Functional Analysis 9: Bounded and Continuous Linear Operators

Definition. Let $X,Y$ be normed spaces and $T:\mathcal{D}(T)\longrightarrow Y$ be a linear operator where $\mathcal{D}(T)\subset X$. $T$ is said to be bounded if there exists $c\in\mathbb{R}$ such that for any $x\in\mathcal{D}(T)$,
$$||Tx||\leq c||x||.$$

Suppose that $x\ne O$. Then
$$\frac{||Tx||}{||x||}\leq c.$$
Let
$$||T||:=\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\
x\ne O\end{array}}\frac{||Tx||}{||x||}.$$
Then $||T||$ is called the norm of the operator $T$. If $\mathcal{D}(T)=\{O\}$ then we define $||T||=0$.

Lemma. Let $T$ be a bounded linear operator. Then

  1. $||T||=\displaystyle\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\||x||=1\end{array}}||Tx||.$
  2. $||\cdot||$ defined on bounded linear operators satisfies (N1)-(N3).

Proof.

  1. \begin{align*}||T||&=\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\x\ne O\end{array}}\frac{||Tx||}{||x||}\frac{||Tx||}{||x||}\\&=\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\x\ne O\end{array}}\left|\left|\frac{||Tx||}{||x||}\right|\right|\\&=\sup_{\begin{array}{c}y\in\mathcal{D}(T)\\||y||=1\end{array}}||Ty||.\end{align*}
  2. \begin{align*}||T||=0&\Longleftrightarrow Tx=0,\ \forall x\in\mathcal{D}(T)\\&\Longleftrightarrow T=0.\end{align*} Since $$\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\||x||=1\end{array}}||(T_1+T_2)x||\leq \sup_{\begin{array}{c}x\in\mathcal{D}(T)\\||x||=1\end{array}}||T_1x||+\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\||x||=1\end{array}}||T_2x||,$$ $$||T_1+T_2||\leq ||T_1||+||T_2||.$$

Examples.

  1. The identity operator $I:X\longrightarrow X$ with $X\ne\{O\}$ is a bounded linear operator with $||I||=1$.
  2. Zero operator $O: X\longrightarrow Y$ is a bounded linear operator with $||O||=0$.
  3. Let $X$ be the normed space of all polynomials on $[0,1]$ with $||x||=\max_{t\in[0,1]}|x(t)|$. Differentiation$$T: X\longrightarrow X;\ Tx(t)=x'(t)$$ is not a bounded operator. To see this, let $x_n(t)=t^n$, $n\in\mathbb{N}$. Then $||x_n||=1$ for all $n\in\mathbb{N}$. $Tx_n(t)=nt^{n-1}$ and $||Tx_n||=n$, for all $n\in\mathbb{N}$. So, $\frac{||Tx_n||}{||x_n||}=n$ and hence $||T||$ is not bounded.
  4. Integral operator $$T:\mathcal{C}[0,1]\longrightarrow\mathcal{C}[0,1];\ Tx=\int_0^1\kappa(t,\tau)x(\tau)d\tau$$ is a bounded linear operator. The function $\kappa(t,\tau)$ is a continuous function on $[0,1]\times[0,1]$ called the kernel of $T$. \begin{align*}||Tx||&=\max_{t\in[0,1]}\left|\int_0^1\kappa(t,\tau)x(\tau)d\tau\right|\\&\leq\max_{t\in[0,1]}\int_0^1|\kappa(t,\tau)||x(\tau)|d\tau\\&\leq k_0||x||,\end{align*}where $k_0=\displaystyle\max_{(t,\tau)\in[0,1]\times[0,1]}\kappa(t,\tau)$.
  5. Let $A=(\alpha_{jk})$ be an $r\times n$ matrix of real entries. The linear map $T:\mathbb{R}^n\longrightarrow\mathbb{R}^r$ given by $Tx=Ax$ for each $x\in\mathbb{R}^n$ is bounded. To see this, Let $x\in\mathbb{R}^n$ and write $x=(\xi_j)$. Then $||x||=\sqrt{\displaystyle\sum_{m=1}^n\xi_m^2}$.\begin{align*}||Tx||^2&=\sum_{j=1}^r\left[\sum_{k=1}^n\alpha_{jk}\xi_k\right]^2\\&\leq\sum_{j=1}^r\left[\left(\sum_{k=1}^n\alpha_{jk}^2\right)^\frac{1}{2}\left(\sum_{m=1}^n\xi_m\right)^\frac{1}{2}\right]^2\\&=||x||^2\sum_{j=1}^r\sum_{k=1}^n\alpha_{jk}^2.\end{align*}By setting $c^2=\displaystyle\sum_{j=1}^r\sum_{k=1}^n\alpha_{jk}^2$, we obtain$$||Tx||^2\leq c^2||x||^2.$$

