Uniform norm .... Garling Section 11.2 Normed Saces .... also Example 11.5.7

In summary: So for example $l^2_2(\Bbb{R})$ is just two-dimensional space with the euclidean distance $\|(x,y)\| = \sqrt{x^2+y^2}$ as the norm.On the face of it the normed space l_\infty( \{ 1, \ ... \ ... \ , d \} ) seems to me to be the set of all functions from the set \{ 1, \ ... \ ... \ , d \} \to \mathbb{R} or \to \mathbb{C} ... does that mean that say that f_1 picks a vector in \mathbb{R}^
  • #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 in order to understand some remarks by Garling on the uniform norm made in Section 11.2 on page 311 and references made and notation used in Example 11.5.7 ... ...

The remarks by Garling made in Section 11.2 on page 311 ... ... read as follows:View attachment 9002... and Example 11.5.7 reads as follows ... ...View attachment 9003In the above remarks by Garling we read the following:

" ... ... We denote \(\displaystyle l_\infty ( \mathbb{N} )\) ( or \(\displaystyle l_\infty ( \mathbb{Z}^{ + } )\) ) by \(\displaystyle l_\infty\), and \(\displaystyle l_\infty( \{ 1, \ ... \ ... \ , d \} )\) by \(\displaystyle l^d_\infty\). ... ... "

My issue/problem is that I'm not certain of the nature of \(\displaystyle l_\infty( \{ 1, \ ... \ ... \ , d \} )\) or \(\displaystyle l^d_\infty\) ... could someone please explain the nature of the normed space \(\displaystyle l_\infty( \{ 1, \ ... \ ... \ , d \} )\) ... ...

On the face of it the normed space \(\displaystyle l_\infty( \{ 1, \ ... \ ... \ , d \} )\) seems to me to be the set of all functions from the set \(\displaystyle \{ 1, \ ... \ ... \ , d \} \to \mathbb{R}\) or \(\displaystyle \to \mathbb{C}\) ... does that mean that say that \(\displaystyle f_1\) picks a vector in \(\displaystyle \mathbb{R}^d\) or \(\displaystyle \mathbb{C}^d\) and \(\displaystyle f_2\) picks (or maps to) a different vector etc ...

My problem in Example 11.5.7 is to make sense of \(\displaystyle l^2_1 ( \mathbb{R} )\) and \(\displaystyle l^2_\infty ( \mathbb{R} )\) ... can someone carefully explain the nature of these spaces ... and then explain the mapping \(\displaystyle T: l^2_1 ( \mathbb{R} ) \to l^2_\infty ( \mathbb{R} )\) ...Hope someone can help ...

Peter==========================================================================================The above post mentions Proposition 11.1.11 ... so I am providing access to the same plus some relevant preliminary remarks ... as follows:
View attachment 9004
View attachment 9005

Hope that helps ... ...

Peter
 

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  • #2
Peter said:
My issue/problem is that I'm not certain of the nature of \(\displaystyle l_\infty( \{ 1, \ ... \ ... \ , d \} )\) or \(\displaystyle l^d_\infty\) ... could someone please explain the nature of the normed space \(\displaystyle l_\infty( \{ 1, \ ... \ ... \ , d \} )\) ... ...
In the notation $l^d_\nu$, the upper index $d$ denotes the dimension of the space. So $l^2$ just means $\Bbb{R}^2$ or $\Bbb{C}^2$, depending on whether the real or complex scalars are being used. The lower index $\nu$ (usually $1$, $2$ or $\infty$) tells you the norm that is being used. So for example $l^2_2(\Bbb{R})$ is just two-dimensional space with the euclidean distance $\|(x,y)\| = \sqrt{x^2+y^2}$ as the norm.

Peter said:
On the face of it the normed space \(\displaystyle l_\infty( \{ 1, \ ... \ ... \ , d \} )\) seems to me to be the set of all functions from the set \(\displaystyle \{ 1, \ ... \ ... \ , d \} \to \mathbb{R}\) or \(\displaystyle \to \mathbb{C}\) ... does that mean that say that \(\displaystyle f_1\) picks a vector in \(\displaystyle \mathbb{R}^d\) or \(\displaystyle \mathbb{C}^d\) and \(\displaystyle f_2\) picks (or maps to) a different vector etc ...
When these spaces are described in terms of functions $f:S\to E$ (where $E = \Bbb{R}$ or $\Bbb{C}$), the set $S$ is just a set of $d$ elements, say $S = \{1,2,\ldots,d\}$. So an element of $l^2$ is a function $f$ from the set $\{1,2\}$ to $E$. If you write $f(1) = x$ and $f(2) = y$ then you can identify $f$ with $(x,y)$. In a finite-dimensional space, it may seem strange to think of points of the space as being functions. But in the case when $d = \infty$ it is natural to think of elements of $l^\infty$ as functions from the natural numbers to $E$.

