.. 1994 Nov 16
.. ===========

Exam 2 (10--16)
=======================

1994 Nov
--------

**Instructions** Masters students: Do any 5 problems. Ph.D. students: Do any 6 problems.

.. index:: complete normed linear space, normed linear space

.. _1994Nov_p1:

.. proof:prob::

   Let :math:`E` be a :term:`normed linear space`. Show that :math:`E` is :term:`complete <complete set>` if and only if, whenever :math:`∑_1^∞ \|x_n \| < ∞`, then :math:`∑_1^∞ x_n` converges to an :math:`s ∈ E`.

.. index:: compact Hausdorff space, Hausdorff space

.. _1994Nov_p2:

.. proof:prob::

   Let :math:`f_n` be a sequence of real-valued :term:`continuous functions <continuous function>` on a :term:`compact <compact set>` :term:`Hausdorff space` :math:`X`. Show that if :math:`f_1 ≥ f_2 ≥ f_3 ≥ \cdots`, and :math:`f_n(x) → 0` for all :math:`x ∈ X`, then :math:`f_n → 0` uniformly.

.. index:: Lebesgue measure

.. _1994Nov_p3:

.. proof:prob::

   Let :math:`f` be :term:`integrable` on the real line with respect to :term:`Lebesgue measure`. Evaluate :math:`\lim\limits_{n→∞} ∫_{-∞}^∞ f(x-n) \left(\frac{x}{1+|x|}\right)\, dx`. Justify all steps.

.. _1994Nov_p4:

.. proof:prob::

   Let :math:`X` and :math:`Y` be :term:`metric spaces <metric space>` and let :math:`f` be a continuous mapping from :math:`X` *onto* :math:`Y`.

   a. Show that if :math:`X` is :term:`compact <compact set>`, then :math:`Y` is compact.

   b. Show that if :math:`X` is compact, and :math:`f` is one-one, then :math:`f` is a :term:`homeomorphism`.

   c. Is the conclusion of (b) true if :math:`Y` is compact, but :math:`X` is not?

.. _1994Nov_p5:

.. proof:prob::

   a. Show :math:`L_p[0,1]⊆ L_q[0,1]` if :math:`p > q`.  Is this true for :math:`L_p[0,∞)` and  :math:`L_q[0,∞)`?

   b. Clearly 

       .. math:: ⋂_{p≥ 1} L_p[0,1]⊇ L_∞[0,1].

       Does equality hold?

.. index:: Lebesgue measure, Borel measure, Radon-Nikodym derivative, absolutely continuous

.. _1994Nov_p6:

.. proof:prob::

   Let :math:`λ` be :term:`Lebesgue measure` on :math:`ℬ(ℝ^2)`, the :term:`Borel sets <Borel set>` of :math:`ℝ^2`.  Define a :term:`Borel measure` :math:`μ` as follows: for each :math:`A ∈ ℬ(ℝ^2)`, 

   .. math:: μ A = ∬_A \frac{dλ(x,y)}{1+x^2y^2}.

   Show that :math:`λ` is :term:`absolutely continuous <absolutely continuous measure>` with respect to :math:`μ` and find :math:`\frac{dλ}{dμ}`, the :term:`Radon-Nikodym derivative`.

.. _1994Nov_p7:

.. proof:prob::

   Let :math:`\mathcal P` be the set of continuous functions on :math:`[0,3]` whose graphs consist of a finite number of straight line segments.  Prove that :math:`\mathcal P` is dense in :math:`L_2[0,3]`.

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

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     **Solution** to :numref:`Problem {number} <1994Nov_p1>`

  (⇒) Suppose :math:`E` is :term:`complete <complete set>`. Let :math:`\{x_n\} ⊆ E` be absolutely convergent; i.e., :math:`∑_n\|x_n\| < ∞`.
  
    We must show

    .. math:: ∑_{n=1}^∞ x_n := \lim_{N→∞}∑_{n=1}^N x_n  = s∈ E.

    Let :math:`S_N = \sum_{n=1}^Nx_n`. Then, for any :math:`j\in \mathbb{N}`,

    .. math::

        \|S_{N+j}-S_N\| = \left\|∑_{n=N+1}^{N+j} x_n \right\| ≤ ∑_{n=N+1}^{N+j}\|x_n\| → 0

    as :math:`N → ∞`, since :math:`∑_n\|x_n\| < ∞`.
    
    Therefore, :math:`\{S_N\}` is a :term:`Cauchy sequence`.
    
    Since :math:`E` is complete, there is an :math:`s ∈ E` such that :math:`∑_{n=1}^∞ x_n  = \lim\limits_{N→∞} S_N = s.`

  (⇐) Suppose whenever :math:`∑_1^∞ \|x_n \| < ∞`, then :math:`∑_1^∞ x_n` converges to an :math:`s ∈ E`.

    Let :math:`\{y_n\} ⊆ E` be a :term:`Cauchy sequence`. That is, :math:`\|y_n - y_m\| → 0` as :math:`n, m → ∞`.

