the norm is induced by the unique inner product as this.
- The reader admits the parallelogram law on any vectors space normed induced by any inner product.
- The reader admits the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space.
- The reader admits the proposition that for any bijective complex-conjugate-linear 'normed vectors space' isometry, if the domain is complete, the codomain is complete.
Target Context
- The reader will have a description and a proof of the proposition that the normed covectors (dual) space of any Hilbert space is a Hilbert 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
\(V\): \(\in \{\text{ the } F \text{ Hilbert spaces }\}\)
\(V^*\): \(= \text{ the normed covectors space of } V\)
\(f\): \(: V \to V^*, v \mapsto \langle \bullet, v \rangle\), \(= \text{ the canonical bijective complex-conjugate-linear isometry }\)
//
Statements:
\(V^* \in \{\text{ the Hilbert spaces }\}\) with the inner product, \(\forall f_v, f_{v'} \in V^* (\langle f_v, f_{v'} \rangle = \overline{\langle v, v' \rangle})\), where \(f_v\) is the image of \(v \in V\) under \(f\)
//
2: Note
\(f\) is the bijective complex-conjugate-linear isometry mentioned in 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.
The reason why "\(f_v\)" is used instead of \(f (v)\) is that as we need to do like \(f (v) (v')\), \(f_v (v')\) seems less confusing.
\(V^*\) has no other option but to take that inner product (after that the inner product has been proved to indeed induce the norm on \(V^*\)), because the norm is determined to be the operator norm and the inner product that induces the norm (if any) is uniquely determined, by the proposition that for any normed vectors space, if the norm satisfies the parallelogram law, the norm is induced by the unique inner product as this: the norm needs to satisfy the parallelogram law if it is induced by any inner product, by the parallelogram law on any vectors space normed induced by any inner product.
3: Proof
Whole Strategy: Step 1: see that \(f\) exists; Step 2: see that the inner product on \(V^*\) is indeed an inner product; Step 3: see that the norm induced by the inner product is the operator norm; Step 4: see that \(V^*\) is complete.
Step 1:
\(f\) exists, by 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.
Step 2:
Let us see that the inner product on \(V^*\) is indeed an inner product.
Let \(f_{v_1}, f_{v_2}, f_{v_3} \in V^*\) and \(r_1, r_2 \in F\) be any.
1) \((0 \le \langle f_{v_1}, f_{v_1} \rangle)\) \(\land\) \((0 = \langle f_{v_1}, f_{v_1} \rangle \iff f_{v_1} = 0)\): \(0 \le \overline{\langle v_1, v_1 \rangle} = \langle f_{v_1}, f_{v_1} \rangle\); \(0 = \langle f_{v_1}, f_{v_1} \rangle = \overline{\langle v_1, v_1 \rangle}\) if and only if \(v_1 = 0\) if and only if \(f_{v_1} = 0\).
2) \(\langle f_{v_1}, f_{v_2} \rangle = \overline{\langle f_{v_2}, f_{v_1} \rangle}\): \(\langle f_{v_1}, f_{v_2} \rangle = \overline{\langle v_1, v_2 \rangle} = \langle v_2, v_1 \rangle = \overline{\overline{\langle v_2, v_1 \rangle}} = \overline{\langle f_{v_2}, f_{v_1} \rangle}\).
3) \(\langle r_1 f_{v_1} + r_2 f_{v_2}, f_{v_3} \rangle = r_1 \langle f_{v_1}, f_{v_3} \rangle + r_2 \langle f_{v_2}, f_{v_3} \rangle\): \(\langle r_1 f_{v_1} + r_2 f_{v_2}, f_{v_3} \rangle = \langle f_{\overline{r_1} v_1 + \overline{r_2} v_2}, f_{v_3} \rangle\), because \(f\) is complex-conjugate-linear, \(= \overline{\langle \overline{r_1} v_1 + \overline{r_2} v_2, v_3 \rangle} = \overline{\overline{r_1} \langle v_1, v_3 \rangle + \overline{r_2} \langle v_2, v_3 \rangle} = r_1 \overline{\langle v_1, v_3 \rangle} + r_2 \overline{\langle v_2, v_3 \rangle} = r_1 \langle f_{v_1}, f_{v_3} \rangle + r_2 \langle f_{v_2}, f_{v_3} \rangle\).
So, the inner product on \(V^*\) is an inner product.
Step 3:
Now, we have the inner product and know that the inner product induces a norm, but we need to know that the norm induced by the inner product is the operator norm.
As \(f\) is an isometry, for the operator norm, \(\Vert f_v \Vert\), \(\Vert f_v \Vert = \Vert v \Vert\).
For the norm induced by the inner product, \(\Vert f_v \Vert\), \(\Vert f_v \Vert = \sqrt{\langle f_v, f_v \rangle} = \sqrt{\overline{\langle v, v \rangle}} = \sqrt{\langle v, v \rangle} = \Vert v \Vert\).
So, the norm induced by the inner product is the operator norm.
The inner product is the unique inner product that induces the operator norm, by the parallelogram law on any vectors space normed induced by any inner product and the proposition that for any normed vectors space, if the norm satisfies the parallelogram law, the norm is induced by the unique inner product as this.
Step 4:
Let us see that \(V^*\) is complete.
\(f\) is a bijective complex-conjugate-linear isometry.
By the proposition that for any bijective complex-conjugate-linear 'normed vectors space' isometry, if the domain is complete, the codomain is complete, \(V^*\) is complete.
