<Home>Qualifying Exams><Analysis>Fall 2016 - Problem 6

Fall 2016 - Problem 6

Banach spaces

Consider the Banach space ℓ1\ell^1 consisting of all sequences u={xi}iu = \set{x_i}_i in R\R ()i.e., xi∈Rx_i \in \R for i∈Ni \in \N) with

∥u∥L1=∑i=1∞∣xi∣<∞\norm{u}_{L^1} = \sum_{i=1}^\infty \abs{x_i} < \infty

and the Banach space ℓ∞\ell^\infty consisting of all sequences v={yi}iv = \set{y_i}_i in R\R with

∥v∥L∞=sup⁡i∈N ∣yi∣<∞.\norm{v}_{L^\infty} = \sup_{i \in \N} \,\abs{y_i} < \infty.

There is a well-defined dual pairing between ℓ1\ell^1 and ℓ∞\ell^\infty given by

⟨u,v⟩=∑i=1∞xiyi\inner{u, v} = \sum_{i=1}^\infty x_iy_i

for u={xi}i∈ℓ1u = \set{x_i}_i \in \ell^1 and v={yi}i∈ℓ∞v = \set{y_i}_i \in \ell^\infty. With this dual pairing, ℓ∞=(ℓ1)∗\ell^\infty = \p{\ell^1}^* is the dual space of ℓ1\ell^1.

  1. Show that there exists no sequence {un}n\set{u_n}_n in ℓ1\ell^1 such that (i) ∥un∥1≥1\norm{u_n}_1 \geq 1 for all n∈Nn \in \N and (ii) ⟨un,v⟩→0\inner{u_n, v} \to 0 for each v∈ℓ∞v \in \ell^\infty.
  2. Show that every weakly convergence sequence {un}n\set{u_n}_n in ℓ1\ell^1 converges in the norm topology of ℓ1\ell^1.
Solution.

  1. By normalizing, we will assume without loss of generality that ∥un∥1=1\norm{u_n}_1 = 1 for all n≥1n \geq 1. Observe that if v=ekv = e_k, which is the sequence with a 11 in the kk-th coordinate and 00 everywhere else, then such a sequence satisfies un(k)→0u_n\p{k} \to 0 as n→∞n \to \infty for all kk, i.e., un→0u_n \to 0 pointwise. Intuitively, if we view the sequence un(k)u_n\p{k} as a grid, then most the mass should lie in the diagonal.

    Fix ε>0\epsilon > 0. Since u1∈ℓ1u_1 \in \ell^1, there exists N1∈NN_1 \in \N such that

    ∑k=N1∞∣u1(k)∣≤ε.\sum_{k=N_1}^\infty \abs{u_1\p{k}} \leq \epsilon.

    Similarly, because un(1),…,un(N1)u_n\p{1}, \ldots, u_n\p{N_1} all tend to 00 as n→∞n \to \infty, there exists M1∈NM_1 \in \N such that

    ∑k=1N1−1∣uM1(k)∣≤ε.\sum_{k=1}^{N_1-1} \abs{u_{M_1}\p{k}} \leq \epsilon.

    Inductively pick Ni,MiN_i, M_i so that Ni<Ni+1N_i < N_{i+1}, Mi<Mi+1M_i < M_{i+1},

    ∑k=Ni∞∣uMi−1(k)∣≤εand∑k=1Ni−1∣uMi(k)∣≤ε,\sum_{k=N_i}^\infty \abs{u_{M_{i-1}}\p{k}} \leq \epsilon \quad\text{and}\quad \sum_{k=1}^{N_i-1} \abs{u_{M_i}\p{k}} \leq \epsilon,

    which gives

    ∑k=NiNi+1−1∣uMi(k)∣=∥uMi∥1−∑k=1Ni−1∣uMi(k)∣−∑k=Ni+1∞∣uMi(k)∣≥1−2ε,\begin{aligned} \sum_{k=N_i}^{N_{i+1}-1} \abs{u_{M_i}\p{k}} &= \norm{u_{M_i}}_1 - \sum_{k=1}^{N_i-1} \abs{u_{M_i}\p{k}} - \sum_{k=N_{i+1}}^\infty \abs{u_{M_i}\p{k}} \\ &\geq 1 - 2\epsilon, \end{aligned}

    i.e., NiN_i and MiM_i are chosen so that most of the mass of uMiu_{M_i} is concentrated in Ni≤k<Ni+1N_i \leq k < N_{i+1}.

    Define v(k)=sgn⁡(u1(k))v\p{k} = \sgn\p{u_1\p{k}} for 1≤k<N11 \leq k < N_1, and v(k)=sgn⁡(uMi−1(k))v\p{k} = \sgn\p{u_{M_{i-1}}\p{k}} for Ni−1≤k<NiN_{i-1} \leq k < N_i, for i≥2i \geq 2. Then

    ∣⟨uMi,v⟩∣=∣∑k=1Ni−1uMi(k)v(k)+∑k=NiNi+1−1∣uMi(k)∣+∑k=Ni+1∞uMi(k)v(k)∣≥∑k=NiNi+1−1∣uMi(k)∣−∣∑k=1Ni−1uMi(k)v(k)+∑k=Ni+1∞uMi(k)v(k)∣≥(1−2ε)−∑k=1Ni−1∣uMi(k)∣−∑k=Ni+1∞∣uMi(k)∣≥1−4ε.\begin{aligned} \abs{\inner{u_{M_i}, v}} &= \abs{\sum_{k=1}^{N_i-1} u_{M_i}\p{k} v\p{k} + \sum_{k=N_i}^{N_{i+1}-1} \abs{u_{M_i}\p{k}} + \sum_{k=N_{i+1}}^\infty u_{M_i}\p{k}v\p{k}} \\ &\geq \sum_{k=N_i}^{N_{i+1}-1} \abs{u_{M_i}\p{k}} - \abs{\sum_{k=1}^{N_i-1} u_{M_i}\p{k} v\p{k} + \sum_{k=N_{i+1}}^\infty u_{M_i}\p{k}v\p{k}} \\ &\geq \p{1 - 2\epsilon} - \sum_{k=1}^{N_i-1} \abs{u_{M_i}\p{k}} - \sum_{k=N_{i+1}}^\infty \abs{u_{M_i}\p{k}} \\ &\geq 1 - 4\epsilon. \end{aligned}

    Thus, if we set ε<14\epsilon < \frac{1}{4}, then ⟨un,v⟩\inner{u_n, v} cannot converge to 00.

  2. Suppose otherwise, and that there is a sequence {un}n⊆ℓ1\set{u_n}_n \subseteq \ell^1 which converges weakly to some u∈ℓ1u \in \ell^1 but does not converge in the norm topology. Then there exists some ε>0\epsilon > 0 such that

    ∥un−u∥ℓ1≥ε  ⟹  ∥un−uε∥≥1.\norm{u_n - u}_{\ell^1} \geq \epsilon \implies \norm{\frac{u_n - u}{\epsilon}} \geq 1.

    But by weak convergence,

    ⟨un−uε,v⟩→n→∞0,\inner{\frac{u_n - u}{\epsilon}, v} \xrightarrow{n\to\infty} 0,

    which contradicts (1). Hence, no such sequence could have existed to begin with.