description/proof of that for normed vectors space, if norm satisfies parallelogram law, norm is induced by unique inner product as this
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of norm induced by inner product on real or complex vectors space.
- The reader knows a definition of metric induced by norm on real or complex vectors space.
- The reader knows a definition of topology induced by metric.
- The reader admits the proposition that for any vectors space with any norm and the topological space induced by the metric induced by the norm, the norm map is continuous.
- The reader admits 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.
- The reader admits the proposition that for any normed vectors space, if the norm is induced by any inner product, the inner product that induces the norm is unique as this.
Target Context
- The reader will have a description and a proof of 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.
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 norm, \(\Vert \bullet \Vert\)
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Statements:
\(\forall v_1, v_2 \in V (\Vert v_1 + v_2 \Vert + \Vert v_1 - v_2 \Vert = 2 \Vert v_1 \Vert + 2 \Vert v_2 \Vert )\)
\(\implies\)
\(\Vert \bullet \Vert\) is induced by \(\langle v_1, v_2 \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert v_1 + i^j v_2 \Vert^2\)
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2: Proof
Whole Strategy: Step 1: suppose that the parallelogram law holds; Step 2: see that \(\langle \bullet, \bullet \rangle\) is an inner product; Step 3: see that \(\Vert \bullet \Vert\) is induced by \(\langle \bullet, \bullet \rangle\); Step 4: see that \(\langle \bullet, \bullet \rangle\) is the unique inner product by which \(\Vert \bullet \Vert\) is induced.
Step 1:
Let us suppose that the parallelogram law holds.
Step 2:
Let us see that \(\langle \bullet, \bullet \rangle\) is an inner product.
Step 2 Strategy: Step 2-1: see that \(0 \le \langle v_1, v_1 \rangle\) and \((0 = \langle v_1, v_1 \rangle \iff v_1 = 0)\); Step 2-2: see that \(\langle v_1, v_2 \rangle = \overline{\langle v_2, v_1 \rangle}\); Step 2-3: see that \(\langle v_1 + v_2, v_3 \rangle = \langle v_1, v_3 \rangle + \langle v_2, v_3 \rangle\); Step 2-4: see that for each \(r_1 \in \mathbb{Z}\), \(\langle r_1 v_1, v_3 \rangle = r_1 \langle v_1, v_3 \rangle\); Step 2-5: see that for each \(r_1 \in \mathbb{Q}\), \(\langle r_1 v_1, v_3 \rangle = r_1 \langle v_1, v_3 \rangle\); Step 2-6: see that for each \(r_1 \in \mathbb{R}\), \(\langle r_1 v_1, v_3 \rangle = r_1 \langle v_1, v_3 \rangle\); Step 2-7: see that for each \(r_1 \in \mathbb{C}\), \(\langle r_1 v_1, v_3 \rangle = r_1 \langle v_1, v_3 \rangle\); Step 2-8: see that for each \(r_1, r_1 \in F\), \(\langle r_1 v_1 + r_2 v_2, v_3 \rangle = r_1 \langle v_1, v_3 \rangle + r_2 \langle v_1, v_3 \rangle\).
Let \(v_1, v_2, v_3 \in V\) and \(r_2, r_2 \in F\) be any, hereafter.
Step 2-1:
Let us see that \(0 \le \langle v_1, v_1 \rangle\) and \((0 = \langle v_1, v_1 \rangle \iff v_1 = 0)\).
\(\langle v_1, v_1 \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert v_1 + i^j v_1 \Vert^2 = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert (1 + i^j) v_1 \Vert^2 = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \vert 1 + i^j \vert^2 \Vert v_1 \Vert^2 = 1 / 4 \Vert v_1 \Vert^2 \sum_{j \in \{0, 1, 2, 3\}} i^j \vert 1 + i^j \vert^2 = 1 / 4 \Vert v_1 \Vert^2 (i^0 \vert 1 + i^0 \vert^2 + i^1 \vert 1 + i^1 \vert^2 + i^2 \vert 1 + i^2 \vert^2 + i^3 \vert 1 + i^3 \vert^2) = 1 / 4 \Vert v_1 \Vert^2 (\vert 1 + 1 \vert^2 + i \vert 1 + i \vert^2 -1 \vert 1 - 1 \vert^2 - i \vert 1 - i \vert^2) = 1 / 4 \Vert v_1 \Vert^2 (\vert 2 \vert^2) = \Vert v_1 \Vert^2\), which is non-negative.