    Let :math:`n_1 < n_2 < \cdots` be a subsequence such that

    .. math:: n, m ≥ n_j \; ⟹ \; \|y_n - y_m\| < 2^{-j}.

    Observe, for :math:`k>1`,

    .. math::

        y_{n_k} = y_{n_1} +  (y_{n_2} - y_{n_1}) + (y_{n_3}-  y_{n_2}) + \cdots + (y_{n_k}-  y_{n_{k-1}}) = y_{n_1}+∑_{j=1}^{k-1} (y_{n_{j+1}}-  y_{n_j}),

    and

    .. math:: ∑_{j=1}^∞ \|y_{n_{j+1}}-  y_{n_j}\| < ∑_{j=1}^∞ 2^{-j} = 1

    By hypothesis, this implies that

    .. math:: y_{n_k} -y_{n_1} = ∑_{j=1}^{k-1} (y_{n_{j+1}}-  y_{n_j}) → s ∈ E, \text{ as } k → \infty.

    We have thus found a subsequence :math:`\{y_{n_k}\} ⊆ \{y_n\}` having a limit in :math:`E`.

    Finally, since :math:`\{y_n\}` is Cauchy, it is easy to verify that :math:`\{y_n\}` must converge to the same limit.
    
    This proves that every Cauchy sequence in :math:`E` converges to a point in :math:`E`, as desired.
    
    ☐

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     **Solution** to :numref:`Problem {number} <1994Nov_p2>`

  Fix :math:`ε > 0`.  We will show there exists :math:`N>0` such that :math:`n ≥ N` implies :math:`\|f_n\|_∞ < ε`.

  For each :math:`x_α ∈ X`, let :math:`N_α` be such that :math:`n ≥ N_α` implies :math:`|f_n(x_α)| < ε/2`.

  Let :math:`V_α` be a :term:`neighborhood` of :math:`x_α` such that :math:`x ∈ V_α` implies :math:`|f_{N_α}(x)-f_{N_α}(x_α)| < ε/2`.

  Then, :math:`|f_{N_α}(x)| ≤ |f_{N_α}(x)-f_{N_α}(x_α)| + |f_{N_α}(x_α)| < ε/2`, and so :math:`f_1 ≥ f_2 ≥ f_3 ≥ \cdots` implies that for all :math:`n≥ N_α` and :math:`x ∈ V_α` we have  :math:`|f_n(x)| < ε`.

  The collection :math:`\{V_α\}_{α∈𝔄}` of these neighborhoods covers :math:`X`. Since :math:`X` is compact, there exists a finite subcover :math:`\{V_{α_1}, \dots, V_{α_M}\}`, so that :math:`X = ⋃_{i=1}^M V_{α_i}`.

  Let :math:`N = \max \{N_{α_i} : 1≤ i ≤ M\}`.  Then for every :math:`x ∈ X`, we have :math:`x ∈ V_{α_i}` for some :math:`i ∈ \{1, \dots, M\}`, and :math:`|f_n(x)|<ε` whenever :math:`n ≥ N`.  Therefore, :math:`\|f_n\|_∞ < ε`.

  ☐

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     **Solution** to :numref:`Problem {number} <1994Nov_p3>`

  Fix :math:`n>0`. Consider the change of variables, :math:`y = x-n`. Then :math:`dy = dx` and :math:`x = y+n`, so

  .. math:: \int_{-\infty}^\infty f(x-n) \frac{x}{1+|x|}\, dx &=\int_{-\infty}^\infty f(y) \frac{y+n}{1+|y+n|}\, dy\\[4pt]
                    &=\int_{-n}^\infty f(y) \frac{y+n}{1+y+n}\, dy +\int_{-\infty}^{-n} f(y) \frac{y+n}{1-(y+n)}\, dy.
     :label: 94-3

  Note that, when :math:`y \geq -n`, :math:`\frac{y+n}{1+y+n} \in [0,1)`, and increases to :math:`1` as :math:`n` tends to infinity. Thus,

  .. math:: 0\leq |f(y)|\,\frac{y+n}{1+y+n} \leq |f(y)|,

  for all :math:`y \geq -n`. Define the function [1]_

  .. math:: g_n(y) = f(y)\, \frac{y+n}{1+y+n} \,\mathbf{1}_{[-n,\infty)}(y).

  Then :math:`|g_n|\leq |f|` and :math:`\lim\limits_{n\rightarrow \infty} g_n = f`. Therefore, by the dominated convergence theorem,

  .. math::

     \lim_{n\rightarrow \infty}\int_{-n}^\infty f(y) \frac{y+n}{1+y+n}\, dy = \lim_{n\rightarrow \infty} \int_{-\infty}^{\infty} g_n(y) \, dy =\int_{-\infty}^{\infty} f(y) \, dy.

  Next, consider the second term in :eq:`94-3`. Define the function

  .. math:: h_n(y) = f(y)\, \frac{y+n}{1-(y+n)} \,\mathbf{1}_{(-\infty,-n]}(y).

