Unleash the Power of Sequence Spaces: Exploring ℓᵖ and ℓ∞ Norms

Here’s a persuasive advertisement for ghostwriting services specifically tailored to the topic of sequence spaces in mathematical analysis:

**Get Expert Help with Sequence Spaces: ℓᵖ and ℓ∞ Made Easy**

Struggling to grasp the intricacies of sequence spaces and their corresponding norms? Our team of expert ghostwriters is here to help! With years of experience in mathematical analysis, we’ll craft high-quality, plagiarism-free content that meets your academic needs.

**Our Services:**

* Customized essays, research papers, and assignments on sequence spaces, ℓᵖ, and ℓ∞
* In-depth explanations of norms, convergence, and completeness
* Clear, concise, and well-structured writing that meets your instructor’s expectations
* Fast turnaround times and flexible deadlines to fit your schedule

**Why Choose Us?**

* Expertise in mathematical analysis, ensuring accuracy and precision in our work
* Confidential and secure service, with your privacy guaranteed
* 24/7 customer support, because we know you’re busy
* Affordable prices without compromising on quality

Don’t let sequence spaces overwhelm you! Let our expert ghostwriters take care of your academic needs. Order now and get back to focusing on your studies!

**Order Now and Get 10% Off Your First Assignment!**

**Definition 65** (Sequence Space).: For \(1\leq p<\infty\) we define the sequence space

\[\ell^{p}=\left\{x=\{x_{i}\}:\sum_{i=1}^{\infty}|x_{i}|^{p}<\infty\right\}.\]

For \(1\leq p<\infty\) we define the norm

\[||x||_{p}=\left(\sum_{i=1}^{\infty}|x_{i}|^{p}\right)^{1/p}.\]

For the space \(\ell^{\infty}\) we define the norm

\[||x||_{\infty}=\sup|x_{n}|.\]

Note that the space \(\ell^{2}\) is of particular importance (why?).

### 2.8 Parallelogram Equality and the Norm

A norm in an inner product space satisfies the important parallelogram equality (Figure 2.8).

**Theorem 79** (Parallelogram Equality).: _For \(x,y\) in an inner product space_

\[||x+y||^{2}+||x-y||^{2}=2(||x||^{2}+||y||^{2}).\]

Proof

: If \(x,y\in X\), then

\[||x+y||^{2}+||x-y||^{2} = \langle x+y,x+y\rangle+\langle x-y,x-y\rangle\] \[= ||x||^{2}+||y||^{2}+\langle x,y\rangle+\langle y,x\rangle\] \[\qquad+||x||^{2}+||y||^{2}-\langle x,y\rangle-\langle y,x\rangle\] \[= 2(||x||^{2}+||y||^{2})\]

as desired.

If a norm does not satisfy the parallelogram equality, it cannot be obtained from an inner product. Thus not all normed spaces are inner product spaces.

Figure 2.8: Parallelogram equality

**Example 77**.: The space \(\ell^{p}\) with \(p\neq 2\) is a normed space but not an inner product space.

Proof

: Recall

\[\ell^{p}=\left\{x=\{x_{i}\}:\sum_{i=1}^{\infty}|x_{i}|^{p}<\infty\right\}.\]

Consider \(x=(1,1,0,0,\dots)\in\ell^{p}\) and \(y=(1,-1,0,0,\dots)\in\ell^{p}\). Then

\[x+y = (2,0,0,0,\dots)\] \[x-y = (0,2,0,0,\dots)\] \[||x+y||_{p}^{p} = \sum_{i=1}^{\infty}|x_{i}|^{p}=|2|^{p}\] \[\implies||x+y||_{p} = (2^{p})^{1/p}=2.\]

Similarly, \(||x-y||_{p}=2\). On the other hand,

\[||x||_{p}^{p}=\sum_{i=1}^{\infty}|x_{i}|^{p}=1^{p}+1^{p}=2\implies||x||_{p}=2^{ 1/p}=||y||_{p}.\]

So the parallelogram equality for this norm does not hold

\[||x+y||_{p}^{2}+||x-y||_{p}^{2} = 2(||x||_{p}^{2}+||y||_{p}^{2})\] \[4+4 \neq 2\left(2^{1/p}+2^{1/p}\right)\]

unless \(p=2\).

### Matrix Norms

We can form norms for matrices the same way that we defined norms for vectors in \(\mathbb{R}^{n}\). After all, the vector space \(M_{mn}\) of all \(m\times n\) matrices is isomorphic to \(\mathbb{R}^{mn}\). Thus the properties a) to c) of a norm will also hold in the setting of matrices. It turns out that, for matrices, the norms that are most useful satisfy an additional property d).

