1048: For Real or Complex Vectors Space with Topology Induced by Metric Induced by Norm Induced by Inner Product, Inner Product with 1 Argument Fixed Is Continuous Map

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description/proof of that for real or complex vectors space with topology induced by metric induced by norm induced by inner product, inner product with 1 argument fixed is continuous map

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Topics

About: vectors space
About: topological space

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The table of contents of this article

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

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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 metric induced by norm on real or complex vectors space.
• The reader knows a definition of topology induced by metric.
• The reader admits the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space.

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Target Context

• The reader will have a description and a proof of the proposition that for any real or complex vectors space with the topology induced by the metric induced by the norm induced by any inner product, the inner product with any 1 argument fixed is a continuous map.

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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
$V$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$⟨\bullet ,\bullet ⟩$: $:V\times V\to F$, $\in \{\text{ the inner products }\}$
$\Vert \bullet \Vert$: $:V\to \mathbb{R}$, $=\text{ the norm induced by }⟨\bullet ,\bullet ⟩$
$dist$: $:V\times V\to \mathbb{R}$, $=\text{ the metric induced by }\Vert \bullet \Vert$
$O$: $=\text{ the topology induced by }dist$
$(V,O)$: $=\text{ the topological space }$
$w$: $\in V$
$f_1$: $:(V,O)\to F,v\mapsto ⟨v,w⟩$
$f_2$: $:(V,O)\to F,v\mapsto ⟨w,u⟩$
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Statements:
$f_1\in \{\text{ the continuous maps }\}$
$\wedge$
$f_2\in \{\text{ the continuous maps }\}$
//

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3: Proof

Whole Strategy: use the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space; Step 1: for $f_1$, for any $v\in V$, take any $ϵ$-'open ball' around $f_1(v)$, $B_{f_1(v),ϵ}$, and take a $\delta$-'open ball' around $v$, $B_{v,\delta }$, such that $f_1(B_{v,\delta })\subseteq B_{f_1(v),ϵ}$; Step 2: for $f_2$, for any $v\in V$, take any $ϵ$-'open ball' around $f_2(v)$, $B_{f_1(v),ϵ}$, and take a $\delta$-'open ball' around $v$, $B_{v,\delta }$, such that $f_2(B_{v,\delta })\subseteq B_{f_1(v),ϵ}$.

Step 1:

Let us see that $f_1$ is continuous.

Let $v\in V$ be any.

Let $B_{f_1(v),ϵ}\subseteq F$ be the $ϵ$-'open ball' around $f_1(v)$.

Let $B_{v,\delta }$ be the $\delta$-'open ball' around $v$.

For each $v^{\prime }\in B_{v,\delta }$, $dist(v,v^{\prime })=\Vert v-v^{\prime }\Vert =\sqrt{⟨v-v^{\prime },v-v^{\prime }⟩}<\delta$.

$f_1(v^{\prime })-f_1(v)=⟨v^{\prime },w⟩-⟨v,w⟩=⟨v-v^{\prime },w⟩$.

By the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space, $|⟨v-v^{\prime },w⟩|\le \sqrt{⟨v-v^{\prime },v-v^{\prime }⟩}\sqrt{⟨w,w⟩}<\delta \sqrt{⟨w,w⟩}$.

So, by taking $\delta =ϵ/\sqrt{⟨w,w⟩}$, $|f_1(v^{\prime })-f_1(v)|=|⟨v-v^{\prime },w⟩|<ϵ/\sqrt{⟨w,w⟩}\sqrt{⟨w,w⟩}=ϵ$, which means that $f_1(B_{v,\delta })\subseteq B_{f_1(v),ϵ}$.

So, $f_1$ is continuous.

Step 2:

Let us see that $f_2$ is continuous.

Let $v\in V$ be any.

Let $B_{f_2(v),ϵ}\subseteq F$ be the $ϵ$-'open ball' around $f_2(v)$.

Let $B_{v,\delta }$ be the $\delta$-'open ball' around $v$.

For each $v^{\prime }\in B_{v,\delta }$, $dist(v,v^{\prime })=\Vert v-v^{\prime }\Vert =\sqrt{⟨v-v^{\prime },v-v^{\prime }⟩}<\delta$.

$f_2(v^{\prime })-f_2(v)=⟨w,v^{\prime }⟩-⟨w,v⟩=⟨w,v-v^{\prime }⟩$.

By the Cauchy-Schwarz inequality for any real or complex inner-producted vectors space, $|⟨w,v-v^{\prime }⟩|\le \sqrt{⟨w,w⟩}\sqrt{⟨v-v^{\prime },v-v^{\prime }⟩}<\sqrt{⟨w,w⟩}\delta$.

So, by taking $\delta =ϵ/\sqrt{⟨w,w⟩}$, $|f_2(v^{\prime })-f_2(v)|=|⟨w,v-v^{\prime }⟩|<\sqrt{⟨w,w⟩}ϵ/\sqrt{⟨w,w⟩}=ϵ$, which means that $f_2(B_{v,\delta })\subseteq B_{f_2(v),ϵ}$.

So, $f_2$ is continuous.

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References

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