.. _2004Apr19:

.. 2004 Apr 19
.. ===========

Exam 8 (57--63)
========================

2004 Apr
---------

**Instructions** Use a separate sheet of paper for each new problem.  Do as many problems as you can. Complete solutions to five problems will be considered as an excellent performance. Be advised that a few complete and well written solutions will count more than several partial solutions.

**Notation**: :math:`f ∈ C(X)` means that :math:`f` is a real-valued, :term:`continuous function` defined on :math:`X`.

.. _2004Apr_p1:

.. proof:prob::

   a. Let :math:`S` be a :term:`Lebesgue measurable set` in :math:`ℝ` and let :math:`f, g: S → ℝ` be :term:`measurable functions <measurable function>`. Prove

      #. :math:`f+g` is measurable and 
      #. if :math:`φ ∈ C(ℝ),` then :math:`φ(f)` is measurable.

   b. Let :math:`f: [a,b] → [-∞, ∞]` be a :term:`measurable function`. Suppose that :math:`f` takes the value :math:`± ∞` only on a set of :term:`Lebesgue measure` zero. Prove that for every :math:`ε > 0` there is a positive number :math:`M` such that :math:`|f| ≤ M,` except on a set of measure less than :math:`ε`.

   .. index:: Egoroff's theorem, Fatou's lemma, dominated convergence theorem

.. _2004Apr_p2:

.. proof:prob::

   a. State :term:`Egoroff's theorem`.
   b. State :term:`Fatou's lemma`.
   c. Let :math:`\{f_n\} ⊂ L_p[0,1],` where :math:`1 ≤ p < ∞`. Suppose that :math:`f_n → f` a.e., where :math:`f ∈ L_p[0,1]`. Prove that :math:`\|f_n - f\|_p → 0` if and only if :math:`\|f_n\|_p → \|f\|_p`.

.. index:: Egoroff's theorem, Minkowski's inequality, Hölder's inequality

.. _2004Apr_p3:

.. proof:prob::

   a. Let :math:`S = [0,1]` and let :math:`\{f_n\} ⊂ L_p(S),` where :math:`1< p< ∞`. Suppose that :math:`f_n → f` a.e. on :math:`S,` where :math:`f ∈ L_p(S)`. If there is a constant :math:`M` such that :math:`\|f_n\|_p ≤ M` for all :math:`n,` prove that for each :math:`g ∈ L_q(S), \frac{1}{p} + \frac{1}{q} = 1,` we have

      .. math:: \lim_{n\to \infty}\int_S f_n g = \int_S fg.

   b. Show by means of an example that this result is false for :math:`p=1`.

      .. index:: closed graph theorem

.. _2004Apr_p4:

.. proof:prob::

   State and prove the :term:`closed graph theorem`.

   .. index:: Hilbert space, Banach space, separable Hilbert space, normed linear space

.. _2004Apr_p5:

.. proof:prob::

   Prove or disprove:

   a. For :math:`1≤ p< ∞`, let :math:`ℓ_p := \{x = \{x_k\} ∣ \|x\|_p = (∑_{k=1}^∞ |x_k|^p)^{1/p} < ∞\}`. Then for :math:`p ≠ 2,` :math:`ℓ_p` is a :term:`Hilbert space`.

   b. Let :math:`X = (C[0,1], \| \cdot \|_1),` where the linear space :math:`C[0,1]` is endowed with the :math:`L_1`-norm: :math:`\|f\|_1 = ∫_0^1 |f(x)|\,dx`. Then :math:`X` is a :term:`Banach space`.

   c. Every real, :term:`separable` :term:`Hilbert space` is :term:`isometrically isomorphic` to :math:`ℓ_2`.

   .. index:: Fubini's theorem, Borel (-measurable) function

.. _2004Apr_p6:

.. proof:prob::

   a. Give a precise statement of some version of :term:`Fubini's theorem` that is valid for nonnegative functions.

   b. Let :math:`f,g ∈ L_1(ℝ)`.

      #. Prove that the integral

         .. math:: h(x) = ∫_ℝ f(x-t) g(t)\, dt

         exists for almost all :math:`x ∈ ℝ` and that :math:`h ∈ L_1(ℝ)`.

      #. Show that :math:`\|h\|_1 ≤ \|f\|_1 \|g\|_1`.

.. index:: Radon-Nikodym theorem

.. _2004Apr_p7:

.. proof:prob::

   a. State the :term:`Radon-Nikodym theorem`.

   b. Let :math:`(X, ℬ(X), μ)` be a :term:`complete measure space`, where :math:`μ` is a :term:`positive measure` defined on the :term:`σ-algebra` :math:`ℬ(X)` of :term:`Borel sets <Borel set>` of :math:`X`.

