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 examples.

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

$\square$

Example. We use $C(-\infty,+\infty)$ to be a set of all the bounded continuous function defined on $\mathbb{R}$, with linear operator the same as $C[a,b]$. Define norm in linear space $C(-\infty,+\infty)$ by

\[ \|y\|=\sup_{-\infty < t < \infty}|y(t)|,\quad y\in (-\infty,+\infty), \]

then we could show that $C(-\infty,+\infty)$ is a Banach Space. Then for $x\in L(-\infty,+\infty)$, we define a operator $T$

\[ y=Tx: y(s)=\int_\mathbb{R} e^{-ist}x(t)dt, \]

So it is a linear operator from $L(-\infty,+\infty)$ to $C(-\infty,+\infty)$, whose boundedness is given by

\[ |(Tx)(s)|=|y(s)|\leq \int_\mathbb{R}|e^{-ist}x(t)|dt=\int_\mathbb{R}|x(t)|dt<\infty. \]

So $Tx$ is bounded and thus $T$ is continuous by Continuity & Boundedness. And $\|T\|\leq 1$.

Example. Lagrangian Formula. As we have shown in Lagrange interpolating polynomial in Numerical Analysis.

Given $x\in C[a,b]$ with $n$ points $(t_1,x_1),\cdots (t_n,x_n)$ passing it, $a<x_1<\cdots<x_n<b$ we have Lagrangian formula

\[ l_k(t)=\prod_{j=1\atop j\neq k}^n \frac{t-t_j}{ t_k-t_j}. \]

then $l_k(t)$ is a basis of $C[a,b]$, and the interpolating polynomial defined by

\[ y=L_nx: y(t)=\sum_{k=1}^n x_k l_k(t). \]

Then $L_n$ is a operator which maps $C[a,b]$ into itself. Show that $L_n$ is a bounded linear operator and its norm satisfies

\[ \|L_n\|=\max_{t\in [a,b]}\sum_{k=1}^n |l_k(t)| \]

Proof

$\square$

Example. Assume $K(t,s)$ is continuous function defined on $[a,b]\times [a,b]$. Define integral operator on $C[a,b]$

\[ y(t)=(Tx)(t)=\int_a^b K(t,s)x(s)ds. \]

show that $T$ is a bounded linear operator mapping $C[a,b]$ to itself and its norm satisfies

\[ \|T\|=\max_{t\in [a,b]}\int_a^b |K(t,s)|ds. \]

Proof

$\square$

Example. Asssume

LO Space¶

Definition of Linear operator space

Assume $E$, $E_1$ are normded linear space. We use $\mathscr{B}(E.E_1)$ to be the set of linear operator which maps $E$ into $E_1$. If we define the following linear calculation

(i) $(T_1+T_2)x=T_1 x+T_2 x$

(ii) $(\alpha T)x=\alpha (Tx)$

where $T_1,T_2,T\in \mathscr{B}(E,E_1)$, $\alpha \in K$. Then $\mathscr{B}(E,E_1)$ is a linear space by the above linear calculation.

Normed linear space

If we use norm defined in the previous part, $\mathscr{B}(E,E_1)$ is a normed linear space.

Proof

(i) Positive & definitive.

(ii) homogeneous.

(iii) Triangle inequation.

Convergence¶

Convergence in operator norm

Assume $\{T_n\},T\subset \mathscr{B}(E,E_1)$, then we call $\{T_n\}$ converges to $T$ in operator norm, or in the uniform operator topology, if

\[ \lim_{n\rightarrow \infty }\|T_n-T\|=0. \]

The above convergence implies some uniform convergence.

Convergence in the uniform operator topology

Assume $\{T_n\},T\subset \mathscr{B}(E,E_1)$, then $\{T_n\}$ converges to $T$ in the uniform operator topology, iff $T_n$ converges to $T$ uniformly on any bounded subset of $E$.