If \(0 = \langle v_1, v_1 \rangle\), \(0 = \langle v_1, v_1 \rangle = \Vert v_1 \Vert^2\) by the previous paragraph, which implies that \(v_1 = 0\).
If \(v_1 = 0\), \(\langle v_1, v_1 \rangle = \Vert v_1 \Vert^2 = 0\).
Step 2-2:
Let us see that \(\langle v_1, v_2 \rangle = \overline{\langle v_2, v_1 \rangle}\).
Note that for any \(v \in V\), \(\Vert v \Vert = \vert i \vert \Vert v \Vert = \Vert i v \Vert\) and \(\Vert v \Vert = \vert -1 \vert \Vert v \Vert = \Vert - v \Vert\).
\(\langle v_1, v_2 \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert v_1 + i^j v_2 \Vert^2 = 1 / 4 (i^0 \Vert v_1 + i^0 v_2 \Vert^2 + i^1 \Vert v_1 + i^1 v_2 \Vert^2 + i^2 \Vert v_1 + i^2 v_2 \Vert^2 + i^3 \Vert v_1 + i^3 v_2 \Vert^2) = 1 / 4 (\Vert v_1 + v_2 \Vert^2 + i \Vert v_1 + i v_2 \Vert^2 - \Vert v_1 - v_2 \Vert^2 - i \Vert v_1 - i v_2 \Vert^2) = 1 / 4 (\Vert v_1 + v_2 \Vert^2 + i \Vert i (v_1 + i v_2) \Vert^2 - \Vert - (v_1 - v_2) \Vert^2 - i \Vert i (v_1 - i v_2) \Vert^2) = 1 / 4 (\Vert v_2 + v_1 \Vert^2 + i \Vert - v_2 + i v_1 \Vert^2 - \Vert v_2 - v_1 \Vert^2 - i \Vert v_2 + i v_1 \Vert^2) = \overline{1 / 4 (\Vert v_2 + v_1 \Vert^2 - i \Vert v_2 - i v_1 \Vert^2 - \Vert v_2 - v_1 \Vert^2 + i \Vert v_2 + i v_1 \Vert^2)} = \overline{1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert v_2 + i^j v_1 \Vert^2} = \overline{\langle v_2, v_1 \rangle}\).
Step 2-3:
Let us see that \(\langle v_1 + v_2, v_3 \rangle = \langle v_1, v_3 \rangle + \langle v_2, v_3 \rangle\).
As \(\langle v_1 + v_2, v_3 \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert v_1 + v_2 + i^j v_3 \Vert^2 = 1 / 4 (i^0 \Vert v_1 + v_2 + i^0 v_3 \Vert^2 + i^1 \Vert v_1 + v_2 + i^1 v_3 \Vert^2 + i^2 \Vert v_1 + v_2 + i^2 v_3 \Vert^2 + i^3 \Vert v_1 + v_2 + i^3 v_3 \Vert^2) = 1 / 4 (\Vert v_1 + v_2 + v_3 \Vert^2 + i \Vert v_1 + v_2 + i v_3 \Vert^2 - \Vert v_1 + v_2 - v_3 \Vert^2 - i \Vert v_1 + v_2 - i v_3 \Vert^2)\), let us massage each term in the parentheses (we will use the parallelogram law here).
