Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 22H (Linear Analysis)

The real Riesz representation theorem states that for every bounded linear functional $f:H\to \mathbb{R}$ on a real Hilbert space there is a unique $y\in H$ such thatand $\Vert f\Vert =\Vert y\Vert$.

If $f=0$, take $y=0$. Otherwise $Y=\ker f$ is a closed proper subspace, so the supplied orthogonal decomposition by a closed subspace gives a nonzero $z\in Y^{\perp }$. Since $f(z)\ne 0$, for every $x\in H$,Taking its inner product with $z$ yieldsand thereforeThus $y=f(z)z/\Vert z{\Vert }^2$ represents $f$. If both $y$ and $y^{\prime }$ represent it, then $⟨x,y-y^{\prime }⟩=0$ for every $x$; taking $x=y-y^{\prime }$ proves uniqueness. Finally, Cauchy-Schwarz inequality gives $\Vert f\Vert \le \Vert y\Vert$, while evaluating at $x=y/\Vert y\Vert$ when $y\ne 0$ gives the reverse inequality. Hence

If $T$ is invertible, the bounded inverse theorem makes $T^{-1}$ bounded, andMoreover $T^{\ast }$ is invertible with inverse $(T^{-1}{)}^{\ast }$, so the same argument shows that $T^{\ast }$ is bounded below.

Conversely, supposefor some $c,d>0$. The first bound makes $T$ injective. It also makes $\mathrm{ran}T$ closed: if $T{x}_n$ is Cauchy, then $x_n$ is Cauchy, and continuity gives its limit in the range. The second bound gives $\ker T^{\ast }=\{0\}$. By image-kernel orthogonality for an adjoint,so the range is dense. Being both closed and dense, it equals $H$. Thus $T$ is bijective, and the first lower bound gives $\Vert T^{-1}\Vert \le c^{-1}$. Therefore

For $x\in X$,The map $S$ defined by $(Sx{)}_n=x_{n-1}$ satisfies $ST=TS=I$, so $T$ is a surjective isometry and hence a unitary operator.

If $Tx=\lambda x$, thenFor $\lambda \ne 0$, this gives $x_n={\lambda }^n{x}_0$ for every integer $n$. When $|\lambda |>1$ the positive tail is not square summable, when $|\lambda |<1$ the negative tail is not square summable, and when $|\lambda |=1$ neither tail is square summable unless $x_0=0$. The case $\lambda =0$ also forces every coordinate to vanish. Thus

The reverse triangle inequality gives, for every $x$,Hence $T-\lambda I$ is bounded below whenever $|\lambda |\ne 1$, so such $\lambda$ do not belong to the approximate point spectrum.

Now let $|\lambda |=1$ and define the unit vectorThe equation $x_{n+1}^{(N)}=\lambda x_n^{(N)}$ fails only at the two endpoints, soTherefore the bilateral shift operator has