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.

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4 Comments »

  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 | Reply

  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 | Reply

  3. Hi! I was surfing and found your blog post… nice! I love your blog. 🙂 Cheers! Sandra. R.

    Comment by sandrar — September 10, 2009 @ 6:04 pm | Reply

  4. this realy a very helpful blog I liked such blog

    Comment by naveen kumar singh — April 11, 2011 @ 1:17 am | Reply


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