Bounded in Norm .... Garling, Section 11.2: Normed Spaces ....

In summary: E$ is both a metric space and a vector space; the metric and the norm are related by$$d(x,y)\ =\ \|x-y\|$$for $x,y\in E$.So if $B\subset E$ is nonempty – say it contains $c$ – and bounded, we have for all $b\in B$,$$\|b\|\ \le\ \|b-c\|+\|c\|\ \le\ \text{diam}(B)+\|c\|$$$\displaystyle\implies\ \sup_{b\in B}\|b\|<
  • #1
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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 with some remarks by Garling concerning a subset being norm bounded of bounded in norm ...

The particular remarks by Garling read as follows:View attachment 8950
Note that the definition of a bounded set by Garling is included in the following text:
View attachment 8951
In the remarks by Garling above we read the following:

" ... ... Then since

\(\displaystyle \mid \mid x - y \mid \mid \le \mid \mid x \mid \mid + \mid \mid y \mid \mid\) and \(\displaystyle \mid \mid y \mid \mid \le \mid \mid y - x \mid \mid + \mid \mid x \mid \mid\)

a subset \(\displaystyle B\) is bounded if and only if \(\displaystyle \text{sup } \{ \mid \mid b \mid \mid \ : \ b \in B \} \lt \infty\) ... ... ... "Can someone please explain/demonstrate how (given Garling;s definition of a bounded subset) that the statements:

\(\displaystyle \mid \mid x - y \mid \mid \le \mid \mid x \mid \mid + \mid \mid y \mid \mid\) and \(\displaystyle \mid \mid y \mid \mid \le \mid \mid y - x \mid \mid + \mid \mid x \mid \mid\)

lead to the statement that:

a subset \(\displaystyle B\) is bounded if and only if \(\displaystyle \text{sup } \{ \mid \mid b \mid \mid \ : \ b \in B \} \lt \infty\) ... ... ... ?Hope someone can help ...

Peter

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It may help some readers of the above post to have access to the start of Garling's section on normed spaces in order to familiarize them with Garling's approach and notation ... so I am providing the same ... as follows:View attachment 8952Hope that helps ...

Peter
 

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  • #2
$E$ is both a metric space and a vector space; the metric and the norm are related by
$$d(x,y)\ =\ \|x-y\|$$
for $x,y\in E$.

So if $B\subset E$ is nonempty – say it contains $c$ – and bounded, we have for all $b\in B$,
$$\|b\|\ \le\ \|b-c\|+\|c\|\ \le\ \text{diam}(B)+\|c\|$$
$\displaystyle\implies\ \sup_{b\in B}\|b\|<\infty$ (since $\text{diam}(B)<\infty$ and $\|c\|$ is a fixed number). Conversely, if $\displaystyle s=\sup_{b\in B}\|b\|<\infty$, then for all $b,b'\in B$,
$$d(b,b')\ =\ \|b-b'\|\ \le\ \|b\|+\|b'\|\ \le\ 2s$$
$\implies\ \text{diam}(B)<\infty$.
 
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  • #3
Olinguito said:
$E$ is both a metric space and a vector space; the metric and the norm are related by
$$d(x,y)\ =\ \|x-y\|$$
for $x,y\in E$.

So if $B\subset E$ is nonempty – say it contains $c$ – and bounded, we have for all $b\in B$,
$$\|b\|\ \le\ \|b-c\|+\|c\|\ \le\ \text{diam}(B)+\|b'\|$$
$\displaystyle\implies\ \sup_{b\in B}\|b\|<\infty$ (since $\text{diam}(B)<\infty$ and $\|c\|$ is a fixed number). Conversely, if $\displaystyle s=\sup_{b\in B}\|b\|<\infty$, then for all $b,b'\in B$,
$$d(b,b')\ =\ \|b-b'\|\ \le\ \|b\|+\|b'\|\ \le\ 2s$$
$\implies\ \text{diam}(B)<\infty$.
Thanks for the help Olinguito ...

