1075: Linear Map from Finite-Dimensional Vectors Space with Norm Induced by Inner Product into Normed Vectors Space Is Bounded

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description/proof of that linear map from finite-dimensional vectors space with norm induced by inner product into normed vectors space is bounded

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof
• 3: Note

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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.

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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.

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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.

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Main Body

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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, $⟨\bullet ,\bullet ⟩$
$V_2$: $\in \{\text{ the }F\text{ vectors spaces }\}$ with any norm, $\Vert \bullet {\Vert }_2$
$\ast f$: $:V_1\to V_2$, $\in \{\text{ the linear maps }\}$
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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, $\{b_1^{\prime },...,b_d^{\prime }\}$, and do the Gram - Schmidt orthonormalization: $b_1:=1/\sqrt{⟨b_1^{\prime },b_1^{\prime }⟩}{b}_1^{\prime }$; $b_2:=1/\sqrt{⟨b_2^{\prime }-⟨b_2^{\prime },b_1⟩b_1,b_2^{\prime }-⟨b_2^{\prime },b_1⟩b_1⟩}(b_2^{\prime }-⟨b_2^{\prime },b_1⟩b_1)$; $b_3=1/\sqrt{⟨b_3^{\prime }-⟨b_3^{\prime },b_1⟩b_1-⟨b_3^{\prime },b_2⟩b_2,b_3^{\prime }-⟨b_3^{\prime },b_1⟩b_1-⟨b_3^{\prime },b_2⟩b_2⟩}(b_3^{\prime }-⟨b_3^{\prime },b_1⟩b_1-⟨b_3^{\prime },b_2⟩b_2)$; ....

Step 2:

Let $v=v^j{b}_j$ be any.

$\Vert v{\Vert }_1^2=⟨v^j{b}_j,v^j{b}_j⟩=\sum _{j\in \{1,...,d\}}|v^j{|}^2$.

$\Vert f(v^j{b}_j){\Vert }_2=\Vert v^{j}f(b_j){\Vert }_2\le |v^j|\Vert f(b_j){\Vert }_2$.

Let $c:=max\{\Vert f(b_j){\Vert }_2\}$.

$\Vert f(v^j{b}_j){\Vert }_2\le \sum _{j\in \{1,...,d\}}|v^j|c$.

As $\sum _{j\in \{1,...,d\}}|v^j{|}^2\le \Vert v{\Vert }_1^2$, $\sum _{j\in \{1,...,d\}}|v^j|\le d\Vert v{\Vert }_1$, 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, $\Vert f(v^j{b}_j){\Vert }_2\le d\Vert v{\Vert }_{1}c=dc\Vert v{\Vert }_1$.

As $dc$ is a constant that does not depend on $v$, $f$ is bounded.

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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.

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References

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