Peter said:
My problem in Example 11.5.7 is to make sense of \(\displaystyle l^2_1 ( \mathbb{R} )\) and \(\displaystyle l^2_\infty ( \mathbb{R} )\) ... can someone carefully explain the nature of these spaces ... and then explain the mapping \(\displaystyle T: l^2_1 ( \mathbb{R} ) \to l^2_\infty ( \mathbb{R} )\) ...
As I explained above, both \(\displaystyle l^2_1 ( \mathbb{R} )\) and \(\displaystyle l^2_\infty ( \mathbb{R} )\) are the space $\Bbb{R}^2$, but with different norms. For $l^2_1$, the norm is given by $\|(x,y)\|_1 = |x| + |y|$. For $l^2_\infty$, it is $\|(x,y)\|_\infty = \max\{|x|, |y|\}$.

In Example 11.5.7, the map $T:l^2_1 \to l^2_\infty$ is given by $T(x,y) = (x+y,x-y)$. The norm of $(x,y)$ in $l^2_1$ is $|x|+|y|$, and the norm of $T(x,y)$ in $l^2_\infty$ is $\max\{|x+y|,|x-y|\}$. Since those two expressions are equal, it follows that $T$ is an isometry.
 
  • #3
Opalg said:
In the notation $l^d_\nu$, the upper index $d$ denotes the dimension of the space. So $l^2$ just means $\Bbb{R}^2$ or $\Bbb{C}^2$, depending on whether the real or complex scalars are being used. The lower index $\nu$ (usually $1$, $2$ or $\infty$) tells you the norm that is being used. So for example $l^2_2(\Bbb{R})$ is just two-dimensional space with the euclidean distance $\|(x,y)\| = \sqrt{x^2+y^2}$ as the norm.When these spaces are described in terms of functions $f:S\to E$ (where $E = \Bbb{R}$ or $\Bbb{C}$), the set $S$ is just a set of $d$ elements, say $S = \{1,2,\ldots,d\}$. So an element of $l^2$ is a function $f$ from the set $\{1,2\}$ to $E$. If you write $f(1) = x$ and $f(2) = y$ then you can identify $f$ with $(x,y)$. In a finite-dimensional space, it may seem strange to think of points of the space as being functions. But in the case when $d = \infty$ it is natural to think of elements of $l^\infty$ as functions from the natural numbers to $E$.As I explained above, both \(\displaystyle l^2_1 ( \mathbb{R} )\) and \(\displaystyle l^2_\infty ( \mathbb{R} )\) are the space $\Bbb{R}^2$, but with different norms. For $l^2_1$, the norm is given by $\|(x,y)\|_1 = |x| + |y|$. For $l^2_\infty$, it is $\|(x,y)\|_\infty = \max\{|x|, |y|\}$.

In Example 11.5.7, the map $T:l^2_1 \to l^2_\infty$ is given by $T(x,y) = (x+y,x-y)$. The norm of $(x,y)$ in $l^2_1$ is $|x|+|y|$, and the norm of $T(x,y)$ in $l^2_\infty$ is $\max\{|x+y|,|x-y|\}$. Since those two expressions are equal, it follows that $T$ is an isometry.
Thanks so much Opalg ... a most helpful post indeed!

But just a clarification ...

You write:

" ... ... As I explained above, both \(\displaystyle l^2_1 ( \mathbb{R} )\) and \(\displaystyle l^2_\infty ( \mathbb{R} )\) are the space $\Bbb{R}^2$, but with different norms. For $l^2_1$, the norm is given by $\|(x,y)\|_1 = |x| + |y|$. For $l^2_\infty$, it is $\|(x,y)\|_\infty = \max\{|x|, |y|\}$. ... ... "I am slightly puzzled as to how you arrived at ... ... " ... For $l^2_\infty$, it is $\|(x,y)\|_\infty = \max\{|x|, |y|\}$ ... "I am trying to match your statement about \(\displaystyle \| . \|_\infty\) with what Garling says about the norm \(\displaystyle \| . \|_\infty\) ...In Section 11.2 on page 311 of Garling we read the following:" ... ... Arguing as in Proposition 11.1.11,