      Suppose :math:`μ X < ∞`. Let :math:`S` be a :term:`closed set` in :math:`ℝ` and let :math:`f ∈ L_1(μ),` where :math:`f: X → [-∞, ∞]` is an extended real-valued function defined on :math:`X`.

      If

      .. math:: A_E(f) = \frac{1}{μ E}∫_E f\, dμ ∈ S

      for every :math:`E ∈ ℬ(X)` with :math:`μ E > 0,` prove that :math:`f(x) ∈ S` for almost all :math:`x ∈ X`.

---------------------------------------------------

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      **Solution** to :numref:`Problem {number} <2004Apr_p1>`

   a. **Proof 1**. Since :math:`f` and :math:`g` are real measurable functions of :math:`S`, and since the mapping :math:`Φ:ℝ × ℝ → ℝ` defined by :math:`Φ(x,y) = x+y` is a :term:`continuous function`, :ref:`Theorem Ru.1.8 <Ru.1.8>` implies that the function :math:`f+g = Φ(f,g)` is :term:`measurable <measurable function>`.
   
        If :math:`φ ∈ C(ℝ),` then :math:`φ(f)` is :term:`measurable <measurable function>` by part b. of :ref:`Theorem Ru.1.7 <Ru.1.7>`. ☐

      **Proof 2**. Let :math:`\{q_i\}_{i=1}^∞` be an enumeration of the rationals. Then, for every :math:`α ∈ ℝ`,

       .. math:: \{x ∈ S ∣ f(x)+g(x) < α\}  = ⋃_{i=1}^∞  \{x ∈ S ∣ f(x) < α - q_i\} ∩ \{x ∈ S ∣ g(x) < q_i\}.

       Since each set on the right is measurable, and since :term:`σ-algebras <σ-algebra>` are closed under countable unions and arbitrary intersections, :math:`\{x ∈ S ∣ f(x)+g(x) < α\}` is a :term:`measurable set`. Since :math:`α` was arbitrary, :math:`f+g` is a :term:`measurable function`.

       The function :math:`φ f` is measurable if and only if for every open subset :math:`U` of :math:`ℝ` the set :math:`(φ f)^{-1}(U)` is measurable.

       Let :math:`U` be open in :math:`ℝ`. Then :math:`φ^{-1}(U)` is open, since :math:`φ \in C(ℝ)`, and so :math:`(φ f)^{-1}(U) = f^{-1}(φ^{-1}(U))` is measurable, since :math:`f` is measurable. Therefore, :math:`φ f` is measurable. ☐

   b. Fix :math:`ε > 0`. For :math:`n ∈ ℕ`, define :math:`A_n = \{x ∈ [a,b] : |f(x)| ≤ n\}`. Then

      .. math:: [a,b] = ⋃_{n=1}^{∞} A_n ∪ A_{∞},
         :label: 2004-b1

      where :math:`A_∞ := \{x: f(x) = ± ∞\}`. [1]_

      Also, :math:`A_1⊆ A_2 ⊆ \cdots`, and since :math:`f` is measurable each :math:`A_n` is measurable.

      Therefore, :math:`μ A_n → μ(⋃_n A_n),` as :math:`n → ∞`. Note that all sets are contained in :math:`[a,b]` and thus have finite measure.

      Let :math:`M ∈ ℕ` be such that :math:`μ(⋃_n A_n) - μ A_M < ε`. Then :math:`|f| ≤ M` except on :math:`[a,b]- A_M,` and by :eq:`2004-b1`,

      .. math:: μ([a,b]- A_M) &= μ(⋃_n A_n ∪ A_∞ - A_M)\\
                              &≤ μ(⋃_n A_n - A_M) + μ A_∞ \\
                              &= μ(⋃_n A_n - A_M) < ε.

      The second equality holds since we assumed :math:`f(x) = ± ∞` only on a set of measure zero; i.e.,\ :math:`μ A_∞ = 0.` ☐

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      **Solution** to :numref:`Problem {number} <2004Apr_p2>`

   a. See :term:`Egoroff's theorem`.
   b. See :term:`Fatou's lemma`.
   c. (⇒) By :term:`Minkowski's inequality`,
   
        .. math:: \|f_n\|_p  = \|f_n-f + f\|_p ≤ \|f_n-f\|_p + \|f\|_p.

        Similarly, :math:`\|f\|_p  ≤ \|f_n-f\|_p + \|f_n\|_p`.