\(\Vert v_1 + v_2 + v_3 \Vert^2 = \Vert (v_1 + v_3) + v_2 \Vert^2 = 2 \Vert v_1 + v_3 \Vert^2 + 2 \Vert v_2 \Vert^2 - \Vert (v_1 + v_3) - v_2 \Vert^2\); \(\Vert v_1 + v_2 + v_3 \Vert^2 = 2 \Vert v_2 + v_3 \Vert^2 + 2 \Vert v_1 \Vert^2 - \Vert (v_2 + v_3) - v_1 \Vert^2\), likewise; so, \(\Vert v_1 + v_2 + v_3 \Vert^2 = 1 / 2 (\Vert v_1 + v_2 + v_3 \Vert^2 + \Vert v_1 + v_2 + v_3 \Vert^2) = 1 / 2 (2 \Vert v_1 + v_3 \Vert^2 + 2 \Vert v_2 \Vert^2 - \Vert (v_1 + v_3) - v_2 \Vert^2 + 2 \Vert v_2 + v_3 \Vert^2 + 2 \Vert v_1 \Vert^2 - \Vert (v_2 + v_3) - v_1 \Vert^2) = \Vert v_1 + v_3 \Vert^2 + \Vert v_2 \Vert^2 - \Vert (v_1 + v_3) - v_2 \Vert^2 / 2 + \Vert v_2 + v_3 \Vert^2 + \Vert v_1 \Vert^2 - \Vert (v_2 + v_3) - v_1 \Vert^2 / 2 = \Vert v_1 + v_3 \Vert^2 + \Vert v_2 \Vert^2 - \Vert v_1 - v_2 + v_3 \Vert^2 / 2 + \Vert v_2 + v_3 \Vert^2 + \Vert v_1 \Vert^2 - \Vert v_1 - v_2 - v_3 \Vert^2 / 2\).
\(- \Vert v_1 + v_2 - v_3 \Vert^2 = - (\Vert v_1 - v_3 \Vert^2 + \Vert v_2 \Vert^2 - \Vert (v_1 - v_3) - v_2 \Vert^2 / 2 + \Vert v_2 - v_3 \Vert^2 + \Vert v_1 \Vert^2 - \Vert (v_2 - v_3) - v_1 \Vert^2 / 2) = - (\Vert v_1 - v_3 \Vert^2 + \Vert v_2 \Vert^2 - \Vert v_1 - v_2 - v_3 \Vert^2 / 2 + \Vert v_2 - v_3 \Vert^2 + \Vert v_1 \Vert^2 - \Vert v_1 - v_2 + v_3 \Vert^2 / 2)\), which is by replacing \(v_3\) in the previous paragraph with \(- v_3\) and taking the whole negative.
\(i \Vert v_1 + v_2 + i v_3 \Vert^2 = i \Vert (v_1 + i v_3) + v_2 \Vert^2 = i (2 \Vert v_1 + i v_3 \Vert^2 + 2 \Vert v_2 \Vert^2 - \Vert (v_1 + i v_3) - v_2 \Vert^2)\); \(i \Vert v_1 + v_2 + i v_3 \Vert^2 = i (2 \Vert v_2 + i v_3 \Vert^2 + 2 \Vert v_1 \Vert^2 - \Vert (v_2 + i v_3) - v_1 \Vert^2)\)\), likewise; \(i \Vert v_1 + v_2 + i v_3 \Vert^2 = 1 / 2 (i \Vert v_1 + v_2 + i v_3 \Vert^2 + i \Vert v_1 + v_2 + i v_3 \Vert^2) = 1 / 2 (i (2 \Vert v_1 + i v_3 \Vert^2 + 2 \Vert v_2 \Vert^2 - \Vert (v_1 + i v_3) - v_2 \Vert^2) + i (2 \Vert v_2 + i v_3 \Vert^2 + 2 \Vert v_1 \Vert^2 - \Vert (v_2 + i v_3) - v_1 \Vert^2)) = i (\Vert v_1 + i v_3 \Vert^2 + \Vert v_2 \Vert^2 - \Vert (v_1 + i v_3) - v_2 \Vert^2 / 2 + \Vert v_2 + i v_3 \Vert^2 + \Vert v_1 \Vert^2 - \Vert (v_2 + i v_3) - v_1 \Vert^2 / 2) = i (\Vert v_1 + i v_3 \Vert^2 + \Vert v_2 \Vert^2 - \Vert v_1 - v_2 + i v_3 \Vert^2 / 2 + \Vert v_2 + i v_3 \Vert^2 + \Vert v_1 \Vert^2 - \Vert v_1 - v_2 - i v_3 \Vert^2 / 2)\).