Proof

\[ \|T_n x-Tx\|\leq \|T_n-T\| \|x\| \]

so for a bounded $x$, necessary condition is apparent.

As for sufficient condition, we have to make use of the scaling transformation. That is, choose a unit sphere $\overline{S}(\theta,1)$. $\forall \varepsilon>0$, $\exists N$, we have

\[ \|T_n x-Tx\|<\varepsilon,\quad \text{whenever } n>N \]

Now by definition of norm of operator,

\[ \|T_n-T\|=\sup_{\|x\|=1}\|(T_n-T)x\|\leq \varepsilon. \]

Completeness¶

Condition for completeness of $\mathscr{B}(E,E_1)$

Assume $E_1$ is a Banach space, then $\mathscr{B}(E,E_1)$ is a Banach space.

Proof

We make use of the completeness of $E_1$ to formulate the limit of Cauchy sequence in $\mathscr{B}(E,E_1)$.

Strong Convergence¶

Definition of strong convergence

Assume $\{T_n\},T\subset \mathscr{B}(E,E_1)$, if for every $x\in E$,

\[ \lim_{n\rightarrow\infty}\|T_n x- Tx\|=0, \]

then we call $\{T_n\}$ converges to $T$ in strong convergence, or in the strong operator topology.

From Proof of Convergence in the uniform operator topology, we could know that convergence in the uniform operator topology implies that in the strong operator topology.

Banach's Open mapping theorem¶

Banach's Open mapping theorem

Assume $E$, $E_1$ are banach space, and a bounded linear operator $T$ maps $E$ into $E_1$. If the range of $T$, denoted by $F=T(E)$, is a set of the second category, then $F=E_1$, i.e. $T$ is surjective, and holds the following estimate: $\exists K>0$, $\forall y\in E_1$, $\exists x\in E$, such that $Tx=y$ and

\[ \|x\|\leq K\|y\|. \]

Moreover, $T$ maps any open set in $E$ into an open set in $E_1$.

Proof

We must use a sequence of ball $O_n=\{x:\|x\|\leq n,x\in E\}$, and $M_n=T(O_n)$. This is the form we depend on.

By definition of set of the second category. Since $E=\bigcup_{n}O_n$, so $F=\bigcup_n M_n$. Since it is not a set of the first category, $\exists n_0$, such that $M_{n_0}$ is not a nowhere dense set, i.e. $\exists $ a closed ball $Q(y_0,r_0)$ in which $M_{n_0}$ is dense.
Move the closed ball to orinin. Show that $M_1$ is dense in $Q(0,\delta)$ where $\delta=r_0/n_0$. This is done by making use of the subtraction of two point in $Q(y_0,r_0)$.
Show that $M_1\supset Q_{\delta/2}$. This is by $T(O_{1/2^n})$ is dense in $Q_{\delta/2^n}$. This is not a closure because the original image of any element in $Q_{\delta/2}$ is constructed by limit of a sequence.
Finally, use scaling transformation to get the result.
For the moreover part, we could make use of the step (iii).

Corollary: open mapping theorem

Assume $E$, $E_1$ are Banach space. If a bounded linear operator $T \in \mathcal{B}(E,E_1)$ is surjective, then it is an open mapping.

Proof

Known from the proof, an open ball $O_\delta$ is mapped into $T(O_\delta)$, which might be of arbitrary shape, but cover a smaller open ball $Q_{\delta/2}$, which means an interior point is mapped into an interior point.

$\square$

Corollary: inverse mapping theorem

Assume $E$, $E_1$ are Banach space. If a bounded linear operator $T \in \mathcal{B}(E,E_1)$ is surjective and injective, then it is inversible, and the inverse mapping is also a linear bounded operator.

Proof

Since $T$ is bijective, then it has an inverse mapping, denoted $T^{-1}$. By the estimate in Banach's Open mapping theorem, we have

\[ \|T^{-1}y\|=\|x\|\leq K \|y\|, \]

which means $T^{-1}$ is bounded. It is easy to show its linearity.