        Together, the two inequalities yield :math:`\bigl| \|f_n\|_p-\|f\|_p \bigr| ≤ \|f_n-f\|_p`. 
         
        Therefore, :math:`\|f_n-f\|_p → 0` implies:math:`\bigl| \|f_n\|_p-\|f\|_p \bigr| → 0`. This proves necessity.

      (⇐) There are at least three proofs of sufficiency.
      
        The second is similar to the first, only much shorter as it exploits the full power of the general version of Lebesgue's :term:`dominated convergence theorem`, whereas the first proof merely relies on :term:`Fatou's lemma`. [2]_

        The third proof uses :term:`Fatou's lemma` and :term:`Egoroff's theorem`. [3]_

        Note that none of the proofs uses the assumption that the measure space is finite, so we may as well work in the more general space :math:`L_p(X, 𝔐, μ)`. Both Proofs 1 and 2 make use of the following basic fact:

        **Lemma**. If :math:`α, β ∈ [0,∞)` and :math:`1 ≤ p < ∞`, then :math:`(α + β)^p ≤ 2^{p-1}(α^p + β^p)`.

        When :math:`p ≥ 1`, :math:`φ x = x^p` is convex on :math:`[0, ∞)`. Thus, for all :math:`α, β ∈ [0,∞)`,

        .. math:: \left(\frac{α+β}{2}\right)^p = φ\left(\frac{α+β}{2}\right) ≤ \frac{1}{2}[φ(α) + φ(β)] = \frac{1}{2}(α^p + β^p).

        When :math:`α, β \in ℝ`, the triangle inequality followed by the lemma yields

        .. math:: |α - β|^p ≤ \bigl||α| + |\beta|\bigr|^p ≤ 2^{p-1}\bigl(|α|^p + |β|^p\bigr).
           :label: convex
  
        ---

        **Proof 1**. By :eq:`convex`, :math:`|f_n - f|^p ≤ 2^{p-1}(|f_n|^p + |f|^p)`. In particular, :math:`f_n - f ∈ L_p`, for each :math:`n ∈ ℕ`. Moreover, the functions

        .. math:: g_n =  2^{p-1}(|f_n|^p + |f|^p) - |f_n - f|^p.
           :label: gnn

        are nonnegative.
         
        Now notice that :math:`\lim \inf g_n = 2^p |f|^p`. Applying :term:`Fatou's lemma` to :eq:`gnn`, then,

        .. math:: ∫ 2^p|f|^p = ∫ \lim \inf g_n ≤ \lim \inf ∫ g_n = \lim \inf ∫ \bigl\{2^{p-1}(|f_n|^p + |f|^p) - |f_n - f|^p\bigr\}.

        Since :math:`\|f_n\|_p → \|f\|_p`, this implies

        .. math:: 2^p ∫ |f|^p ≤ 2^p ∫ |f|^p - \lim \sup ∫ |f_n - f|^p.

        Equivalently :math:`0 ≤ - \lim \sup ∫ |f_n - f|^p`. This proves :math:`\|f_n - f\|^p → 0`. ☐

        ---

        **Proof 2**. By :eq:`convex`,

        .. math:: |f_n - f|^p ≤ 2^{p-1}(|f_n|^p + |f|^p).
           :label: fnn

        In particular, :math:`f_n - f ∈ L_p`, for each :math:`n ∈ ℕ`.
         
        Define the functions

        .. math:: g_n =  2^{p-1}(|f_n|^p + |f|^p) \quad \text{ and } \quad g =  2^p|f|^p.

        Then :math:`g_n → g` a.e., and :math:`\|f_n\|_p → \|f\|_p` implies :math:`∫ g_n \to ∫ g`.
         
        Also, :math:`g_n ≥ |f_n - f|^p → 0` a.e., by :eq:`fnn`. Therefore, the :term:`dominated convergence theorem` implies :math:`∫ |f_n - f|^p → 0`. ☐

        ---

        **Proof 3**. Since :math:`f ∈ L_p`, for all :math:`ε > 0`, there is a number :math:`δ >0` and a set :math:`B ∈ 𝔐` of finite measure such that :math:`f` is bounded on :math:`B`, :math:`∫_{X - B}|f|^p \, dμ < ε/2`, and :math:`∫_{E}|f|^p \, d\mu < ε/2`, for all :math:`E ∈ 𝔐` with :math:`μ E < δ`.

