description/proof of that linear map from finite-dimensional vectors space with norm induced by inner product into normed vectors space is bounded
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 bounded map between normed vectors spaces.
- The reader admits the proposition that when the sum of any \(n\) squared non-negative numbers is equal to or smaller than any squared non-negative number, the sum of the numbers is equal to or smaller than \(n\) times the number.
Target Context
- The reader will have a description and a proof of the proposition that any linear map from any finite-dimensional vectors space with the norm induced by any inner product into any normed vectors space is bounded.
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
\(V_1\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\) with the norm, \(\Vert \bullet \Vert_1\), induced by any inner product, \(\langle \bullet, \bullet \rangle\)
\(V_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\) with any norm, \(\Vert \bullet \Vert_2\)
\(*f\): \(: V_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
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Statements:
\(f \in \{\text{ the bounded maps }\}\)
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2: Proof
Whole Strategy: Step 1: take any orthonormal basis for \(V_1\), \(\{b_1, ..., b_d\}\); Step 2: evaluate \(\Vert f (v^j b_j) \Vert_2\).
Step 1:
As \(V_1\) has the inner product, there is an orthonormal basis for \(V_1\), \(\{b_1, ..., b_d\}\): take any basis, \(\{b'_1, ..., b'_d\}\), and do the Gram - Schmidt orthonormalization: \(b_1 := 1 / \sqrt{\langle b'_1, b'_1 \rangle} b'_1\); \(b_2 := 1 / \sqrt{\langle b'_2 - \langle b'_2, b_1 \rangle b_1, b'_2 - \langle b'_2, b_1 \rangle b_1 \rangle} (b'_2 - \langle b'_2, b_1 \rangle b_1)\); \(b_3 = 1 / \sqrt{\langle b'_3 - \langle b'_3, b_1 \rangle b_1 - \langle b'_3, b_2 \rangle b_2, b'_3 - \langle b'_3, b_1 \rangle b_1 - \langle b'_3, b_2 \rangle b_2 \rangle} (b'_3 - \langle b'_3, b_1 \rangle b_1 - \langle b'_3, b_2 \rangle b_2)\); ....
Step 2:
Let \(v = v^j b_j\) be any.
\(\Vert v \Vert_1^2 = \langle v^j b_j, v^j b_j \rangle = \sum_{j \in \{1, ..., d\}} \vert v^j \vert^2\).
\(\Vert f (v^j b_j) \Vert_2 = \Vert v^j f (b_j) \Vert_2 \le \vert v^j \vert \Vert f (b_j) \Vert_2\).
Let \(c := max \{\Vert f (b_j) \Vert_2\}\).
\(\Vert f (v^j b_j) \Vert_2 \le \sum_{j \in \{1, ..., d\}} \vert v^j \vert c\).
As \(\sum_{j \in \{1, ..., d\}} \vert v^j \vert^2 \le \Vert v \Vert_1^2\), \(\sum_{j \in \{1, ..., d\}} \vert v^j \vert \le d \Vert v \Vert_1\), by the proposition that when the sum of any \(n\) squared non-negative numbers is equal to or smaller than any squared non-negative number, the sum of the numbers is equal to or smaller than \(n\) times the number.
So, \(\Vert f (v^j b_j) \Vert_2 \le d \Vert v \Vert_1 c = d c \Vert v \Vert_1\).
As \(d c\) is a constant that does not depend on \(v\), \(f\) is bounded.
3: Note
This proposition does not require \(\Vert \bullet \Vert_2\) to be induced by any inner product, as Proof shows.
This proposition requires \(\Vert \bullet \Vert_1\) to be induced by an inner product, because otherwise, orthonormal-ness would not make sense.