\(\displaystyle \| f \|_\infty = d_\infty ( f, 0 ) = \text{sup} \{ \| f(s) \|_\infty \ : \ s \in S \}\)

... ... ... ... ... "... and then going back to Proposition 11.1.11 we note that " ... ... \(\displaystyle d_\infty ( f, g ) = \text{sup} \{ d( f(s) , g(s) ) \ : \ s \in S \}\) ... ... "Can you please explain how/why \(\displaystyle \| f \|_\infty\) becomes \(\displaystyle \max\{|x|, |y|\}\) ... I am guessing it has to do with "identifying" \(\displaystyle f\) with \(\displaystyle (x,y)\) ... and given that there are only two values sup becomes max ... but not sure ...

Hope you can help ... Just by the way I can see how it helps and why it is plausible to "identify" \(\displaystyle f\) with \(\displaystyle (x,y)\) ... but is this rigorous ... and logically/analytically the case .( that is true ... ) ... can you comment ... ?Peter
 
  • #4
Peter said:
In Section 11.2 on page 311 of Garling we read the following:

" ... ... Arguing as in Proposition 11.1.11,
\(\displaystyle \| f \|_\infty = d_\infty ( f, 0 ) = \text{sup} \{ \| f(s) \|_\infty \ : \ s \in S \}\)
... ... ... ... ... "

... and then going back to Proposition 11.1.11 we note that

" ... ... \(\displaystyle d_\infty ( f, g ) = \text{sup} \{ d( f(s) , g(s) ) \ : \ s \in S \}\) ... ... "

Can you please explain how/why \(\displaystyle \| f \|_\infty\) becomes \(\displaystyle \max\{|x|, |y|\}\) ...
Formally and rigorously, an element of \(\displaystyle l^2_\infty ( \mathbb{R} )\) is a function $f:S\to\Bbb{R}$, where $S = \{1,2\}$. The norm of $f$ is $\|f\|_\infty = \sup\{|f(s)|:s\in S\} = \max\{|f(1)|,|f(2)|\}$ (because the sup of a finite set of real numbers is just the maximum of its elements). If $f(1) = x$ and $f(2) = y$ then this becomes $\|f\|_\infty = \max\{|x|,|y|\}$.

Since the function $f$ is completely specified by its two values $f(1)$ and $f(2)$, it is convenient to identify $f$ with the ordered pair $(x,y)$.
 
  • #5
Opalg said:
Formally and rigorously, an element of \(\displaystyle l^2_\infty ( \mathbb{R} )\) is a function $f:S\to\Bbb{R}$, where $S = \{1,2\}$. The norm of $f$ is $\|f\|_\infty = \sup\{|f(s)|:s\in S\} = \max\{|f(1)|,|f(2)|\}$ (because the sup of a finite set of real numbers is just the maximum of its elements). If $f(1) = x$ and $f(2) = y$ then this becomes $\|f\|_\infty = \max\{|x|,|y|\}$.

Since the function $f$ is completely specified by its two values $f(1)$ and $f(2)$, it is convenient to identify $f$ with the ordered pair $(x,y)$.
Thanks for all your guidance on the above issues/problems ...

Your help and assistance is much appreciated ...

Peter
 

FAQ: Uniform norm .... Garling Section 11.2 Normed Saces .... also Example 11.5.7

1. What is the uniform norm in mathematics?

The uniform norm, also known as the sup norm, is a mathematical concept used to measure the size or magnitude of a function. It is defined as the maximum absolute value of the function over its entire domain.

2. How is the uniform norm used in normed spaces?

In normed spaces, the uniform norm is used to define a metric, or distance function, between points. This metric is then used to define the topology, or the notion of convergence, in the normed space.

3. What is the significance of Example 11.5.7 in Garling's Section 11.2?

Example 11.5.7 in Garling's Section 11.2 is a specific example that illustrates the concept of uniform norm and its application in normed spaces. It helps to understand how the uniform norm is used to define a metric and topology in a normed space.

4. Can you give an intuitive explanation of the uniform norm?

Think of the uniform norm as a way to measure the "size" of a function. Just like a ruler measures the length of an object, the uniform norm measures the "length" of a function by looking at its maximum absolute value.

5. How does the uniform norm relate to other norms in mathematics?

The uniform norm is just one type of norm in mathematics. It is often used in functional analysis, while other norms such as the Euclidean norm and the p-norm are used in other areas of mathematics. However, all norms share some common properties, such as being positive, homogeneous, and satisfying the triangle inequality.

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