Past exam of the analysis course of the University of Oxford / 2010 / A

For $\alpha ,\beta \in {\ell }_{1,2}$ and real scalars $c,d$, the scalar triangle inequality givesThe zero sequence belongs to the space, so it is a real vector space. The weighted sum is nonnegative and homogeneous, and the same inequality with $c=d=1$ proves its triangle inequality. Since every weight $j^2$ is strictly positive, zero weighted sum forces every coordinate to vanish. Therefore $\Vert \cdot {\Vert }_{1,2}$ is a norm.

Define $U:{\ell }_{1,2}\to {\ell }_1$ by $(U\alpha {)}_j=j^2{\alpha }_j$. This is a linear bijection, with inverse $(U^{-1}\gamma {)}_j=j^{-2}{\gamma }_j$, andThus $U$ is an isometric isomorphism of normed spaces onto the absolutely summable sequence space, which is a Banach space. Explicitly, a Cauchy sequence ${\alpha }^{(n)}$ maps to a Cauchy sequence $U{\alpha }^{(n)}\to \gamma \in {\ell }_1$; setting $\alpha =U^{-1}\gamma$ gives convergence in the weighted norm. Hence ${\ell }_{1,2}$ is a Banach space, an instance of weighted summable sequence space isometry.

Let $D$ be the set of finitely supported sequences with rational coordinates. It is countable: it is the union over $N$ of copies of ${\mathbb{Q}}^N$. Given $\alpha \in {\ell }_{1,2}$ and $\epsilon >0$, choose $N$ with ${\sum }_{j>N}{j}^2|{\alpha }_j|<\epsilon /2$. Choose rational numbers $q_j$ for $1\le j\le N$ so that ${\sum }_{j=1}^N{j}^2|q_j-{\alpha }_j|<\epsilon /2$, and put $q_j=0$ thereafter. Then $q\in D$ and $\Vert q-\alpha {\Vert }_{1,2}<\epsilon$. Thus $D$ is countable and dense, so the space is separable.

For the first linear map, take $(\alpha ,\beta )=(0,e_n)$. Its input norm is one, while $T_1(0,e_n)=(e_n,0)$ has norm $n^2$. Thus $T_1$ is unbounded.

For the second map, since $j^2\ge 1$,Equality holds at $(0,e_1)$, soBoth formulas map $c_{00}\times c_{00}$ into itself because adding two finitely supported sequences preserves finite support.

Equip the continuous dual space $X^{\prime }$ with the operator normContinuity of a linear functional makes this finite; homogeneity, definiteness and the triangle inequality follow directly from the supremum formula.

For completeness of the dual space, let $(f_n)$ be Cauchy in this norm. For each $x\in X$, $|f_n(x)-f_m(x)|\le \Vert f_n-f_m\Vert \Vert x\Vert$, so $(f_n(x))$ is Cauchy in $\mathbb{R}$. Define $f(x)={\lim }_n{f}_n(x)$. Passing to the limit in the linearity identities makes $f$ linear. Since a Cauchy sequence is bounded, ${\sup }_n\Vert f_n\Vert \le C$ and $|f(x)|\le C\Vert x\Vert$, so $f\in X^{\prime }$.

Given $\epsilon >0$, choose $N$ with $\Vert f_n-f_m\Vert \le \epsilon$ for $n,m\ge N$. Let $m\to \infty$ pointwise; then $|f_n(x)-f(x)|\le \epsilon \Vert x\Vert$ and therefore $\Vert f_n-f\Vert \le \epsilon$. Thus $X^{\prime }$ is a Banach space. This proof in fact works even if $X$ itself is not complete.

If a bilinear map is continuous, continuity at $(0,0)$ gives $\delta >0$ such that $\Vert F(x,y)\Vert \le 1$ whenever $\Vert x\Vert ,\Vert y\Vert <\delta$. For nonzero $x,y$, apply this to $\delta x/(2\Vert x\Vert )$ and $\delta y/(2\Vert y\Vert )$. Bilinearity givesIf either input is zero, the inequality follows from bilinearity directly.

Conversely, suppose the product bound holds with $C\ge 0$. At a fixed point $(x,y)$,Its norm is at most $C(\Vert h\Vert \Vert y\Vert +\Vert x\Vert \Vert k\Vert +\Vert h\Vert \Vert k\Vert )$, which tends to zero as $(h,k)\to (0,0)$. Thus the product bound is equivalent to joint continuity. This is boundedness criterion for a bilinear map.

