# Abstract Nonsense

## Linear Operators and the Operator Norm (Pt. I)

Point of Post: In this post we discuss the notion of linear operators on normed spaces (in doing so prove some of the fundamental results concerning continuity/boundedness) and discuss the operator norm of a linear operator and some its properties.

$\text{ }$

Motivation

$\text{ }$

The concept of linear operators pop up everywhere in discrete mathematics: pure linear algebra, graph theory, etc. That said, we have recently seen that the generalization of the derivative is itself a linear operator. It makes sense then that we should then study the properties of the algebraically defined linear operator with the topology on the space given. In particular, we shall see that linear operators between normed vector spaces has the interesting property that ‘boundedness’ (in a sense defined below) and continuity are synonymous. After proving this fact we shall show how to define a norm on the set of bounded operators on a space. We shall then show that for finite dimensional spaces (such as Euclidean space) the landscape is much simpler since every linear operator is bounded.

$\text{ }$

Continuity is Equivalent to Boundedness

$\text{ }$

We first show that for linear operators between normed vector spaces continuity and boundedness are synonymous.  By bounded we mean a linear operator $T:\mathscr{V}\to\mathscr{W}$ with the property that there exists some $M\in\mathbb{R}^+$ such that $\|T(v)\|\leqslant M\|v\|$ for every $v\in\mathscr{V}$. Indeed:

$\text{ }$

Theorem: Let $\left(\mathscr{V},\|\cdot\|_\mathscr{V}\right)$ and $\left(\mathscr{W},\|\cdot\|_\mathscr{W}\right)$ be two normed (real or complex) vector spaces and let $T\in\text{Hom}\left(\mathscr{V},\mathscr{W}\right)$. Then, $T$ is continuous (in the topology induced by norms) if and only if $T$ is bounded.

Proof: Suppose first that $T$ is bounded with bounding constant $M$. We note then that $\|T(x)-T(y)\|_\mathscr{W}=\|T(x-y)\|_\mathscr{W}\leqslant M\|x-y\|_\mathscr{V}$ for every $x,y\in\mathscr{V}$ and so $T$ is Lipschitz and so trivially continuous.

$\text{ }$

Conversely, suppose that $T$ is continuous. Since $T$ is continuous at zero we know there exists an open ball $B_{\delta}(\bold{0})$ for which $\|T(x)\|_\mathscr{W}\leqslant 1$ for $x\in B_{\delta}(\bold{0})$ and so in particular for $\|x\|_\mathscr{V}<\delta$ we have that $\|T(x)\|_\mathscr{W}\leqslant 1$ and so for any $v\in\mathscr{V}-\{\bold{0}\}$ we have that

$\text{ }$

$\displaystyle \|T(v)\|_\mathscr{W}=\left\|T\left(\frac{2\delta \|v\|v}{2\delta\|v\|}\right)\right\|_\mathscr{W}=\frac{2\|v\|}{\delta }\left\|T\left(\frac{\delta v}{2\|v\|}\right)\right\|_\mathscr{W}\leqslant \frac{2}{\delta}\|v\|_\mathscr{V}$

$\text{ }$

(where we made note of the fact that $\displaystyle \left\|\frac{\delta v}{2\|v\|}\right\|_\mathscr{V}=\frac{\delta}{2}<\delta$) and so the conclusion follows. $\blacksquare$

$\text{ }$

$\text{ }$

Every Linear Operator on a Finite Dimensional Space is Bounded

$\text{ }$

So, what we’d now like to show is that for finite dimensional normed spaces all linear operators are bounded. We admittedly leave out one big detail in the following proof. Namely, that any two norms on a finite dimensional vector space are equivalent (for the reader who hasn’t seen the statement that all norms are equivalent on finite dimensional vector spaces can see here). With that small caveat we proceed:

$\text{ }$

Theorem: All linear operators between two finite dimensional normed spaces $\left(\mathscr{V},\|\cdot\|_\mathscr{V}\right)$ and $\left(\mathscr{W},\|\cdot\|_\mathscr{W}\right)$ are bounded.

Proof: We may evidently assume that $\mathscr{V}\cong\mathbb{F}^n$ and $\mathscr{W}\cong\mathbb{F}^m$ (where $\mathbb{F}$ is $\mathbb{C}$ or $\mathbb{R}$ according to whether or not $\mathscr{V},\mathscr{W}$ are real or complex vector spaces and $n=\dim_\mathbb{F}\mathscr{V}$, $m=\dim_\mathbb{F}\mathscr{W}$). Since all norms are equivalent it also suffices to show this is tru assuming we’ve given both spaces the taxicab norm. In that case, let $M=\text{max}\{\|T(e_1)\|,\cdots,\|T(e_n)\|\}$ (where $\{e_1,\cdots,e_n\}$ is the canonical basis for $\mathbb{R}^n$) we then have then for every $(a_1,\cdots,a_n)\in\mathbb{R}^n$ that

$\text{ }$

\displaystyle \begin{aligned}\left\|T(a_1,\cdots,a_n)\right\| &=\left\|a_1 T(e_1)+\cdots+a_n T(e_n)\right\|\\ &\leqslant |a_1|\|T(e_1)\|+\cdots+|a_n| \|T(e_n)\|\\ &\leqslant M\left(|a_1|+\cdots+|a_n|\right)\\ &=M\|(a_1,\cdots,a_n)\|\end{aligned}

$\text{ }$

Since $(a_1,\cdots,a_n)$ was arbitrary the conclusion follows. $\blacksquare$

$\text{ }$

$\text{ }$

References:

1. Kreyszig, Erwin. Introductory Functional Analysis with Applications. New York: Wiley, 1978. Print.