$\square$

Corollary: Equivalence of norm

We recall that norms $\|\cdot\|_1$, $\|\cdot\|_2$ of $E$ are equivalent iff there exists $K_1,K_2$ such that $\forall x \in E$

\[ K_2\|x\|_1 \leq \|x\|_2 \leq K_1\|x\|_1. \]

by Open mapping theorem, we have the following theorem.

If there exsits $K$ such that

\[ \|x\|_2 \leq K\|x\|_1, \]

then the two norms are equivalent.

Proof

This is by utilizing the inverse mapping theorem. That is, let $I$ to be the identity mapping from $E$ to $E$, with norm $\|\cdot\|_1$ and $\|\cdot\|_2$ respectively.

so $I$ is bijective, by condition $\|x\|_2=\|Ix\|_2\leq K\|x\|_1$, we have $I$ is bounded. So it has an inverse mapping, which is also bounded, i.e. there exists $K'$

\[ \|x\|_1=\|I^{-1}x\|_1\leq K'\|x\|_2. \]

$\square$

Closed image theorem¶

Definition of image

For a linear operator $T\in \mathcal{L}(E,E_1)$, define its direct sum of $E$ and $E_1$, and its norm

\[ \|(x,y)\|=\|x\|+\|y\|,\quad x\in E, y\in E_1. \]

by definition $E \oplus E_1$ is a normed linear space. For a linear operator T which maps $D\subset E$ to $E_1$, we call $G_t:=\{(x,Tx): x\in D\} \subset E\oplus E_1$ to be the image of $T$, and we call $T$ is closed, if $G_T$ is a closed subspace.

Description for closed linear operator

A linear operator $T\in \mathcal{L}(D, E_1)$ is closed, iff for a given $(x,y)\in E\oplus E_1$, there exists sequence $\{x_n\}\subset D$ and $x_n\rightarrow x$, $\{Tx_n\}\rightarrow y$, then $x\in D$ and $Tx=y$.

Proof

Necessary. $G_T$ is closed, so given $x_n\rightarrow x$, $Tx_n \rightarrow y$, $(x_n,Tx_n)\rightarrow (x,y)$, so $(x,y) \in G_T$, so $x\in D$, $y=Tx$.
Sufficient. For $(x,y)\in \overline{G_T}$, $\exists x_n \subset D$, $x_n\rightarrow x$ and $Tx_n\rightarrow y$, by condition, $x\in D$, and $y=Tx$, so $(x,y)\in G_T$.

The following theorem answers when a closed linear operator becomes bounded. In the proof we could check that property closedness is weaker then boundedness.

Closed image theorem

If a linear operator $T$ maps $E$ to $E_1$, where $E$ and $E_1$ are Banach space, then $T$ is bounded iff $T$ is closed.

Proof

Sufficient.

We still make use of inverse mapping theorem. So if $T$ is closed, $G_T$ is by definition a closed subspace of $E \oplus E_1$. Then define projection mapping $\tilde{T}$ from $G_T$ to $E$ to be $\tilde{T}(x, Tx)=x$. $\tilde{T}$ is injective also surjective apparently. Now we show that it is bounded, i.e.

\[ \|\tilde{T}(x, Tx)\|=\|x\|\leq \|x\|+\|Tx\|=\|(x,Tx)\|. \]

so by inverse mapping theorem, $\exists \tilde{T}^{-1}$ such that $(x, Tx)=\tilde{T}^{-1}x$, and bounded $\|(x, Tx)\|\leq K\|x\|$, which means $\|Tx\|\leq K\|x\|$.

Necessary.

We make use of the continuity of $T$ when it is bounded. For $(x,y)\in E\oplus E_1$, $\exists x_n\subset E$, $x_n \rightarrow x$, $Tx_n \rightarrow y$. By continuity of $T$, $Tx_n\rightarrow Tx$, so by uniqueness of limit, $Tx=y$, so $T$ is bounded.