In general, if a normed space $X$ is finite dimensional, then every linear operator on $X$ is bounded. Before we discuss this, we first introduce the following lemma without proof.

Lemma. Let $\{x_1,\cdots,x_n\}$ be a linearly independent set of vectors in a normed space $X$. Then there exist a number $c>0$ such that for any scalars $\alpha_1,\cdots,\alpha_n$, we have the inequality
$$||\alpha_1x_1+\cdots+\alpha_nx_n||\geq c(|\alpha_1|+\cdots+|\alpha_n|).$$

Theorem. If a normed space $X$ is finite dimensional, then every linear operator on $X$ is bounded.

Proof. Let $\dim X=n$ and $\{e_1,\cdots,e_n\}$ be a basis for $X$. Let $x=\displaystyle\sum_{j=1}^n\xi_je_j\in X$. Then
\begin{align*}
||Tx||&=||\sum_{j=1}^n\xi_jTe_j||\\
&\leq\sum_{j=1}^n||\xi_j|||Te_j||\\
&\leq\max_{k=1,\cdots,n}||Te_k||\sum_{j=1}^n|\xi_j|.
\end{align*}
By Lemma, there exists a number $c>0$ such that
$$||x||=||\xi_1e_1+\cdots+\xi_ne_n||\geq c(|\xi_1|+\cdots+|\xi_n|)=c\sum_{j=1}^n|\xi_j|.$$
So, $\displaystyle\sum_{j=1}^n|\xi_j|\leq\frac{1}{c}||x||$ and hence
$$||Tx||\leq M||x||,$$ where
$M=\frac{1}{c}\max_{k=1,\cdots,n}||Te_k||$.

What is really nice about linear operators from a normed space into a normed space is that a linear operator being bounded is equivalent to it being continuous.

Theorem. Let $X,Y$ be normed spaces and $T:\mathcal{D}(T)\subset X\longrightarrow Y$ a linear operator. Then

  1. $T$ is continuous if and only if $T$ is bounded.
  2. If $T$ is continuous at a single point, it is continuous.

Proof.

  1. If $T=O$, then we are done. Suppose that $T\ne O$. Then $||T||\ne 0$. Assume that $T$ is bounded and $x_0\in\mathcal{D}(T)$. Let $\epsilon>0$ be given. Choose $\delta=\frac{\epsilon}{||T||}$. Then for any $x\in\mathcal{D}(T)$ such that $||x-x_0||<\delta$, $$||Tx-Tx_0||=||T(x-x_0)||\leq ||T||||x-x_0||<\epsilon.$$ Conversely, assume that $T$ is continuous at $x_0\in\mathcal{D}(T)$. Then given $\epsilon>0$ there exists $\delta>0$ such that $||Tx-Tx_0||<\epsilon$ whenever $||x-x_0||\leq\delta$. Take $y\ne 0\in\mathcal{D}(T)$ and set $$x=x_0+\frac{\delta}{||y||}y.$$ Then $x-x_0=\frac{\delta}{||y||}y$ and $||x-x_0||=\delta$. So,\begin{align*}||Tx-Tx_0||&=||T(x-x_0)||\\&=\left|\left|T\left(\frac{\delta}{||y||}y\right)\right|\right|\\&=\frac{\delta}{||y||}||Ty||\\&<\epsilon.\end{align*}Hence, for any $y\in\mathcal{D}(T)$, $||Ty||\leq\frac{\epsilon}{\delta}||y||$ i.e. $T$ is bounded.
  2. In the proof of part (a), we have shown that if $T$ is continuous at a point, it is bounded. If $T$ is bounded, then it is continuous by part (a).

