$\newcommand{\L}{\mathcal{L}}$ $\renewcommand{\C}{\mathbb{C}}$ $\newcommand{\R}{\mathbb{R}}$ # Lecture 4: Dual Space, Adjoint, Convergence of Bounded Operators ###### tags: 224a 2020 ## Dual Space A *linear functional* on $H$ is a linear map $\ell:H\rightarrow \mathbb{C}$. The *norm* of a linear functional is $$\|\ell\| = \sup_{x\in H\setminus \{0\}} \|\ell(x)|/\|x\|,$$ and it is called bounded if it has finite norm. **Proposition.** A linear function is continuous (with respect to convergence in $H$) if and only if it is bounded. (proof on the homework) Note that whenever we mention linear functions that are continuous, we must be mindful of which notion of convergence they are continuous with respect to; we will consider various other notions later in the course. Though we will not explicitly mention topological spaces, they are essentially an abstract way of specifying a notion of convergence. The set of bounded linear functions on $H$ forms a vector space with respect to pointwise addition and multiplication, caled the dual space $H^*$. In fact, there is an isometry (norm-preserving bijection) between $H$ and $H^*$. **Theorem.**(Riesz Lemma) For every $\ell\in H^*$ there is a unique $y\in H$ such that $\ell(x)=(y,x)$ for all $x\in H$. This map is an *isometry*, i.e., a bijection satisfying $\|\ell\|_{H^*}=\|y\|_H$. The proof is given in Reed and Simon Theorem II.4. In particular, this implies that $H^*$ is also a Hilbert space. We now begin to develop the spectral theory of bounded operators on Hilbert space. Besides giving a great deal of insight into the structure of these operators, it is an indispensible tool in solving "infinite dimensional linear equations" (such as differential equations), due to the absence of algorithms like Gaussian elimination. ## Adjoints and Positivity We will use $\L(H)$ to denote bounded operators from a Hilbert space to itself. **Theorem.** If $A\in \L(X,Y)$ for complex separable Hilbert spaces $X,Y$, there is a unique operator $A^*\in\L(Y,X)$ called its *adjoint* which satisfies $$(y,Ax)_Y =(A^*y,x)_X\quad (*)$$ for all $x\in X, y\in Y$. *Proof.* Given $y\in Y$, consider the linear functional $\ell_{A^*y}(x):=(y,Ax)$ on $X$. This is bounded by Cauchy-Schwartz, so by the Riesz theorem there is a vector $z$ such that $\ell_{A^*y}(x)=(z,x)$ for all $x\in X$; define $A^*y=z$ for this $z$. By construction, the property $(*)$ is satisfied for all $x\in X$. Linearity is then implied by the properties of the inner product, as $$(A^*(ay+by'),x) = (ay+by',Ax) = \bar{a}(y,Ax)+\bar{b}(y',Ax)$$ $$= \bar{a}(A^*y,x)+\bar{b}(A^*y',x)=(aA^*y+bA^*y',x)$$ for all $x\in X$, so by uniqueness in the Riesz theorem we must have $A^*(ay+by')=aA^*y+bA^*y'$. Boundedness follows by observing that $$\|A^*y\|^2=|(A^*y,A^*y)|=\|(y,AA^*y)|\le \|y\|\|A\|\|A^*y\|$$ for all $y$, whence $\|A^*\|\le \|A\|$. Uniqueness of $A^*$ follows because if $(Cy,x)=(Dy,x)$ for all $x,y$ then we must have $Cy=Dy$ for all $y$ or $C=D$. $\square$ The properties of the adjoint operation with respect to other operations like addition, multiplication, and norm are listed in Reed and Simon Theorem VI.3. An operator $A\in \L(H)$ is called *self-adjoint* if $A=A^*$. Self-adjointness is an important property both from a mathematical (becuase it allows a much better understanding of the spectrum) and physical (because such operators correspond to observables in quantum mechanics) point of view. A homework problem asks you to prove that $$\|A\|=\sup_{\|x\|=1} (x,Ax)$$ whenever $A=A^*$. An operator $A\in \L(H)$ is called *positive*, denoted $A\ge 0$, if $(x,Ax)\ge 0$ for all $x\in H$. On the homework it is shown in a complex Hilbert space, a positive operator is always self-adjoint. ## Norms and Convergence A linear transformation $T:X\rightarrow Y$ for Banach spaces $X,Y$ is called *bounded* if there is a constant $C$ such that $$\|Tx\|_Y\le C\|x\|_X\quad\forall x\in X,$$ and the least such constant is called its *operator norm* $\|T\|$. We will denote the set of bounded operators $X\rightarrow Y$ by $\L(X,Y)$. It is shown on the homework that an operator $T$ is continuous with respect to convergence in norm if and only if it is bounded. The fact that $\|T\|$ satisfies the triangle inequality follows from linearity and the triangle inequality for $\|\cdot\|_Y$. Here are some examples of bounded operators: 1. $X=Y=\ell^2(\mathbb{Z})$ with $(Tx)_n = x_{n-1} (shift operator). Check that$\|Tx\|=\|x\|$for all$x$. 2.$X=L^1(\R), Y=\C$and$Tf = \int_{\R} f(x)dx.$The triangle inequality shows that$\|T\|\le 1$and this is achieved for$f$any nonnegative function. 3.$X=Y=L^2(0,1)$,$(Vf)(x):=\int_0^x f(y)dy,$the *Volterra operator*. A simple Cauchy-Schwartz argument reveals that the norm is bounded by$1/\sqrt{2}$. The true answer is$2/\pi,$though this is harder to show. 4. Any bounded linear functional$\ell_y:H\rightarrow \C$. By the Riesz theorem this is always equal to$(y,x)$for some$y$, and it is easy to see that$\|\ell_y\|=\|y\|$. We will often define operators in$\L(X,Y)$as absolutely convergent power series in other operators, so it is important to know that this space contains enough limit points. Let$T_n\rightarrow T$denote$\|T_n-T\|\rightarrow 0$as$n\rightarrow \infty$. **Theorem.** If$X$is a normed linear space and$Y$is a Banach space,$\L(X,Y)$is complete with respect to the operator norm. *Proof.* Suppose$T_n$is a sequence in$\L(X,Y)$satisfying$\|T_n-T_m\|\rightarrow 0$as$n,m\rightarrow\infty$. For a fixed vector$x\in X$, the sequence$T_nx\in Y$satisfies$\|T_nx-T_mx\|\le \|T_n-T_m\|\|x\|\rightarrow 0$so it must have a limit point; define$Tx$to be equal to this limit point. Linearity of$T$can be checked easily. We will now show$\|T_n-T\|\rightarrow 0$. Fix$\epsilon>0$, and choose$n$such that$\|T_n-T_m\|\le \epsilon/2$whenever$m/ge n$. Let$\|x\|=1$and observe that for every$m\ge n$we have $$\|T_nx-Tx\|\le \|T_n x-T_mx\|+\|T_mx-Tx\|\le \epsilon/2+\epsilon/2,$$ by taking$m$sufficiently large and using$\|T_mx-Tx\|\rightarrow 0$(note that$m$can depend on$x$). Since$n$depended only on$\epsilon$we have shown that$\sup_{\|x\|=1} \|T_nx-T_x\|<\epsilon$, as desired. To see that$T$is bounded, observe that for every$n$one has$\|T\|\le \|T_n-T\|+\|T_n\|$, and choose$n$such that$\|T_n-T\|\le 1$(say).$\square$Besides convergence in norm, there are two other useful notions of convergence for sequences of operators, called *strong* and *weak* convergence; for definitions and examples, see Sec VI.1 of Reed and Simon. Roughly speaking, weak convergence corresponds to "entrywise" convergence of matrices, and strong convergence corresponds to "columnwise" convergence. For the purposes of this course, the word "topology" just means notion of convergence. As seen from the examples, these notions of convergence are very different, and it is important to be mindful of which one is meant. We remark that$\L(X,Y)\$ is also closed with respect to weak convergence, and this is proven in Reed and Simon Theorem VI.1.