$\square$

Dual Space¶

For a bounded linear operator space $\mathcal{B}(E,E_1)$ is complete, if $E_1$ is complete. So Let $E_1=K$, the number domain, then $\mathcal{B}(E,K)$, space composed by all the bounded linear functional, is naturally complete.

Dual Space

Denote the dual space of $E$ to be $E^*=\mathcal{B}(E,K)$.

Now we first dive into the relationship between $E$ and $E^{**}$.

Assume $x\in E$, $f\in E^*$. Define $x^{**}$ to be $x^{**}(f)=f(x)$, then by linearity on $E^*$, $x^{**}$ is also a linear functional. It is also bounded, since

\[ \begin{align} |x^{**}(f)|=|f(x)|\leq \|x\|\|f\|\label{boudnedness-1} \end{align} \]

so $x^{**}\in E^{**}$. For each $x\in E$, we could get a corresponding $x^{**}$. We could discuss about the space spanned by $x^{**}$.

Properties of $x\mapsto x^{**}$, canonical mapping

(i) Linearity. For $x_1^{**}$, $x_2^{**}$, we have

\[ (\alpha x_1+\beta x_2)^{**}=\alpha x_1^{**} + \beta x_2^{**}. \]

(ii) Isometry. $\|x^{**}\|=\|x\|$, then $x^{**}$ is injective.

Proof

(i) Easy to show.

(ii) From the estimation $\ref{boudnedness-1}$, we have $\|x^{**}\|\leq \|x\|$. Then on the other hand, we choose $f_0\in E^*$, such that

\[ \|f_0\|=1,|f_0(x)|=\|x\|. \]

So the proposition holds.

$\square$

If $T: x\mapsto x^{**}$ is surjective, then we call $E$ is a reflexive Space.

We could say $T(E)$ is a subspace of $E^{**}$, if $T(E)=E^{**}$, then $E$ is a reflexive space.

Example.

(i) The dual space of $L^p[a,b](1 < p < \infty)$ is $L^q[a,b]$.

(ii) The dual space of $l^p(1 < p < \infty)$ is $l^q$.

Proof

For

Example. (Riesz Representation Theorem) The dual space of Hilbert space $\mathscr{U}$ is itself. That is, for any bounded linear functional $f$ on $\mathscr{U}$, there exists a unique element $u\in \mathscr{U}$ such that

\[ f(x)= \langle x,u \rangle, \quad \|f\|=\|u\|. \]

Conversely, for a given $u\in \mathscr{U}$, $\langle x,u \rangle$ defines a linear bounded functional on $\mathscr{U}$, and satisfies the isometric property.

Proof

Sufficient. If $f=\theta$, then choose $x=\theta$ and we are done. Suppose $f\neq \theta$, then define its kernel

\[ L=\{x: f(x)=\theta\}. \]

which is a nonempty, pure and closed subset. (why?)

Choose $x_0\neq \theta\in L^\perp$, and let $\|x_0\|=1$. Then by a lemma, we have a direct product decomposition,

\[ \mathscr{U}=L \oplus \{\alpha x_0\}. \]

which is actually a orthogonal decomposition with $x=y+\langle x, x_0\rangle x_0$ where $\langle x, x_0\rangle x_0\in L^\perp$ is the orthogonal projection of $x$ on $\{alpha x_0\}$. Then apply $f$ we have

\[ f(x)=f(y)+\langle x, x_0\rangle f(x_0)=\langle x, \overline{f}(x_0) x_0\rangle. \]

so choose $u=\overline{f}(x_0)x_0$.

Prove uniqueness.

Assume $u'\in \mathscr{U}$ also satisfies the condition, then $f(x)=\langle x, u'\rangle=\langle x, u\rangle$ which implies $\langle x, u'-u\rangle=0$, so $u'-u=\theta$.

Prove isometry.