Corollary. Let $T$ be a bounded linear operator. Then

  1. If $x_n\to x$ then $Tx_n\to Tx$.
  2. $\mathcal{N}(T)$ is closed.

Proof.

  1. If $T$ is bounded, it is continuous and so the statement is true.
  2. Let $x\in\overline{\mathcal{N}(T)}$. Then there exists a sequence $(x_n)\subset\mathcal{N}(T)$ such that $x_n\to x$. Since $Tx_n=0$ for each $n=1,2,\cdots$, $Tx=0$. Hence, $x\in\mathcal{N}(T)$.

Theorem. Let $X$ be a normed space and $Y$ a Banach space. Let $T:\mathcal{D}(T)\subset X\longrightarrow Y$ be a bounded linear operator. Then $T$ has an extension $\tilde T:\overline{\mathcal{D}(T)}\longrightarrow Y$ where $\tilde T$ is a bounded linear operator of norm $||\tilde T||=||T||$.

Proof. Let $x\in\overline{\mathcal{D}(T)}$. Then there exists a sequence $(x_n)\subset\mathcal{D}(T)$ such that $x_n\to x$. Since $T$ is bounded and linear,
\begin{align*}
||Tx_m-Tx_n||&=||T(x_m-x_n)||\\
&\leq||T||||x_m-x_n||,
\end{align*}
for all $m,n\in\mathbb{N}$. Since $(x_n)$ is convergent, it is Cauchy so given $\epsilon>0$ there exists a positive integer $N$ such that for all $m,n\geq N$, $||x_m-x_n||<\frac{\epsilon}{||T||}$. Hence, for all $m,n>N$,
\begin{align*}
||Tx_m-Tx_n||&\leq ||T||||x_m-x_n||\\
&<\epsilon.
\end{align*}
That is, $(Tx_n)$ is a Cauchy sequence in $Y$. Since $Y$ is a Banach space, there exists $y\in Y$ such that $Tx_n\to y$. Define $\tilde T:\overline{\mathcal{D}(T)}\longrightarrow Y$ by $\tilde Tx=y$. In order for $\tilde T$ to be well- defined, its definition should not depend on the choice $(x_n)$. Suppose that there is a sequence $(z_n)\subset\mathcal{D}(T)$ such that $z_n\to x$. Then $x_n-z_n\to 0$. Since $T$ is bounded, it is continuous so $T(x_n-z_n)\to 0$. This means that $\displaystyle\lim_{n\to\infty}Tz_n=\lim_{n\to\infty}Tx_n=y$. $\tilde T$ is linear and $\tilde T|_{\mathcal{D}(T)}=T$. To show that $\tilde T$ is bounded, let $x\in\overline{\mathcal{D}(T)}$. Then there exists a sequence $(x_n)\subset\mathcal{D}(T)$ such that $x_n\to x$ as before. Since $T$ is bounded, for each $n=1,2,\cdots$,
$$||Tx_n||\leq ||T||||x_n||.$$ Since the norm $x\longmapsto||x||$ is continuous, as $n\to\infty$ we obtain
$$||\tilde Tx||\leq ||T||||x||.$$ Hence, $\tilde T$ is bounded and $||\tilde T||\leq ||T||$. On the other hand, since $\tilde T$ is an extension of $T$, $||T||\leq||\tilde T||$. Therefore, $||\tilde T||=||T||$.

Leave a Reply

Your email address will not be published. Required fields are marked *