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 $\mathbb {F}$ with two operations defined on $V$, addition and scalar multiplication. Let $x,y\in V$ and $\alpha ,\beta \in \mathbb {F}$. 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 $\mathbb {F} =\mathbb {R}$, we call this a real linear space. Similarly, if the field $\mathbb {F} =\mathbb {C}$, 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 $\mathbb {R} ^{n}$ is the set of all n-tuples of real numbers. So for some $x\in \mathbb {R} ^{n}$, $x=(x_{1},x_{2},\dots ,x_{n})$ where $x_{i}\in \mathbb {R}$, $1\leq i\leq n$. Say $x,y\in \mathbb {R} ^{n}$ and $\alpha \in \mathbb {R}$. 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 $\mathbb {R} ^{n}$ 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 $\mathbb {R} ^{n}$) that is given by

$\langle x,y\rangle =x\cdot y=\sum _{i=1}^{n}x_{i}y_{i}$.

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 $\alpha ,\beta \in \mathbb {R}$ we have that $\alpha x+\beta y\in W$(i.e., $W$ is closed under addition and scalar multiplication).

Example

Consider $V=\mathbb {R} ^{3}$ and $A=\{(2,0,0),(0,43,0)\}\subset V$. Then the span of $A$ is the set $W=\mathrm {span} A=\{(x_{1},x_{2},0):x_{1},x_{2}\in \mathbb {R} \}$. 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, $W\neq \emptyset$. Now, suppose $x,y\in W$and $\alpha ,\beta \in \mathbb {R}$. We see

${\begin{aligned}\alpha x+\beta y&=\alpha (x_{1},x_{2},0)+\beta (y_{1},y_{2},0)\\&=(\alpha x_{1},\alpha x_{2},0)+(\beta y_{1},\beta y_{2},0)\\&=(\alpha x_{1}+\beta y_{1},\alpha x_{2}+\beta y_{2},0)\\&=(z_{1},z_{2},0)\\&=z\end{aligned}}$.

Since $z_{1},z_{2}\in \mathbb {R}$ 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 $\mathbb {R} ^{3}$.

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 $\alpha _{i}\in \mathbb {R}$ for $1\leq i\leq n$. 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 $\mathrm {span} A$. 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 $\mathrm {span} A=V$ 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 $\mathrm {dim} V$.

Example

The standard basis of the vector space $\mathbb {R} ^{n}$ 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 $\mathbb {R} ^{3}$ our basis is the set

$\{e_{1},e_{2},e_{3}\}=\{(1,0,0),(0,1,0),(0,0,1)\}$.

Hence, any vector in $\mathbb {R} ^{3}$ 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, $\mathrm {dim} \mathbb {R} ^{3}=3$.

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 $\|\cdot \|:V\to \mathbb {R}$ such that for $x,y\in V$ and $\alpha \in \mathbb {R}$ the following properties hold.

1. $\|x\|\geq 0$
2. $\|x\|=0\iff x=0$
3. $\|\alpha x\|=|\alpha |\|x\|$
4. $\|x+y\|\leq \|x\|+\|y\|$ (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 ($\|x-y\|$), 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

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

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 $\mathbb {R} ^{n}$ has a norm given by

$\|x\|=\left(\sum _{i=1}^{n}x_{i}^{2}\right)^{\frac {1}{2}}={\sqrt {x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}}}$.

This should be familiar from the Pythagorean theorem. Since $\mathbb {R}$ is complete, we have that $\mathbb {R} ^{n}$ 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

$\|x\|=\left(\sum _{i=1}^{n}x_{i}^{2}\right)^{\frac {1}{2}}=\langle x,x\rangle ^{\frac {1}{2}}$.

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