<Home>Qualifying Exams><Analysis>Spring 2019 - Problem 6

Spring 2019 - Problem 6

Hilbert spaces

Let H\mathcal{H} be a Hilbert space and {ξn}n\set{\xi_n}_n a sequence of vectors in H\mathcal{H} such that ∥ξn∥=1\norm{\xi_n} = 1 for all nn.

  1. Show that if {ξn}n\set{\xi_n}_n converges weakly to a vector ξ∈H\xi \in \mathcal{H} with ∥ξ∥=1\norm{\xi} = 1, then lim⁡n→∞∥ξn−ξ∥=0\lim_{n\to\infty} \norm{\xi_n - \xi} = 0.
  2. Show that if lim⁡n,m→∞∥ξn+ξm∥=2\lim_{n,m\to\infty} \norm{\xi_n + \xi_m} = 2, then there exists a vector ξ∈H\xi \in \mathcal{H} such that lim⁡n→∞∥ξn−ξ∥=0\lim_{n\to\infty} \norm{\xi_n - \xi} = 0.
Solution.

  1. Notice that

    ∥ξn−ξ∥2=∥ξn∥2+∥ξ∥2−2Re⁡⟨ξn,ξ⟩=2−2Re⁡⟨ξn,ξ⟩,\begin{aligned} \norm{\xi_n - \xi}^2 &= \norm{\xi_n}^2 + \norm{\xi}^2 - 2\Re{\inner{\xi_n, \xi}} \\ &= 2 - 2\Re{\inner{\xi_n, \xi}}, \end{aligned}

    so by weak convergence,

    lim⁡n→∞ ∥ξn−ξ∥2=lim⁡n→∞ (2−2Re⁡⟨ξn,ξ⟩)=2−2∥ξ∥2=0.\begin{aligned} \lim_{n\to\infty} \,\norm{\xi_n - \xi}^2 &= \lim_{n\to\infty} \,\p{2 - 2\Re{\inner{\xi_n, \xi}}} \\ &= 2 - 2\norm{\xi}^2 \\ &= 0. \end{aligned}

  2. We have

    ∥ξn+ξm∥2=2+2Re⁡⟨ξn,ξm⟩\norm{\xi_n + \xi_m}^2 = 2 + 2\Re{\inner{\xi_n, \xi_m}}

    and so

    ∥ξn−ξm∥2=2−(∥ξn+ξm∥2−2)=4−∥ξn+ξm∥2→n,m→∞0.\begin{aligned} \norm{\xi_n - \xi_m}^2 &= 2 - \p{\norm{\xi_n + \xi_m}^2 - 2} \\ &= 4 - \norm{\xi_n + \xi_m}^2 \xrightarrow{n,m\to\infty} 0. \end{aligned}

    Thus, {ξn}n\set{\xi_n}_n is Cauchy, so it converges to some ξ∈H\xi \in \mathcal{H} by completeness.