# Mathematics Prelims

## July 18, 2008

### Linear Operators

Filed under: Analysis,Functional Analysis — cjohnson @ 4:22 pm

If $X$ and $Y$ are normed spaced and $\mathcal{D}_T \subseteq X$, then a mapping $T : \mathcal{D}_T \to Y$ is called a linear operator if for every $x,y \in \mathcal{D}_T$ and $\alpha, \beta \in K$ (the underlying field),

$\displaystyle T(\alpha x + \beta y) = \alpha Tx + \beta Ty$

If $T$ is a linear operator, then $T$ is injective if and only if $Tx = 0$ implies $x = 0$.

Proof: Suppose $T$ is injective.  Then $Tx = Ty \Rightarrow x = y$, since $T$ is linear, it must map zero to zero, so $Tx = 0 \Rightarrow x = 0$.  Now suppose that $Tx = 0$ only for $x = 0$.  If $x,y \in \mathcal{D}_T$ and $Tx = Ty$, then $Tx - Ty = 0$ and so $T(x - y) = 0$ (by linearity), and this means that $x - y = 0$, so $x = y$.

Now, we say that $T : \mathcal{D}_T \to Y$ is bounded if there exists a $c \in \mathbb{R}$ such that for every $x \in \mathcal{D}_T$ we have $\| Tx \| \leq c \|x\|$.  If such a $c$ exists (i.e., if $T$ is bounded), then we say that the least such $c$ is the operator norm of $T$, and write $\|T\| = c$.  We can then find $c$ as follows.

$\displaystyle \|T\| = \sup_{x \in \mathcal{D}_T, \, x \neq 0} \frac{\|Tx\|}{\|x\|}$

(This follows from rewriting the above earlier inequality as $\frac{\|Tx\|}{\|x\|} \leq c$.)  Note that by letting $c = \|T\|$, we have that $\|Tx\| \leq \|T\|\|x\|$.

In a finite dimensional normed space, every linear operator is bounded.  Suppose that $\dim X = n$ and that $\{ e_1, ..., e_n \}$ is a basis for $X$.  If we let $x \in X$ with $x = \sum_{i=1}^n \alpha_i e_i$, then we have the following.

$\displaystyle \|Tx\| = \left\| T\left( \sum_{i=1}^n \alpha_i e_i \right) \right\|$

$\displaystyle \qquad = \left\|\sum_{i=1}^n \alpha_i Te_i \right\|$

$\displaystyle \qquad \leq \sum_{i=1}^n | \alpha_i | \|Te_i\|$

$\displaystyle \qquad \leq \frac{k}{c} \|x\|$

Where $k = \max \{ \|T e_1\|, ..., \|T e_n \| \}$ and $c$ is the value such that $\| \alpha_1 e_1 + ... + \alpha_n e_n \| \geq c (|\alpha_1| + ... + |\alpha_n|)$.

Another important property is that a linear operator is bounded if and only if it is continuous.

Proof: Suppose $T$ is bounded and let $\epsilon > 0$ be given.  Let $\delta = \frac{\epsilon}{\|T\|}$.  Let $x_0$ be a fixed point in $\mathcal{D}_T$, and $x$ some other point in $\mathcal{D}_T$ such that $\| x - x_0 \| < \delta$.

$\displaystyle \|Tx - Tx_0\| = \|T(x - x_0)\|$

$\displaystyle \qquad \leq \|T\| \|x - x_0\|$

$\displaystyle \qquad \leq \|T\| \delta$

$\displaystyle \qquad \leq \|T\| \frac{\epsilon}{\|T\|}$

$\displaystyle \qquad = \epsilon$

This shows that $T$ is continuous at $x_0$.  However, our $x_0$ was arbitrary, and $\delta$ didn’t depend on our choice of $x_0$, so $T$ is uniformly continuous.

For the converse I’m going to copy the proof from Wikipedia, since I think it’s a bit clearer than Kreyszig’s proof.

Suppose that $T$ is continuous.  Since $\mathcal{D}_T$ is a subspace of $X$, it contains the zero vector.  Since $T$ is continuous, it’s continuous at the zero vector.  This means there exists a $\delta > 0$ such that for all $x \in \mathcal{D}_T$ satisfying $\|x\| \leq \delta$ we have that $\|Tx\| = \|T(x - 0)\| = \|Tx - T0\| < 1$.  Now let $y$ be any point in $\mathcal{D}_T$.

$\displaystyle \|Ty\| = \left\| \frac{\|y\|}{\delta} T\left( \frac{\delta}{\|y\|} y\right )\right\|$

$\displaystyle \qquad = \frac{\|y\|}{\delta} \left\| T \left( \frac{\delta}{\|y\|} y \right) \right\|$

$\displaystyle \qquad \leq \frac{1}{\delta} \| y \|$

And so $T$ is bounded.

1. Rudin’s approach is to define the operator norm (equivalently) as $\|T\|=\sup_{\{x: \|x\|\leq 1\}} \|Tx\|$. Which turns out to be equivalent to just checking the sup over $\{x: \|x\|=1\}$ which is equivalent to what you did.

Note: when I post random stuff like this in the comments, it isn’t because I think you don’t know them, it is because my prelim is coming up, too, so it is a nice refresher for me.

Comment by hilbertthm90 — July 18, 2008 @ 5:00 pm

2. […] assume that is either or .  Properties and theorems associated with “traditional” linear operators apply since and can be thought of as normed spaces with the “traditional” norms […]

Pingback by The Algebraic Dual Space « Mathematics Prelims — July 18, 2008 @ 10:27 pm

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