        By :term:`Egoroff's theorem`, there is a set :math:`A ⊆ B` such that :math:`μ (B - A) < δ` and :math:`f_n → f` uniformly on :math:`A`. Therefore,

        .. math:: \int_X |f|^p &= \int_{X\setminus B} |f|^p + \int_{B\setminus A} |f|^p + \int_{A} |f|^p \\
            &< \epsilon/2 + \epsilon/2 + \int_{A} |f|^p\\
            &\leq \epsilon + \varliminf \int_{A} |f_n|^p,
            :label: eq1

        since, by :term:`Fatou's lemma`, :math:`\int_{A} |f|^p = \int_{A} \varliminf |f_n|^p \leq \varliminf \int_{A} |f_n|^p`. By hypothesis, :math:`\|f_n\|_p \rightarrow \|f\|_p`. Therefore,

        .. math:: \varliminf \int_{A} |f_n|^p = \varliminf \left\{\int_{X} |f_n|^p - \int_{X\setminus A} |f_n|^p\right\} =\int_{X} |f|^p - \varlimsup \int_{X\setminus A} |f_n|^p.

        By :eq:`eq1`, then,

        .. math:: \int_{X} |f|^p < \epsilon + \int_{X} |f|^p - \varlimsup \int_{X\setminus A} |f_n|^p.

        Therefore, :math:`\varlimsup \int_{X\setminus A} |f_n|^p < \epsilon` (since :math:`f\in L_p`). Finally, note that

        .. math:: \|f_n-f\|^p &= \|(f_n - f) χ_A + (f_n - f) χ_{X-A}\|_p &\\
                              &≤ \|(f_n - f) χ_A \|_p + \|(f_n - f) χ_{X- A}\|_p & (∵ \text{Minkowsky})\\
                              &≤ \|(f_n - f) χ_A \|_p + \|f_nχ_{X- A}\|_p + \|f χ_{X-A}\|_p. &

        Therefore, :math:`\varlimsup \|f_n-f\|^p` is bounded above by

        .. math:: \varlimsup\{|f_n(x) - f(x)|: x∈ A\} μ(A)^{1/p} + \varlimsup \left(∫_{X-A} |f_n|^p\right)^{1/p} + \left(∫_{X-A} |f|^p\right)^{1/p}.

        The first term on the right goes to zero since :math:`f_n → f` uniformly on :math:`A`. The other terms are bounded by :math:`2 ε^{1/p}`. ☐

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     **Solution** to :numref:`Problem {number} <2004Apr_p3>`

  a. Since :math:`g\in L_q(S)`, for all :math:`\epsilon > 0` there exists :math:`\delta>0` such that if :math:`A` is a measurable set with :math:`\mu A < \delta` then

     .. math:: \int_A |g|^q \, d\mu < \epsilon.

     Let :math:`B_0\subset S` denote the set on which :math:`f_n` does not converge to :math:`f`. Let :math:`B\subseteq S` be such that :math:`f_n \rightarrow f` uniformly in :math:`S\setminus B`, and such that :math:`\mu B< \delta`. (Such a set exists by :term:`Egoroff's theorem` since :math:`\mu S < \infty`.) Now throw the set :math:`B_0` in with :math:`B` (i.e. redefine :math:`B` to be :math:`B\cup B_0`). Then,

     .. math:: D_n &:= \int_S |f_n g - f g | \, d\mu = \int_S |f_n - f| |g | \, d\mu \\
               &=\int_{S\setminus B} |f_n - f| |g | \, d\mu + \int_{B} |f_n - f| |g | \, d\mu \\
               &\leq \| (f_n - f) \chi_{S\setminus B}\|_p \|g\|_q + \|f_n - f\|_p \|g \chi_B\|_q.

     The final inequality holds because :math:`f_n, f \in L_p` implies :math:`|f_n - f| \in L_p` and by :term:`Hölder's inequality`.

     By :term:`Minkowski's inequality`, :math:`\|f_n - f\|_p \leq \|f_n\|_p + \|f\|_p`, so

     .. math:: D_n \leq \sup\{| f_n(x) - f(x)| : x\notin B\}\mu(S\setminus B)^{1/p} \|g\|_q + (\|f_n\|_p + \|f\|_p)\left(\int_B |g|^q\, d\mu\right)^{1/q}.

     Now we are free to choose the :math:`\delta>0` above so that

     .. math:: \left(\int_B |g|^q\, d\mu\right)^{1/q} < \frac{\epsilon}{2(M+\|f\|_p)}

     holds whenever :math:`\mu B < \delta`. Also, :math:`f_n \rightarrow f` uniformly on :math:`S\setminus B`, so let :math:`N` be such that

     .. math:: \sup\{|f_N(x) - f(x)| : x ∉ B\} < \frac{ε}{2μ(S-B)^{1/p}\|g\|_q}.