\(- i \Vert v_1 + v_2 - i v_3 \Vert^2 = - i (\Vert v_1 - i v_3 \Vert^2 + \Vert v_2 \Vert^2 - \Vert (v_1 - i v_3) - v_2 \Vert^2 / 2 + \Vert v_2 - i v_3 \Vert^2 + \Vert v_1 \Vert^2 - \Vert (v_2 - i v_3) - v_1 \Vert^2 / 2) = - i (\Vert v_1 - i v_3 \Vert^2 + \Vert v_2 \Vert^2 - \Vert v_1 - v_2 - i v_3 \Vert^2 / 2 + \Vert v_2 - i v_3 \Vert^2 + \Vert v_1 \Vert^2 - \Vert v_1 - v_2 + i v_3 \Vert^2 / 2)\), which is by replacing \(v_3\) in the previous paragraph with \(- v_3\) and taking the whole negative
So, \(\langle v_1 + v_2, v_3 \rangle = 1 / 4 (\Vert v_1 + v_3 \Vert^2 + \Vert v_2 \Vert^2 - \Vert v_1 - v_2 + v_3 \Vert^2 / 2 + \Vert v_2 + v_3 \Vert^2 + \Vert v_1 \Vert^2 - \Vert v_1 - v_2 - v_3 \Vert^2 / 2 - (\Vert v_1 - v_3 \Vert^2 + \Vert v_2 \Vert^2 - \Vert v_1 - v_2 - v_3 \Vert^2 / 2 + \Vert v_2 - v_3 \Vert^2 + \Vert v_1 \Vert^2 - \Vert v_1 - v_2 + v_3 \Vert^2 / 2) + i (\Vert v_1 + i v_3 \Vert^2 + \Vert v_2 \Vert^2 - \Vert v_1 - v_2 + i v_3 \Vert^2 / 2 + \Vert v_2 + i v_3 \Vert^2 + \Vert v_1 \Vert^2 - \Vert v_1 - v_2 - i v_3 \Vert^2 / 2) - i (\Vert v_1 - i v_3 \Vert^2 + \Vert v_2 \Vert^2 - \Vert v_1 - v_2 - i v_3 \Vert^2 / 2 + \Vert v_2 - i v_3 \Vert^2 + \Vert v_1 \Vert^2 - \Vert v_1 - v_2 + i v_3 \Vert^2)) = 1 / 4 (\Vert v_1 + v_3 \Vert^2 + \Vert v_2 + v_3 \Vert^2 - (\Vert v_1 - v_3 \Vert^2 + \Vert v_2 - v_3 \Vert^2 ) + i (\Vert v_1 + i v_3 \Vert^2 + \Vert v_2 + i v_3 \Vert^2) - i (\Vert v_1 - i v_3 \Vert^2 + \Vert v_2 - i v_3 \Vert^2)) = 1 / 4 (\Vert v_1 + v_3 \Vert^2 + i \Vert v_1 + i v_3 \Vert^2 - \Vert v_1 - v_3 \Vert^2 - i \Vert v_1 - i v_3 \Vert^2) + 1 / 4 (\Vert v_2 + v_3 \Vert^2 + i \Vert v_2 + i v_3 \Vert^2 - \Vert v_2 - v_3 \Vert^2 - i \Vert v_2 - i v_3 \Vert^2) = \langle v_1, v_3 \rangle + \langle v_2, v_3 \rangle\).
Step 2-4:
Let us see that for each \(r_1 \in \mathbb{Z}\), \(\langle r_1 v_1, v_3 \rangle = r_1 \langle v_1, v_3 \rangle\).
