Real Analysis/Normed Linear Spaces

We will give a brief review of concepts from linear algebra regarding linear spaces and their properties. This is not an exhaustive discussion, so the reader is advised to consult a linear algebra text for more details if these topics are unfamiliar.

Linear Space

A linear space (also called a vector space) is a set $V$over a field $𝔽$ with two operations defined on $V$, addition and scalar multiplication. Let $x,y\in V$ and $\alpha ,\beta \in 𝔽$. The following eight properties define the structure of a linear space.

1.	$x+y=y+x$ (Symmetry of addition)
2.	$(x+y)+z=x+(y+z)$ (Associativity of addition)
3.	There exists a unique element $0\in V$ such that $x+0=x$ (Additive identity element)
4.	For each $x\in V$ there exists $(-x)\in V$ such that $x+(-x)=0$ (Additive inverse)
5.	$1x=x$ (Scalar multiplication by multiplicative identity)
6.	$\alpha (\beta x)=(\alpha \beta )x$ (Associativity of scalar multiplication)
7.	$(\alpha +\beta )x=\alpha x+\beta x$
8.	$\alpha (x+y)=\alpha x+\alpha y$

If the field $𝔽=ℝ$, we call this a real linear space. Similarly, if the field $𝔽=ℂ$, we call this a complex linear space. We will restrict our study to the case of real linear spaces. Elements of a linear space (or vector space) are called vectors.

Example

Euclidean Space

The set ${ℝ}^n$ is the set of all n-tuples of real numbers. So for some $x\in {ℝ}^n$, $x=(x_1,x_2,\dots ,x_n)$ where $x_i\in ℝ$, $1\le i\le n$. Say $x,y\in {ℝ}^n$ and $\alpha \in ℝ$. This set is a linear space. The addition property and scalar multiplication are shown by

$x+y=(x_1,x_2,\dots ,x_n)+(y_1,y_2,\dots ,y_n)=(x_1+y_1,x_2+y_2,\dots ,x_n+y_n)$

and

$\alpha x=\alpha (x_1,x_2,\dots ,x_n)=(\alpha x_1,\alpha x_2,\dots ,\alpha x_n).$

The reader should verify ${ℝ}^n$ satisfies the above eight properties of a linear space. Recall, the Euclidean space is equipped with an inner product (often called the dot product in ${ℝ}^n$) that is given by

$⟨x,y⟩=x\cdot y=\sum _{i=1}^n{x}_i{y}_i$.

This will come in handy in a following example.

Subspace

A subspace of a vector space $V$ is a nonempty subset $W\subseteq V$ such that $W$ is also a linear space. So for any $x,y\in W$ and $\alpha ,\beta \in ℝ$ we have that $\alpha x+\beta y\in W$(i.e., $W$ is closed under addition and scalar multiplication).

Example

Consider $V={ℝ}^3$ and $A=\{(2,0,0),(0,43,0)\}\subset V$. Then the span of $A$ is the set $W=\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}A=\{(x_1,x_2,0):x_1,x_2\in ℝ\}$. We will show that the set $W$ is a subspace of $V$.

Proof

First, we need to show that the zero element is in $W$ (otherwise, it could not be a linear space). It follows that

$0(2,0,0)+0(0,43,0)=(0,0,0)+(0,0,0)=(0,0,0)\in W$.

Therefore, $W\ne \mathrm{\varnothing }$. Now, suppose $x,y\in W$and $\alpha ,\beta \in ℝ$. We see

$\begin{array}{cc}\alpha x+\beta y& =\alpha (x_1,x_2,0)+\beta (y_1,y_2,0)\\ & =(\alpha x_1,\alpha x_2,0)+(\beta y_1,\beta y_2,0)\\ & =(\alpha x_1+\beta y_1,\alpha x_2+\beta y_2,0)\\ & =(z_1,z_2,0)\\ & =z\end{array}$.

Since $z_1,z_2\in ℝ$ by the field properties of the reals, we have that $z\in W$. Hence, this set is closed under the vector space operations. Therefore, $W$ is a subspace of ${ℝ}^3$.

The astute reader should notice that this subspace is a plane in three-dimensional space.