     Then :math:`D_N < ε/2 + ε/2`.   ☐

  b. The result is not true for :math:`p=1` as the following example shows.

     Let :math:`f_n = n χ_{[0,1/n]}, \, n=1, 2, \dots`.

     First note that :math:`f_n → 0` a.e. For if :math:`x ∈ (0,1]`, then there exists :math:`N>0` such that :math:`1/N < x` and thus :math:`f_n(x) = 0` for all :math:`n≥ N`. Therefore, :math:`\{x∈ [0,1] : f_n(x) ↛ 0 \} = \{0\}`.

     Now, if :math:`g` is the constant function :math:`g = 1`, then :math:`∫ f_n g \, dμ = 1` for all :math:`n=1, 2, \dots`. Therefore, :math:`∫ f_n g \, dμ → 1`, while :math:`∫ fg \, dμ = ∫ 0g\, dμ = 0`.

     Finally, note that :math:`\|f_n\|_1 = n \, μ [0,\frac{1}{n}] = 1` for :math:`n=1, 2, \dots`, so :math:`\{f_n\}` satisfies the hypothesis :math:`\|f_n\|_1 ≤` some constant :math:`M`. ☐

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     **Solution** to :numref:`Problem {number} <2004Apr_p4>`

  **The Closed Graph Theorem** If :math:`X` and :math:`Y` are Banach spaces and :math:`T : X \to Y` is a linear mapping, then the graph :math:`Γ(T) := \{(x,y) ∈ X × Y: y = T x\}` is closed if and only if :math:`T` is bounded.

  *Proof*. (⇒) Suppose :math:`\overline{Γ(T)} ⊆ Γ(T)` (i.e., :math:`Γ(T)` is closed).

    Then, since :math:`T` is linear, :math:`T(X)` is a subspace of :math:`Y`, and :math:`\Gamma(T) = X \times T(X)` is a subspace of the Banach space :math:`X \oplus Y`.

    Endowed with the (usual) norm, :math:`\|(x, Tx)\| = \|x\| + \|Tx\|`, and the subspace :math:`\Gamma(T) = X \oplus T(X)` is a Banach space (since we assume :math:`\Gamma(T)` is closed, so :math:`(x_n, Tx_n) \to (x, Tx) \in \Gamma(T)`).

    Consider the continuous projection mappings, :math:`\pi_1, \pi_2`, defined on :math:`\Gamma(T)` by :math:`\pi_1(x, Tx) = x` and :math:`\pi_2(x, Tx) = Tx`.

    Then :math:`\pi_1` is a continuous bijection of :math:`\Gamma(X)` onto :math:`X`. Therefore, by the :term:`inverse mapping theorem`, :math:`\pi^{-1}` is continuous.

    Since :math:`\pi_1\colon \Gamma(T) \to X` and :math:`\pi_2\colon \Gamma(T) \to T(X)`  and :math:`\pi_1^{-1}\colon X \to \Gamma(T)`, we can write :math:`T = \pi_2 \circ \pi_1^{-1}` as a mapping from :math:`X` to :math:`\Gamma(T)` to :math:`T(X) \subseteq Y`.

    Since :math:`\pi_1^{-1}` and :math:`\pi_2` are both continuous, so is :math:`T`.

  (⇐) Suppose :math:`T` is bounded (equivalently, continuous).

    Let :math:`(x,y)` be a limit point of :math:`\Gamma(T)` in :math:`X \oplus Y`.  That is, :math:`\exists \{(x_n, Tx_n)\} \subseteq X \times Y` such that :math:`\|(x_n, T x_n) - (x,y)\| \to 0`.

    To see that :math:`(x,y) \in \Gamma(T)` (so :math:`\Gamma(T)` is closed), observe that :math:`x_n \to x` implies :math:`T x_n \to T x`, , since :math:`T` is continuous.

    Therefore, :math:`T x = y`, so :math:`(x,y) = (x, Tx) \in \Gamma(T)`, as desired.

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     **Solution** to :numref:`Problem {number} <2004Apr_p5>`

  a. First we recall some well-known facts about normed linear spaces.

     | *Fact 1*. If :math:`X^∗` is :term:`separable`, then :math:`X` is separable.
     | *Fact 2*. :math:`ℓ_1` is a separable.
     | *Fact 3*. :math:`ℓ_∞` is not separable.
     | *Fact 4*. :math:`ℓ_1^∗ ≅ ℓ_∞`.
     | *Fact 5*. If :math:`H` is a :term:`separable` :term:`Hilbert space`, then :math:`H^{∗∗} ≅ H`.