**Definition 66**.: A _matrix norm_ on \(M_{nn}\) is a mapping that associates with each \(n\times n\) matrix \(A\) a real number \(||A||\), called the norm of \(A\), such that the following hold for all \(n\times n\) matrices \(A\) and \(B\) and all scalars \(c\).

1. \(||A||\geq 0\) and \(||A||=0\) if and only if \(A=0\).
2. \(||cA||=|c|||A||\).
3. \(||A+B||\leq||A||+||B||\).
4. \(||AB||\leq||A||||B||\).

A matrix norm on \(M_{nn}\) is said to be _compatible_ with the norm of a vector \(||x||\) on \(\mathbb{R}^{n}\) if

\[||Ax||\leq||A||||x||\]

holds true for all \(n\times n\) matrices \(A\) and for all \(x\in\mathbb{R}^{n}\).

**Example 78**.: The _Frobenius norm_\(||A||_{F}\) of a matrix \(A\) is obtained by taking the square root of the sum of the squares of the entries of \(A\). Thus, if \(A=[a_{ij}]\), then

\[||A||_{F}=\sqrt{\sum_{i,j=1}^{n}a_{i,j}^{2}}.\]

Suppose the given matrix \(A\) is equal to

\[\begin{pmatrix}3&1\\ -2&5\end{pmatrix}.\]

Then its Frobenius norm is \(||A||_{F}=\sqrt{3^{2}+1^{2}+(-2)^{2}+5^{2}}=\sqrt{39}\). Now observe that if \(r_{1}=(3,1)\) and \(r_{2}=(-2,5)\) are the row vectors of \(A\), then the Euclidean norm of these vectors is \(||r_{1}||=\sqrt{3^{2}+1^{2}}=\sqrt{10}\) and \(||r_{2}||=\sqrt{(-2)^{2}+5^{2}}=\sqrt{29}\). So,

\[||A||_{F}=\sqrt{||r_{1}||^{2}+||r_{2}||^{2}}.\]

Similarly one can see if \(c_{1}\) and \(c_{2}\) are the column vectors of \(A\), then

\[||A||_{F}=\sqrt{||c_{1}||^{2}+||c_{2}||^{2}}.\]

It is easy to see that these facts extends to \(n\times n\) matrices in general.

**Proposition 28**.: _The Frobenius norm is a matrix norm and it is compatible with the Euclidean norm._

Proof

: We first show that it is compatible with the Euclidean norm \(||\cdot||_{E}\). Let us start by writing

\[A=\begin{pmatrix}A_{1}\\ A_{2}\\ \vdots\\ A_{n}\end{pmatrix},\]where \(A_{i}\) is the \(i\)th row of the matrix \(A\). Then

\[||Ax||_{E} = \left|\left|\begin{pmatrix}A_{1}\\ A_{2}\\ \vdots\\ A_{n}\end{pmatrix}x\right|\right|_{E}\] \[= \sqrt{||A_{1}x||_{E}^{2}+\cdots+||A_{n}x||_{E}^{2}}\] \[\leq \sqrt{||A_{1}||_{E}^{2}||x||_{E}^{2}+\cdots+||A_{n}||_{E}^{2}||x|| _{E}^{2}}\] \[= \sqrt{||A_{1}||_{E}^{2}+\cdots+||A_{n}||_{E}^{2}}\cdot||x||_{E}\] \[= ||A||_{F}||x||_{E}.\]

Note that in the above we used the CSB inequality to conclude that \(||A_{i}x||_{E}\)\(\leq||A_{i}||_{E}||x||_{E}\) for \(i=1,\ldots,n.\) Hence the Frobenius norm is compatible with the Euclidean norm. Let us now demonstrate that \(||\cdot||_{F}\) is a matrix norm. Let \(b_{i}\) denote the \(i\)th column of \(B\). Using the representation for the product (matrix-column representation)

\[||AB||_{F}=||[Ab_{1}\cdots Ab_{n}]||_{F},\]

we get:

\[||AB||_{F} = \sqrt{||Ab_{1}||_{E}^{2}+\cdots+||Ab_{n}||_{E}^{2}}\] \[\leq \sqrt{||A||_{F}^{2}||b_{1}||_{E}^{2}+\cdots+||A||_{F}^{2}||b_{n}|| _{E}^{2}}\] \[= ||A||_{F}\sqrt{||b_{1}||_{E}^{2}+\cdots+||b_{n}||_{E}^{2}}\] \[= ||A||_{F}||B||_{F}.\]