By Schwarz inequation we have

\[ |f(x)|=|\langle x, u\rangle|\leq \|x\|\|u\|, \]

so $\|f\|\leq \|u\|$. On the other hand, we have choose $x=u$, we achieve

\[ |f(u)|=|\langle u,u \rangle|=\|u\|\|u\| \]

so $\|f\|\geq \|u\|$.

Prove Necessary.

We only need to prove $f(x)=\langle x,u\rangle$ is a linear bounded operator, which is by the lineariry of inner product and Schwarz inequation.

$\square$

Existence of functional¶

Extension of functional

Assume $E$ is a vector space, and linear functionals $f_1,f_2\in E^*$ defined on subspace $G_1, G_2\subset E$ respectively. If $G_1\subset G_2$ and

\[ f_1(x)=f_2(x),\quad \forall x\in G_1 \]

then we call $f_2$ is an extension of $f_1$.

subadditive and positive-homogeneous functional (SP functional)

A functional $f\in E^*$ is subadditive if for $x,y\in E$

\[ f(x+y)\leq f(x)+f(y). \]

It is positive-homogeneous if for positive real number $\alpha$,

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

Theorem for extension of SP functional

Assume $G$ is a subspace of linear space $E$, $f$ is a real functional on $G$, and $p(x)$ is a SP functional on $E$, and thay satisfies

\[ f(x)\leq p(x),\quad \forall x\in G. \]

Then there exists a real functional $f_0$ on $E$ which is an extension of $f$ and holds the boundedness of $p$, i.e.

\[ f_0(x)\leq p(x),\quad x\in E. \]

Proof

Prove extension of one dimension.

define $G_1=\{\alpha x\}\oplus G$ for $x\in E-G$,

\[ f(y+\alpha x)=f(y)+c\alpha \]

where $c$ satisfies

\[ \sup_{x'\in G} {-p(-x'-x)-f(x')}\leq c\leq \inf_{x''\in G} \{p(x''+x)-f(x'')\} \]

this is because we want to let the following inequation holds

\[ \begin{align*} f(y+\alpha x)&=f(y)+c\alpha \leq p(y+\alpha x)\\ \alpha c&\leq p(y+\alpha x)-f(y) \end{align*} \]

if $\alpha =0$ it holds naturally. If $\alpha>0$, we have to let $c\leq p(y/\alpha +x)-f(y/\alpha)$. If $\alpha<0$, then we have to let $-p(-y/\alpha -x)-f(y/\alpha)\leq c$. The above two gives the estimation. We have to show that the estimation holds because $\forall y_1,y_2\in G$,

\[ \begin{align*} f(y_2)-f(y_1)&=f(y_2-y_1)\leq p(y_2-y_1)\\ &= p(y_2+x-y_1-x)\\ &\leq p(y_2+x)+p(-y_1-x)\\ \Rightarrow \quad -p(-y_1-x)-f(y_1)&\leq p(y_2+x)-f(y_2). \end{align*} \]

Use Zorn's Lemma. It states that a partially ordered set containing upper bound for every totally ordered subset necessarily contains at least one maximal element.

Hahn-Banach Theorem¶

Lemma: Complexification of functional

Assume $f$ is a complex bounded linear functional on $E$, define

\[ \varphi(x)=\text{Re}(f(x)),\quad x\in E, \]

then $\varphi$ is a real bounded linear functional on $E$ and satisfies

\[ f(x)=\varphi(x)-i\varphi(ix). \]

Proof

Assume $f(x)=\varphi(x)+i\psi(x)$, then

\[ \varphi(ix)+i\psi(ix)=f(ix)=if(x)=i(\varphi(x)+i\psi(x))=-\psi(x)+i\varphi(x). \]

so we have $\psi(x)=-\varphi(ix)$. And for boundedness

\[ |\varphi(x)|\leq |f(x)| \]

so $\|\varphi\|\leq \|f\|$.

Hahn-Banach Theorem