     We will use these facts to prove that :math:`ℓ_1` is not a Hilbert space (so :math:`ℓ_1` is a counterexample that disproves assertion a.)

     If :math:`ℓ_1` is a Hilbert space, then :math:`ℓ_∞^∗ ≅ ℓ_1`, by Fact 5.  Therefore, :math:`ℓ_∞^∗` is separable since :math:`ℓ_1` is (Fact 2). But then Fact 1 implies :math:`ℓ_∞` is separable, which contradicts Fact 3. Therefore, :math:`ℓ_∞^∗ ≇ ℓ_1`, so :math:`ℓ_1` is not a Hilbert space.

  b. **Claim**. :math:`X = (C[0,1], \| \; \|_1)` is not a Banach space.

     *Proof*. For :math:`n = 4, 5, 6, \dots`, let

       .. math:: f_n(x) = \begin{cases}
                             0, & 0≤ x < \frac{1}{4}-\frac{1}{n},\\
                             1 - \frac{n}{4} + nx, & \frac{1}{4}-\frac{1}{n} ≤ x < \frac{1}{4},\\
                             1, & \frac{1}{4} ≤ x < \frac{3}{4}\\
                             1 + \frac{3n}{4} - nx, & \frac{3}{4}≤ x < \frac{3}{4}+\frac{1}{n}\\
                             0, & \frac{3}{4}+\frac{1}{n}\leq x ≤ 1,\\
                          \end{cases}

       Thus :math:`f_n ∈ C[0,1]` is constant except on the interval :math:`A_n := [\frac{1}{4}-\frac{1}{m}, \frac{1}{4}]`, where it is linearly increasing from 0 to 1, and the interval :math:`B_n := [\frac{3}{4}, \frac{3}{4}+\frac{1}{m}]`, where it is linearly decreasing from 1 to 0.

       We show that :math:`\{f_n\}` is a :term:`Cauchy sequence` in :math:`X = (C[0,1], \| \; \|_1)` that does not converge to a function in :math:`C[0,1]`.

       For all :math:`n > m ≥ 4`, :math:`f_n` and :math:`f_m` are zero outside the interval :math:`[\frac{1}{4}-\frac{1}{m}, \frac{3}{4}+\frac{1}{m}]` and equal to :math:`1` in the interval :math:`[\frac{1}{4}, \frac{3}{4}]`.

       So the only region where these functions may differ is :math:`A_m ∪ B_m`, and over this region we have :math:`\sup \{|f_n(x) - f_m(x)| : x ∈ A_m ∪ B_m\} ≤ 1`.  Therefore,

       .. math:: \|f_n - f_m\|_1 ≤  \sup \{|f_n(x) - f_m(x)| : x ∈ A_m ∪ B_m\} ⋅ m(A_m ∪ B_m) ≤ m(A_m \cup B_m) ≤ \frac{1}{m},

       which tends to 0 as :math:`m` tends to infinity.  Therefore, :math:`\{f_n\}` is a Cauchy sequence in :math:`X`.

       On the other hand, :math:`f_n` converges to the characteristic function :math:`χ_{[1/4,3/4]}` which is not continuous, so :math:`\lim_{n → ∞} f_n ∉ C[0,1]`.

  c. We adopt the convention that the term :term:`separable` applies only to infinite Hilbert spaces.

     **Claim**. Every separable Hilbert space is isometrically isomorphic to :math:`\ell_2`.

     *Proof*. Let :math:`ℋ` be a separable Hilbert space and let :math:`\{x_n\}` be a countable :term:`dense set` in :math:`ℋ`.

       By throwing away elements if necessary, we can assume :math:`\{x_n\}` is a set of linearly independent vectors whose span is dense in :math:`ℋ`.

       Applying the Gram-Schmidt procedure to :math:`\{x_n\}`, we can assume it is a set of orthonormal vectors whose span is dense in :math:`ℋ`.

       Define the mapping :math:`U: ℋ → ℂ^ω` (from :math:`ℋ` to the sequence space :math:`ℂ^ω`) as follows:

       .. math:: U x = \{⟨x, x_n⟩\}_{n=0}^∞

       Then :math:`U` is a :term:`unitary operator` from :math:`ℋ` onto :math:`ℓ_2`.

       To see this, let :math:`x∈ ℋ`.  We first show :math:`U x ∈ ℓ_2`.