When \(r_1 = 0\), \(\langle 0 v_1, v_3 \rangle = \langle 0, v_3 \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert 0 + i^j v_3 \Vert^2 = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert i^j v_3 \Vert^2 = 1 / 4 (i^0 \Vert i^0 v_3 \Vert^2 + i^1 \Vert i^1 v_3 \Vert^2 + i^2 \Vert i^2 v_3 \Vert^2 + i^3 \Vert i^3 v_3 \Vert^2) = 1 / 4 (\Vert v_3 \Vert^2 + i \Vert i v_3 \Vert^2 - \Vert - v_3 \Vert^2 - i \Vert - i v_3 \Vert^2) = 0 = 0 \langle v_1, v_3 \rangle\).
When \(r_1\) is positive, \(\langle r_1 v_1, v_3 \rangle = \langle v_1 + ... + v_1, v_3 \rangle = \langle v_1, v_3 \rangle + ... + \langle v_1, v_3 \rangle\), by Step 2-3, \(= r_1 \langle v_1, v_3 \rangle\).
When \(r_1 = -1\), \(\langle r_1 v_1, v_3 \rangle = \langle - v_1, v_3 \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert - v_1 + i^j v_3 \Vert^2 = 1 / 4 (i^0 \Vert - v_1 + i^0 v_3 \Vert^2 + i^1 \Vert - v_1 + i^1 v_3 \Vert^2 + i^2 \Vert - v_1 + i^2 v_3 \Vert^2 + i^3 \Vert - v_1 + i^3 v_3 \Vert^2) = 1 / 4 (\Vert - v_1 + v_3 \Vert^2 + i \Vert - v_1 + i v_3 \Vert^2 - \Vert - v_1 - v_3 \Vert^2 - i \Vert - v_1 - i v_3 \Vert^2) = 1 / 4 (\Vert v_1 - v_3 \Vert^2 + i \Vert v_1 - i v_3 \Vert^2 - \Vert v_1 + v_3 \Vert^2 - i \Vert v_1 + i v_3 \Vert^2) = - 1 / 4 (\Vert v_1 + v_3 \Vert^2 + i \Vert v_1 + i v_3 \Vert^2 - \Vert v_1 - v_3 \Vert^2 - i \Vert v_1 - i v_3 \Vert^2) = - 1 \langle v_1, v_3 \rangle = r_1 \langle v_1, v_3 \rangle\).
When \(r_1\) is negative, \(\langle r_1 v_1, v_3 \rangle = \langle (-1) (- r_1) v_1, v_3 \rangle = -1 \langle (- r_1) v_1, v_3 \rangle = -1 (- r_1) \langle v_1, v_3 \rangle\), because \(- r_1\) is a positive integer, \(= r_1 \langle v_1, v_3 \rangle\).
Step 2-5:
Let us see that for each \(r_1 \in \mathbb{Q}\), \(\langle r_1 v_1, v_3 \rangle = r_1 \langle v_1, v_3 \rangle\).
\(r_1 = r'_1 / r''_1\) where \(r'_1, r''_1 \in \mathbb{Z}\) where \(r''_1 \neq 0\).
\(\langle r'_1 v_1, v_3 \rangle = r'_1 \langle v_1, v_3 \rangle\), by Step 2-4.
But \(\langle r'_1 v_1, v_3 \rangle = \langle r_1 r''_1 v_1, v_3 \rangle = r''_1 \langle r_1 v_1, v_3 \rangle\), by Step 2-4.
So, \(r''_1 \langle r_1 v_1, v_3 \rangle = r'_1 \langle v_1, v_3 \rangle\), so, \(\langle r_1 v_1, v_3 \rangle = r'_1 / r''_1 \langle v_1, v_3 \rangle = r_1 \langle v_1, v_3 \rangle\).
Step 2-6:
Let us see that for each \(r_1 \in \mathbb{R}\), \(\langle r_1 v_1, v_3 \rangle = r_1 \langle v_1, v_3 \rangle\).
