Bounded Linear Operator

We talk about the mapping from a space to another space. Because linear operator helps analysis, we fucos on normed linear space.

Basic Concepts

Assume \(E\) and \(E_1\) are linear sapce (number domain \(K\)), \(T\) is a mapping from \(D\) (a subspace of \(E\)) to \(E_1\). If for any \(x,y\in D\), we have

\[ T(x+y)=Tx+Ty, \]

we call \(T\) is additive. If for all number \(\alpha\in K\), we have

\[ T(\alpha x)=\alpha Tx, \]

we call \(T\) is homogeneous.

An additive and homogeneous mapping is called linear mapping or linear operator. If \(E_1\) is number domain, then the mapping is called functional. And it is called linear functional if it is additive andhomogeneous.

Now we give some important concepts about continuity and boundedness.

continuity and boundedness

Assume \(D\) is a subspace of \(E\), \(T\) is a linear operator from \(D\) to \(E_1\).

If \(T\) maps a bounded set in \(D\) into a bounded set in \(E_1\), then we call \(T\) is a bounded linear operator. Otherwise, there exsits a bounded subset \(A\) of \(D\), it is mapped to a unbounded set in \(E_1\) by \(T\), then \(T\) is called an unbounded linear operator.

This result is directly from the proof in Parallelogram Law.

Theorem

Assume \(E\), \(E_1\) are real normed linear spaces, \(D\) is a subspace of \(E\), and \(T\) is a continuous additive operator from \(D\) to \(E_1\). Then \(T\) is homogeneous.

Proof

For any \(x\in D\), define \(f(\alpha)=T(\alpha x)\), so \(f\) is continuous and for all \(\alpha_1\), \(alpha_2\)

\[ f(\alpha_1+\alpha_2)=f(\alpha_1)+f(\alpha_2). \]

By lemma in Parallelogram Law, we have for all \(\alpha \in \mathbb{R}\), \(f(\alpha)=\alpha f(1)\), i.e.

\[ T(\alpha x)=\alpha T(x). \]

\(\square\)

Boundedness & Continuity for LO

The following result is an equivalent proposition for bounded linear operator. It is the linearity that bridges the difference between these two.

Lemma: Equivalent proposition for bounded linear operator

Assume \(E\), \(E_1\) are normed linear spaces, \(D\) is a subspace of \(E\), and \(T\) is a linear operator from \(D\) to \(E_1\). Then \(T\) is bounded iff \(\exists M>0\), such that

\[ \|Tx\|\leq M\|x\| \]

Proof

• Necessary. This is a general method. Choose a unit sphere \(S=\{x: \|x\|=1\}\), and we have \(S\) is bounded set, so \(T(S)\) is also bounded by assumption. So \(\exists M>0\), such that for all \(x\in S\), \(\|Tx\|\leq M\). Then \(\forall x\in D\), \(\frac{x}{\|x\|}\in S\), so

\[ \left|T\left(\frac{x}{\|x\|}\right)\right|\leq M \]

By linear properties, we have \(\|Tx\|\leq M \|x\|\).

• Sufficient.

This is easy. Since we choose an bounded set \(A\subset D\), \(\exists K>0\), we have \(\|x\|\leq K\) for all \(x\in A\), so by assumption, \(\|Tx\|\leq M\|x\|\leq MK\), which shows that \(T(A)\) is bounded in \(E_1\).

Now we give a clear connection between continuity and boundedness.

Continuity & Boundedness

Assume \(E\), \(E_1\) are normed linear spaces, \(D\) is a subspace of \(E\), and \(T\) is a linear operator from \(D\) to \(E_1\). Then the following statement is equivalent.

(i) \(T\) is continuous.

(ii) \(T\) is continuous at \(\theta\).

(iii) \(T\) is bounded.

Proof

• (i) \(\Rightarrow\) (ii) is apparent.

• (ii) \(\Rightarrow\) (i). Choose \(\varepsilon-\delta\) language. Choose \(\varepsilon=1\), then \(\exists \delta\) such that whenever \(\|x\|\leq \delta\), we have

\[ \|Tx\|\leq 1 \]

Now we consider any \(x\neq \theta\in D\), then \(\frac{\delta x}{\|x\|}\leq \delta\), so

\[ \begin{align*} \|T\frac{\delta x}{\|x\|}\|\leq 1\\ \Rightarrow \quad \|Tx\|\leq \frac{1}{\delta}\|x\|. \end{align*} \]

by Lemma: Equivalent proposition for bounded linear operator, we complete the proof. Actually the abouve inequation is Lipschitz continuity at the origin.

• (iii) \(\Rightarrow\) (i). Choose a sequence \(x_n\rightarrow x\), because it is bounded, we have

\[ \|Tx_n-Tx\|\leq \|T(x_n-x)\|\leq M\|x_n-x\|\rightarrow 0. \]

Norm of Bounded linear Operator

Definition of Norm

Assume \(E,E_1\) are normed linear space, \(D\) is a subspace of \(E\), \(T\) is a bounded linear mapping from \(D\) to \(E_1\). Define the norm of \(T\)

\[ \|T\|=\sup_{x\neq \theta\atop x\in D}\frac{\|Tx\|}{\|x\|}. \]

Actually, \(\|T\|\) is also given by

\[ \|T\|=\inf\{M: \|Tx\|\leq M\|x\|,\forall x\in D\}. \]

Corollary

(i) for all \(x\in D\), \(\|Tx\|\leq \|T\|\|x\|\).

(ii) Normalized definition.

\[ \|T\|=\sup_{\|x\|\leq 1\atop x\in D}\|Tx\|=\sup_{\|x\|=1 \atop x\in D}\|Tx\|. \]

That is, we only need to consider the vector on the unit sphere.

It is usually hard to consider the exact magnitude of norm of an operator. Check the following example.

Example. Assume \((a_{ij}) (i,j=1,\cdots n)\) is a given \(n\times n\) matrix, \(a_{ij}\in \mathbb{R}\), define

\[ \eta_i=\sum_{j=1}^n a_{ij} \xi_j,\quad i=1,\cdots,n \]

then \(x=(\xi_1,\cdots,x_n), y=(\eta_1,\cdots, \eta_n) \in\mathbb{R}^n\), define \(T: \mathbb{R}^n\mapsto \mathbb{R}^n\) by \(Tx=y\). Show that \(T\) is a bounded linear operator.

Proof