Put $\rho =r-\sqrt{r^2-1}$. Since $r>1$, $0<\rho <1$ and ${\rho }^2-2r\rho +1=0$. The nonzero square-summable sequence ${\beta }_j={\rho }^{j-1}$ defines a continuous linear functional on ${\ell }_2$ byThe Cauchy-Schwarz inequality proves continuity. For every $k$,Hence $f$ vanishes on the closed linear span $M$, but $f(e_1)=1$. ThereforeMore precisely, any vector in the orthogonal complement of $M$ satisfies the recurrence ${\beta }_{k+2}-2r{\beta }_{k+1}+{\beta }_k=0$. Its roots are $r\pm \sqrt{r^2-1}$, one less than one and the other greater than one. Square summability eliminates the growing solution, so $M^{\perp }$ is exactly the span of $({\rho }^{j-1})$. Thus $M=\ker f$ is a closed hyperplane of codimension one. This is annihilator of a second-difference span in l2.

The spectrum of a bounded operator is $\sigma (T)=\{\lambda \in \mathbb{C}:T-\lambda I\text{ has no bounded inverse}\}$. The point spectrum consists of those $\lambda$ for which $\ker (T-\lambda I)\ne 0$, namely the eigenvalues. The spectral radius is $r(T)={\sup }_{\lambda \in \sigma (T)}|\lambda |$.

If $\Vert T\Vert <1$, the Neumann series $S={\sum }_{n=0}^{\infty }(-T{)}^n$ converges in the Banach space of bounded linear operators, since its term norms are bounded by a convergent geometric series. Multiplying the finite sums on either side by $I+T$ gives $I-(-T{)}^{N+1}$, which tends in operator norm to $I$. Therefore

If $|\lambda |>\Vert T\Vert$, write $T-\lambda I=-\lambda (I-T/\lambda )$. The Neumann series applies to $T/\lambda$, so $\lambda$ is outside the spectrum of a bounded operator. Thus $\sigma (T)\subseteq \{\lambda :|\lambda |\le \Vert T\Vert \}$ and $r(T)\le \Vert T\Vert$.

The polynomial spectral mapping theorem gives $\sigma (T^m)=\{{\lambda }^m:\lambda \in \sigma (T)\}$, and henceTaking $m$th roots and then the limit, whose existence is allowed, provesOnly this inequality is required; the reverse inequality is the additional content of the full spectral radius formula.

For the multiplication operator, $\Vert Tf{\Vert }_{\infty }\le \Vert g{\Vert }_{\infty }\Vert f{\Vert }_{\infty }=\Vert f{\Vert }_{\infty }$. Taking $f=1$ gives equality, so $\Vert T\Vert =1$.

An eigenfunction obeys $(g(t)-\lambda )f(t)=0$ at every $t$. If $\lambda \in [0,1)$, its level set is the single point $t=\lambda$, and a continuous function supported at a single point is zero. If $\lambda \notin [0,1]$, that level set is empty. For $\lambda =1$, every continuous function vanishing on $[0,1]$ is an eigenfunction, and nonzero examples exist, such as $f(t)=\max \{t-1,0\}$. Thus ${\sigma }_p(T)=\{1\}$.

If $\lambda \notin [0,1]$, multiplication by $(g-\lambda {)}^{-1}$ is a bounded inverse of $T-\lambda I$. If $\lambda \in [0,1]$, choose $t_0$ with $g(t_0)=\lambda$. Every function in the range of $T-\lambda I$ vanishes at $t_0$, so the range omits the constant function one. Hence it is not onto. Therefore

The unilateral shift operator $R$ is not onto, since its outputs have first coordinate zero. The left shift operator $L$ is not injective, since $L{e}_1=0$. Both have operator norm one, and $LR=I$. ConsequentlyIf $\Vert A\Vert <1$, then $\Vert LA\Vert <1$, so the Neumann series makes $I+LA$ invertible. If $R+A$ were invertible too, the identity above would givea product of invertible operators, contradicting the nontrivial kernel of $L$. ThusThis is norm stability of noninvertibility for the unilateral right shift.