       Observe that :math:`x = ∑_{n=0}^∞ ⟨ x, x_n⟩ x_n`, so

       .. math:: \|x\|^2_{ℋ} = ⟨ x, ∑_{n=0}^∞ ⟨ x, x_n⟩ x_n⟩= ∑_{n=0}^∞ \overline{⟨ x, x_n⟩} ⟨ x, x_n⟩ = ∑_{n=0}^∞ |⟨ x, x_n⟩| = \|U x\|_{ℓ_2}^2.
         :label: 2004Apr19_5.1

       Since :math:`x ∈ ℋ`, :math:`\|x\|_2 < ∞`, so :math:`\|Ux\|_2 < ∞`. Thus, :math:`U x ∈ ℓ_2` and :eq:`2004Apr19_5.1` shows :math:`U` is unitary.

       To complete the proof, we must show that :math:`U` maps :math:`ℋ` *onto* :math:`ℓ_2`.

       Fix :math:`y = \{y_i\}_{i=0}^∞ ∈ ℓ_2`.  Define :math:`x = ∑_{i=0}^∞ y_i x_i`.  Then,

       .. math:: \{⟨ x, x_n⟩\}_{i=0}^∞ = U x = U(∑_{i=0}^∞ y_i x_i) = ∑_{i=0}^∞ y_i Ux_i = \{y_i\}_{i=0}^∞.

       The last equality holds since :math:`U x_i = \{⟨ x_i, x_j⟩\}_{j=0}^∞` which is the function from :math:`ω` to :math:`ℂ` that takes :math:`i` to 1 and every other :math:`j ∈ ω` to 0.

       We have thus shown that there exists :math:`x ∈ ℋ` such that :math:`U x = y`, as desired. ☐

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     **Solution** to :numref:`Problem {number} <2004Apr_p6>`

  a. See :term:`Fubini's theorem` in the appendix.

  b. (Note: It's likely that this solution supplies many more of the technical details than most examiners would expect one to produce while sitting for a timed examination.)

     #. In order to apply :term:`Fubini's theorem` to the function :math:`F(x,y):= f(x-y)g(y)`, we should first observe that :math:`F(x,y)` is :math:`(ℬ(ℝ) \otimes ℬ(ℝ))`-measurable, assuming :math:`f` and :math:`g` are :math:`ℬ(ℝ)`-measurable.

        The details are not hard, but it's instructive to work through them at least once in a lifetime.

        The function :math:`φ(x,y) = x-y` is a continuous mapping from :math:`ℝ^2` to :math:`ℝ`, so it is a :term:`Borel measurable function`.

        Consider the composition :math:`(f ∘ φ)(x,y)= f(x-y)`.  If :math:`U⊆ ℝ` is :term:`open <open set>`, then :math:`(f ∘ φ)^{-1}(U)= φ^{-1}f^{-1}(U)`.

        Assume for now that :math:`f` is a :term:`Borel function`, so :math:`f^{-1}(U) ∈ ℬ(ℝ)`.

        Let :math:`Ω := \{V ⊆ ℝ : φ^{-1}(V) ∈ ℬ(ℝ) ⊗ ℬ(ℝ)\}`.

        Clearly :math:`Ω` contains the open sets. In fact, it's not hard to show that :math:`Ω` is a :term:`σ-algebra`, so it must contain :math:`ℬ(ℝ)`.

        Therefore, :math:`f^{-1}(U) ∈ Ω` for every open set :math:`U ⊆ ℝ`.

        We have thus observed that if :math:`f: ℝ → ℝ` is Borel and :math:`φ: ℝ^2 → ℝ` is continuous, then :math:`f ∘ φ` is Borel.

        Next, let :math:`g: ℝ → ℝ` be a function and note that :math:`g(y) = (g ∘ π_2)(x,y)`, where :math:`π_2(x,y) = y` is the (continuous) second projection.

        Thus, if :math:`g` is Borel, then :math:`g ∘ π_2: ℝ^2 → ℝ` is Borel.

        Finally, we note that :math:`f(x-y)g(y)` is the product :math:`f(x-y)g(y) = (f ∘ φ)(x,y) (g ∘ π_2)(x,y)` of two Borel functions, which is also Borel, as expected.

        The final technical detail needed to make this observation useful in the present context is the fact that for every Lebesgue :term:`measurable function` :math:`f` there exists a :term:`Borel function` :math:`f̃` such that :math:`f̃ = f` a.e. (See, e.g., Chapter 9 of Rudin :cite:`Rudin:1987`.)

        So we assume without loss of generality that the functions involved in this problem are Borel.