The map, \(\langle r v_1, v_3 \rangle: \mathbb{R} \to \mathbb{C}, r \mapsto \langle r v_1, v_3 \rangle\), is the composition of \(f_1: \mathbb{R} \to V, r \mapsto r v_1\), before \(f_2: V \to \mathbb{C}, v \mapsto \langle v, v_3 \rangle\).
Let \(V\) have the topology induced by the norm.
\(f_1\) is continuous: for each \(r \in \mathbb{R}\), take any open ball around \(f_1 (r)\), \(B_{f_1 (v), \epsilon} \subseteq V\); take \(\delta := \epsilon / \Vert v \Vert\) and the open ball around \(r\), \(B_{r, \delta}\); for each \(r' \in B_{r, \delta}\), \(\Vert f_1 (r') - f_1 (r) \Vert = \Vert r' v - r v \Vert = \Vert (r' - r) v \Vert = \vert r' - r \vert \Vert v \Vert \lt \delta \Vert v \Vert = \epsilon\), which means that \(f_1 (B_{r, \delta}) \subseteq B_{f_1 (v), \epsilon}\).
\(f_2\) is continuous: \(\langle v, v_3 \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert v + i^j v_3 \Vert^2\), but the translation, \(v \mapsto v + i^j v_3\), is obviously continuous, the norm map is continuous, by the proposition that for any vectors space with any norm and the topological space induced by the metric induced by the norm, the norm map is continuous, the squaring map is continuous, multiplying \(i^j\) is continuous, and taking the sum is continuous.
So, the map, \(\langle r v_1, v_3 \rangle: \mathbb{R} \to \mathbb{C}, r \mapsto \langle r v_1, v_3 \rangle\), is continuous.
Let us take a sequence, \(s: \mathbb{N} \setminus \{0\} \to \mathbb{Q}\), that converges to \(r_1\), which is possible because for each \(j \in \mathbb{N} \setminus \{0\}\), take the open ball around \(r_1\), \(B_{r_1, 1 / j}\), and take an \(s (j) \in \mathbb{Q} \cap B_{r_1, 1 / j}\).
Then, \(\langle s (j) v_1, v_3 \rangle = s (j) \langle v_1, v_3 \rangle\), and \(lim_{j \to \infty} \langle s (j) v_1, v_3 \rangle = lim_{j \to \infty} s (j) \langle v_1, v_3 \rangle\), but \(lim_{j \to \infty} \langle s (j) v_1, v_3 \rangle = \langle lim_{j \to \infty} s (j) v_1, v_3 \rangle\), by 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, \(= \langle r_1 v_1, v_3 \rangle\), while \(lim_{j \to \infty} s (j) \langle v_1, v_3 \rangle = r_1 \langle v_1, v_3 \rangle\).
Step 2-7:
When \(F = \mathbb{R}\), we are finished for Step 2.
When \(F = \mathbb{C}\), let us see that for each \(r_1 \in \mathbb{C}\), \(\langle r_1 v_1, v_3 \rangle = r_1 \langle v_1, v_3 \rangle\).
Let \(r_1 = i\).
