# Normed and Inner Product Spaces .... Garling, Corollary 11.3.2 ....

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In summary, Corollary 11.3.2 states that for two vectors x and y in a normed space, the condition that \mathscr{R} \ \langle x, y \rangle = \| x \| \cdot \| y \| is equivalent to either y = 0 or x = \alpha y with \alpha \ge 0. This can be rigorously demonstrated by considering the two cases, y = 0 and y \neq 0, and using the fact that the Cauchy-Schwarz Inequality only allows equality when x and y are linearly dependent. I hope this helps clarify things for you!
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I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help to fully understand the proof of Corollary 11.3.2 ... (Corollary to the Cauchy-Schwarz Inequality ... )

Garling's statement and proof of Corollary 11.3.2 reads as follows:View attachment 8960
View attachment 8961In the above text from Garling we read the following:

" ... ... Equality holds if and only if $$\displaystyle \mathscr{R} \ \langle x, y \rangle = \| x \| . \| y \|$$, which is equivalent to the condition stated. ... ... "
Can someone please rigorously demonstrate that the condition that ... $$\displaystyle \mathscr{R} \ \langle x, y \rangle = \| x \| . \| y \|$$ ...

... is equivalent to ... either $$\displaystyle y = 0$$ or $$\displaystyle x = \alpha y$$ with $$\displaystyle \alpha \ge 0$$ ... ...Help will be appreciated ...

Peter

#### Attachments

• Garling - 1 - Corollary 11.3.2 ... ... PART 1 ... .png
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• Garling - 2 - Corollary 11.3.2 ... ... PART 2 ... .png
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Peter said:
Can someone please rigorously demonstrate that the condition that ... $$\displaystyle \mathscr{R} \ \langle x, y \rangle = \| x \| . \| y \|$$ ...

... is equivalent to ... either $$\displaystyle y = 0$$ or $$\displaystyle x = \alpha y$$ with $$\displaystyle \alpha \ge 0$$ ... ...
${\frak R}\langle x,y\rangle \leqslant |\langle x,y\rangle| \leqslant \|x\|\|y\|$, and (by Cauchy-Schwarz) equality can only hold in the second of those inequalities if $x$ and $y$ are linearly dependent. In that case, either $y$ is zero or $x$ must be a scalar multiple of $y$.

To see that $\alpha \geqslant0$, if $x = \alpha y$ then $\langle x,y\rangle = \langle \alpha y,y\rangle = \alpha \|y\|^2$. Also, $\|x\|\|y\| = |\alpha|\|y\|^2$. So if ${\frak R}\langle x,y\rangle = \|x\|\|y\|$ (and $y\ne0$) it follows that ${\frak R}\alpha = |\alpha|$. For a complex number $\alpha$, that implies that $\alpha$ is real and non-negative.

Sure, I can try to help explain this for you. First, let's break down the statement and proof of Corollary 11.3.2. The corollary is a result that follows from the Cauchy-Schwarz Inequality, which states that for any two vectors x and y in a normed space, we have:

| \langle x, y \rangle | \le \| x \| \cdot \| y \|

with equality holding if and only if x and y are linearly dependent. This means that either x = \alpha y or y = \alpha x for some scalar \alpha.

Now, let's look at Garling's statement and proof. He says that equality in the Cauchy-Schwarz Inequality holds if and only if \mathscr{R} \ \langle x, y \rangle = \| x \| \cdot \| y \|, which is equivalent to the condition stated. This condition is that either y = 0 or x = \alpha y with \alpha \ge 0.

To see why this is true, we can consider two cases.

Case 1: y = 0. In this case, it is clear that \mathscr{R} \ \langle x, y \rangle = \| x \| \cdot \| y \|, since both sides are equal to 0.

Case 2: y \neq 0. In this case, we can write y as a scalar multiple of x, i.e. y = \alpha x for some scalar \alpha. Then, we have:

\mathscr{R} \ \langle x, y \rangle = \mathscr{R} \ \langle x, \alpha x \rangle = \alpha \mathscr{R} \ \langle x, x \rangle = \alpha \| x \|^2

On the other hand, \| x \| \cdot \| y \| = \| x \| \cdot \| \alpha x \| = \| x \| \cdot | \alpha | \| x \| = | \alpha | \| x \|^2

Since x \neq 0 (otherwise y would be 0), we have \| x \| \neq 0, which means that | \alpha | = \alpha. Therefore, we have:

\mathscr{R} \ \langle x, y \rangle = \| x \| \cdot \| y \| if and

## 1. What is a normed space?

A normed space is a mathematical concept that refers to a vector space where each vector has a corresponding non-negative length or magnitude. This length is defined by a function called a norm, which satisfies certain properties such as the triangle inequality.

## 2. What is an inner product space?

An inner product space is a vector space equipped with an additional structure called an inner product. This inner product is a function that takes in two vectors and outputs a scalar, and it satisfies properties such as linearity and symmetry. Inner product spaces are used to define notions of length and angle in a vector space.

## 3. What is the significance of Garling, Corollary 11.3.2 in normed and inner product spaces?

Garling, Corollary 11.3.2 is a result in functional analysis that states that every normed space can be embedded into an inner product space. This is significant because it allows us to use the additional structure of inner product spaces to study and analyze normed spaces.

## 4. How are normed and inner product spaces related?

Normed and inner product spaces are closely related mathematical structures. In fact, every inner product space is also a normed space, but not every normed space is an inner product space. This means that inner product spaces have additional structure and properties that normed spaces do not necessarily have.

## 5. What are some applications of normed and inner product spaces?

Normed and inner product spaces have many applications in mathematics and other fields such as physics, engineering, and computer science. They are used to study and analyze vector spaces, which have wide-ranging applications in areas such as optimization, data analysis, and quantum mechanics.

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