        Now that we have completed the tedious task of showing that :math:`f(x-y)g(y)` is a measurable function on :math:`ℝ^2`, we can apply :term:`Fubini's theorem`, as follows:

        .. math:: ∬ |f(x-y) g(y)| \, dx\, dy &= ∫  |g(y)| \left(∫ |f(x-y)|\, dx\right)\, dy &\\
                     &= ∫  |g(y)| \left(∫ |f(x)|\, dx\right)\, dy & (∵ \text{ invariance of Leb. meas.}) \\
                     &= ∫  |g(y)| \, dy ∫ |f(x)|\, dx < ∞ & (∵ f, g ∈ L_1(ℝ)).
           :label: fubi

        Therefore, :term:`Fubini's theorem` part b. implies :math:`f(x-y)g(y) ∈ L_1(dx × dy)`, so :term:`Fubini's theorem` (Part c) implies

        .. math:: F(x_0,y) = f(x_0-y) g(y) ∈ L_1(dy) \text{ for almost every } x_0 ∈ ℝ,

        and

        .. math:: h(x) = ∫ f(x-t) g(t) \, dt ∈ L_1(dx).

     #. Having proved that :math:`h` exists for almost every :math:`x ∈ ℝ`, we now we show :math:`\|h\|_1 ≤ \|f\|_1 \|g\|_1`.

        This follows immediately from :term:`Fubini's theorem` (Part a). Indeed,

        .. math:: ∫ |h| \, dm &= ∫ \left| ∫ f(x-y) g(y)\, dy\right|\, dx \\
                 & ≤  ∬  |f(x-y) g(y)| \, dy \, dx\\
                 & = ∬ | f(x-y) g(y)|\, dx\, dy\\
                 & = \|f\|_1 \|g\|_1,

        where the last equality follows from :eq:`fubi`.

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     :premium:`Solution to` :numref:`Problem {number} <2004Apr_p7>`

  a. See the :term:`Radon-Nikodym theorem` in the :ref:`appendix of theorems <appendix-theorems>`.

  b. (coming soon)

--------------------------

.. rubric:: Footnotes

.. [1]
   Since :math:`f` is an *extended* real valued function, we must not forget to include :math:`A_\infty`, without which the union in :eq:`2004-b1` would not be all of :math:`[a,b]`.

.. [2]
   Disclaimer: you should check the first proof carefully for yourself and decide whether you believe it.

.. [3]
   Judging from Parts a and b of this problem, the third proof may be closer to what the examiners had in mind.


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.. Let :math:`(X, \mathscr{B}, \mu)` be a complete measure space, where :math:`\mu` is a finite, positive measure defined on the :math:`\sigma`-algebra :math:`\mathscr{B}` of subsets of :math:`X`.

.. Let :math:`S` be a closed subset of :math:`ℝ` and let :math:`f\in L_1(\mu),` where :math:`f` is an extended real-valued function defined on :math:`X`. If

.. .. math:: A_E(f) = \frac{1}{\mu(E)}\int_E f\, d\mu \in S

.. for every :math:`E\in \mathscr{B}` with :math:`\mu(E) > 0,` prove that :math:`f(x)\in S` for almost all :math:`x\in X`.

.. Define the measure :math:`\nu` on :math:`(X, \mathscr{B})` by :math:`\nu(E) = \int_E f \, d\mu`.

.. Then :math:`\nu` is a finite measure since :math:`f\in L_1(\mu)` and :math:`\mu` is finite.

.. Also, if :math:`\mu(E) = 0`, then :math:`\nu(E) = \int_E f\, d\mu = 0`, so :math:`\nu \ll \mu`.

.. Therefore, :math:`\int h\, d\nu = \int h f \, d\mu`.

.. Since :math:`S` is closed, its complement :math:`S^c` is open.

.. Since :math:`f\in L_1(\mu),` :math:`f` is measurable, so :math:`U:= f^{-1}(S^c) = \{x \in X : f(x) \notin S\}` is a measurable set.

.. If :math:`\mu(U) > 0`, then

.. .. math:: A_U(f) = \frac{1}{\mu(U)}\int_U f\, d\mu \in S

.. Let :math:`s_0 := \frac{1}{\mu(U)}\int_U f\, d\mu \in S`.  Then,

.. .. math:: \int_U f\, d\mu = s_0 \mu(U) = s_0 \int_U d\mu,

.. so,

.. .. math:: \int_U (f-s_0) \, d\mu = 0.

.. But this is impossible because :math:`s_0 \in S` and so :math:`|f(x)-s_0|>0` for all :math:`x\in U`, and :math:`\mu(U) > 0`.