Basis

A linear combination of vectors is an expression

${\alpha }_1{x}_1+{\alpha }_2{x}_2+\cdots +{\alpha }_n{x}_n$,

where $x_i\in V$and ${\alpha }_i\in ℝ$ for $1\le i\le n$. This can be expressed in more concise notation as

$\sum _{i=1}^n{\alpha }_i{x}_i$.

Given a nonempty subset of a linear space $A\subseteq V$ the set of all linear combinations of elements of $A$ is called the span of $A$, which we denote with $\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}A$. The span of $A$ will generate a subspace $W$ of $V$.

A collection of vectors $x_1,\dots ,x_k$ from $V$ is said to be linearly independent when

$\sum _{i=1}^k{\alpha }_i{x}_i={\alpha }_1{x}_1+{\alpha }_2{x}_2+\cdots +{\alpha }_k{x}_k=0$

only when ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_k=0$. If any nonzero ${\alpha }_i$ satisfies this equation, then the set of vectors is called linearly dependent.

A basis of a linear space $V$ is a linearly independent set of vectors which spans $V$. That is, the subset $A\subseteq V$ is a basis if $\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{n}A=V$ and $A$ is a linearly independent set of vectors. The number of linearly independent vectors it takes to span a vector space defines the dimension of that vector space. A vector space $V$ is finite-dimensional if it can be spanned by a finite set of basis vectors. If a vector space is not finite-dimensional, it is infinite-dimensional. We denote the dimension of a vector space as $\mathrm{d}\mathrm{i}\mathrm{m}V$.

Example

The standard basis of the vector space ${ℝ}^n$ is the set

$\{e_1,e_2,\dots ,e_n\}=\{(1,0,\dots ,0),(0,1,\dots ,0),\dots ,(0,0,\dots ,1)\}$,

where the $i$-th basis vector has a one in the $i$-th entry and zeroes elsewhere.

So, in three-dimensional space ${ℝ}^3$ our basis is the set

$\{e_1,e_2,e_3\}=\{(1,0,0),(0,1,0),(0,0,1)\}$.

Hence, any vector in ${ℝ}^3$ can be expressed as a linear combination of these basis vectors. A suggested exercise for the reader is to prove that the expression of a vector as a linear combination of basis vectors is unique. Note that the basis set has 3 elements, so the space it spans has a dimension of 3, i.e, $\mathrm{d}\mathrm{i}\mathrm{m}{ℝ}^3=3$.

Norms

In a linear space, we often want to have a concept of the "size" of elements or of the distance between elements. A norm is a function $\Vert \cdot \Vert :V\to ℝ$ such that for $x,y\in V$ and $\alpha \in ℝ$ the following properties hold.

1. $\Vert x\Vert \ge 0$
2. $\Vert x\Vert =0⟺x=0$
3. $\Vert \alpha x\Vert =|\alpha |\Vert x\Vert$
4. $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$ (Triangle inequality)

The norm serves as a way to describe the size of individual elements. Now, if the norm is used to describe the size of a difference between vectors ($\Vert x-y\Vert$), it measures the distance between the two. So, we find that the norm induces a metric on the space $V$. Hence, we describe a metric function on $V$ by

$d(x,y)=\Vert x-y\Vert$.

The reader is encouraged to verify that this function satisfies the metric properties.

A linear space with a norm on it is called a normed linear space. A normed linear space that is complete is called a Banach space.

Example

Euclidean space ${ℝ}^n$ has a norm given by

$\Vert x\Vert ={\left(\sum _{i=1}^n{x}_i^2\right)}^{\frac{1}{2}}=\sqrt{x_1^2+x_2^2+\cdots +x_n^2}$.

This should be familiar from the Pythagorean theorem. Since $ℝ$ is complete, we have that ${ℝ}^n$ is also complete. Thus, Euclidean space would also be an example of a Banach space. We should also note that the norm here can be expressed as

$\Vert x\Vert ={\left(\sum _{i=1}^n{x}_i^2\right)}^{\frac{1}{2}}=⟨x,x{⟩}^{\frac{1}{2}}$.

So, in this case, the inner product induces a norm on our space. Template:BookCat