\(\langle r_1 v_1, v_3 \rangle = \langle i v_1, v_3 \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert i v_1 + i^j v_3 \Vert^2 = 1 / 4 (i^0 \Vert i v_1 + i^0 v_3 \Vert^2 + i^1 \Vert i v_1 + i^1 v_3 \Vert^2 + i^2 \Vert i v_1 + i^2 v_3 \Vert^2 + i^3 \Vert i v_1 + i^3 v_3 \Vert^2) = 1 / 4 (\Vert i v_1 + v_3 \Vert^2 + i \Vert i v_1 + i v_3 \Vert^2 - \Vert i v_1 - v_3 \Vert^2 - i \Vert i v_1 - i v_3 \Vert^2) = 1 / 4 (\Vert i (i v_1 + v_3) \Vert^2 + i \Vert i (i v_1 + i v_3) \Vert^2 - \Vert i (i v_1 - v_3) \Vert^2 - i \Vert i (i v_1 - i v_3) \Vert^2) = 1 / 4 (\Vert -1 v_1 + i v_3 \Vert^2 + i \Vert - v_1 - v_3 \Vert^2 - \Vert - v_1 - i v_3 \Vert^2 - i \Vert - v_1 + v_3 \Vert^2) = 1 / 4 (\Vert v_1 - i v_3 \Vert^2 + i \Vert v_1 + v_3 \Vert^2 - \Vert v_1 + i v_3 \Vert^2 - i \Vert v_1 - v_3 \Vert^2) = i (-i) 1 / 4 (\Vert v_1 - i v_3 \Vert^2 + i \Vert v_1 + v_3 \Vert^2 - \Vert v_1 + i v_3 \Vert^2 - i \Vert v_1 - v_3 \Vert^2) = i 1 / 4 (- i \Vert v_1 - i v_3 \Vert^2 + \Vert v_1 + v_3 \Vert^2 + i \Vert v_1 + i v_3 \Vert^2 - \Vert v_1 - v_3 \Vert^2) = i \langle v_1, v_3 \rangle\).
Let \(r_1 = r'_1 + r''_1 i\) where \(r'_1, r''_1 \in \mathbb{R}\).
\(\langle r_1 v_1, v_3 \rangle = \langle (r'_1 + r''_1 i) v_1, v_3 \rangle = \langle r'_1 v_1 + r''_1 i v_1, v_3 \rangle = \langle r'_1 v_1, v_3 \rangle + \langle r''_1 i v_1, v_3 \rangle\), by Step 2-3, \(= r'_1 \langle v_1, v_3 \rangle + r''_1 \langle i v_1, v_3 \rangle\), by Step 2-6, \(= r'_1 \langle v_1, v_3 \rangle + r''_1 i \langle v_1, v_3 \rangle = (r'_1 + r''_1 i) \langle v_1, v_3 \rangle = r_1 \langle v_1, v_3 \rangle\).
Step 2-8:
Let us see that for each \(r_1, r_1 \in F\), \(\langle r_1 v_1 + r_2 v_2, v_3 \rangle = r_1 \langle v_1, v_3 \rangle + r_2 \langle v_1, v_3 \rangle\).
\(\langle r_1 v_1 + r_2 v_2, v_3 \rangle = \langle r_1 v_1, v_3 \rangle + \langle r_2 v_2, v_3 \rangle\), by Step 2-3, \(= r_1 \langle v_1, v_3 \rangle + r_2 \langle v_2, v_3 \rangle\), by Step 2-7.
So, \(\langle \bullet, \bullet \rangle\) is an inner product.
Step 3:
Let us see that \(\Vert \bullet \Vert\) is induced by \(\langle \bullet, \bullet \rangle\).
For each \(v \in V\), \(\langle v, v \rangle = 1 / 4 \sum_{j \in \{0, 1, 2, 3\}} i^j \Vert v + i^j v \Vert^2 = 1 / 4 (i^0 \Vert v + i^0 v \Vert^2 + i^1 \Vert v + i^1 v \Vert^2 + i^2 \Vert v + i^2 v \Vert^2 + i^3 \Vert v + i^3 v \Vert^2) = 1 / 4 (\Vert v + v \Vert^2 + i \Vert v + i v \Vert^2 - \Vert v - v \Vert^2 - i \Vert v - i v \Vert^2) = 1 / 4 (\Vert 2 v \Vert^2 + i \Vert (1 + i) v \Vert^2 - \Vert 0 \Vert^2 - i \Vert (1 - i) v \Vert^2) = 1 / 4 (4 \Vert v \Vert^2 + i \vert 1 + i \vert^2 \Vert v \Vert^2 - i \vert 1 - i \vert^2 \Vert v \Vert^2) = \Vert v \Vert^2\).
Step 4:
\(\langle \bullet, \bullet \rangle\) is the unique inner product by which \(\Vert \bullet \Vert\) is induced, by the proposition that for any normed vectors space, if the norm is induced by any inner product, the inner product that induces the norm is unique as this.