But ... just a clarification ...

You write: $$\|b\|\ \le\ \|b-c\|+\|c\|\ \le\ \text{diam}(B)+\|b'\|$$

Did you mean
$$\|b\|\ \le\ \|b-c\|+\|c\|\ \le\ \text{diam}(B)+\|c\|$$ ... ... ?
Thanks again ...

Peter
 
  • #4
Yes, I did. I’ve fixed the post now.
 
  • #5
Olinguito said:
Yes, I did. I’ve fixed the post now.
Thanks again for your help ...

Peter
 
  • #6
Olinguito said:
$E$ is both a metric space and a vector space; the metric and the norm are related by
$$d(x,y)\ =\ \|x-y\|$$
for $x,y\in E$.

So if $B\subset E$ is nonempty – say it contains $c$ – and bounded, we have for all $b\in B$,
$$\|b\|\ \le\ \|b-c\|+\|c\|\ \le\ \text{diam}(B)+\|c\|$$
$\displaystyle\implies\ \sup_{b\in B}\|b\|<\infty$ (since $\text{diam}(B)<\infty$ and $\|c\|$ is a fixed number). Conversely, if $\displaystyle s=\sup_{b\in B}\|b\|<\infty$, then for all $b,b'\in B$,
$$d(b,b')\ =\ \|b-b'\|\ \le\ \|b\|+\|b'\|\ \le\ 2s$$
$\implies\ \text{diam}(B)<\infty$.

Just returned to this thread with another question ...

It has been shown above that \(\displaystyle \text{diam} (B) \lt \infty\) follows from \(\displaystyle \mid \mid y \mid \mid \le \mid \mid y - x \mid \mid + \mid \mid x \mid \mid\) alone ...

... so why does Garling mention \(\displaystyle \mid \mid x - y \mid \mid \le \mid \mid x \mid \mid + \mid \mid y \mid \mid\) Help will be appreciated ...

Peter*** EDIT ***

Oh! The other inequality is needed for the converse ... sorry ... should have read your post more carefully ...

Peter
 
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FAQ: Bounded in Norm .... Garling, Section 11.2: Normed Spaces ....

1. What is a normed space?

A normed space is a mathematical concept that refers to a vector space equipped with a norm, which is a function that assigns a positive length or size to each vector in the space. Normed spaces are important in functional analysis and are used to study the convergence of sequences and the continuity of functions.

2. How is a norm defined in a normed space?

In a normed space, a norm is defined as a function that satisfies three properties: non-negativity, homogeneity, and the triangle inequality. Non-negativity means that the norm of a vector is always greater than or equal to zero. Homogeneity means that the norm of a scalar multiple of a vector is equal to the absolute value of that scalar multiplied by the norm of the vector. The triangle inequality states that the norm of the sum of two vectors is less than or equal to the sum of their individual norms.

3. What is the purpose of a bounded set in a normed space?

A bounded set in a normed space is a set of vectors whose norms are all less than or equal to a certain value. Bounded sets are important because they allow us to define the concept of convergence in a normed space. If a set is bounded, then any sequence of vectors in that set will also be bounded, which is necessary for the concept of convergence to make sense.

4. What is the difference between a bounded set and a bounded sequence in a normed space?

A bounded set is a set of vectors whose norms are all less than or equal to a certain value, while a bounded sequence is a sequence of vectors whose norms are all less than or equal to a certain value, regardless of the number of vectors in the sequence. In other words, a bounded set is a static collection of vectors, while a bounded sequence is a dynamic list of vectors.

5. How are normed spaces used in real-world applications?

Normed spaces are used in a variety of real-world applications, including physics, engineering, and economics. They are particularly useful in optimization problems, where the norm can be used to measure the distance between a solution and the optimal solution. Normed spaces are also used in the study of differential equations, where they can be used to define the concept of a solution and determine the convergence of numerical methods.

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