  It is not hard to check that

  .. math:: \frac{|y+n|}{|1-(y+n)|} \,\mathbf{1}_{(-\infty,-n]}(y)\in [0,1),

  from which it follows that :math:`|h_n| \leq |f|`. Also, it is clear that, for all :math:`y`,

  .. math::

     \lim_{n\rightarrow \infty} h_n(y) = f(y)\,\lim_{n\rightarrow \infty}\frac{y+n}{1-(y+n)} \,\mathbf{1}_{(-\infty,-n]}(y) = 0.

  Therefore, the dominated convergence theorem implies that

  .. math:: \lim_{n\rightarrow \infty}\int_{-\infty}^{-n} f(y) \frac{y+n}{1-(y+n)}\, dy = 0.

  Combining the two results above, we see that :math:`\lim\limits_{n\rightarrow \infty} \int_{-\infty}^\infty f(x-n) \left(\frac{x}{1+|x|}\right)\, dx = \int_{-\infty}^{\infty}f(x) \, dx`.

  ☐

  **Remark** Intuitively, this is the result we expect because the translation :math:`f(x-n) = T_n f(x)` is merely shifting the support of :math:`f` to the right tail of the measure :math:`d\mu := \frac{x}{1+|x|}\, dx`, and in the tail this measure looks like :math:`dx`.

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     **Solution** to :numref:`Problem {number} <1994Nov_p4>`

  a. Let :math:`\{V_α\}_{α∈𝔄} ` be an :term:`open covering` of :math:`Y`.
  
     Since :math:`Y` is a :term:`metric space`, this cover can be reduced to a countable subcover :math:`\{V_{α_i}\}_{i=1}^∞`.

     Define :math:`U_{α_i} = f^{-1}(V_{α_i})`.  Then :math:`X = ⋃_{i=1}^∞ U_{α_i}`.

     Since :math:`X` is :term:`compact <compact set>`, there is a finite subcover :math:`\{U_{α_{i_1}}, \dots, U_{α_{i_M}}\}` so that :math:`X = ⋃_{j=1}^M U_{α_{i_j}}`.

     Now, :math:`f(U_{α_{i_j}}) = f f^{-1}(V_{α_{i_j}}) ⊆ V_{α_{i_j}}.`  Therefore,

     .. math:: Y = f(X) = f(⋃_{j=1}^M U_{α_{i_j}}) = ⋃_{j=1}^M f(U_{α_{i_j}}) ⊆ ⋃_{j=1}^M V_{α_{i_j}}.

     Thus, we have reduced an arbitrary cover of :math:`Y` to a finite subcover, thereby showing that :math:`Y` is compact.
     
     ☐
  
  b. Suppose :math:`X` is :term:`compact <compact set>` and :math:`f` is a bijection.
  
     We must show that :math:`f^{-1}` exists and is :term:`continuous <continuous function>`.

     That :math:`f^{-1}` exists is obvious since :math:`f` is bijective.

     To show :math:`f^{-1}` is continuous, we show that for each open set :math:`U ⊆ X` the set :math:`[f^{-1}]^{-1} (U) = f(U)` is open in :math:`Y`.

     Let :math:`U` be an open subset of :math:`X`.  Let :math:`U'` denote the complement of :math:`U` in :math:`X`, which is closed, thus compact (since :math:`X` is compact).

     Since :math:`f ∈ C(X, Y)`, the argument in Part a implies :math:`f(U')` is a compact subset of :math:`Y`. Therefore, :math:`f(U')` is closed (and bounded), so :math:`[f(U')]'` is open in :math:`Y`.

     Now, it's not hard to show that if :math:`f` is one-one, then :math:`f(U) = f(X)- f(U')`, from which :math:`f(U) = [f(U')]'` is open in :math:`Y`, as desired.
     
     ☐

  c. Finally, suppose :math:`f: X → Y` is a continuous bijection and :math:`Y` is :term:`compact <compact set>`.
  
     If :math:`f` is a :term:`homeomorphism`, then :math:`f^{-1}` is a continuous mapping from :math:`Y` onto :math:`X` and, by a, :math:`X` is compact.  Thus, if :math:`X` is not compact, then :math:`f` cannot be a homeomorphism.
     
     ☐

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     **Solution** to :numref:`Problem {number} <1994Nov_p5>`

  a. Fix :math:`f\in L_p[0,1]`.  Let :math:`\alpha = p/q` (which is greater than :math:`1`, since :math:`p>q`).

     By :term:`Hölder's inequality`, with :math:`\frac{1}{\alpha}+\frac{1}{\beta} = 1`,

     .. math:: \int |f|^q \, d\mu = \int |f|^q \chi_{[0,1]}\, d\mu \leq \| |f|^q\|_\alpha \|\chi_{[0,1]}\|_\beta = \| |f|^q \|_\alpha.
        :label: 5.1

     Since :math:`f \in L_p[0,1]`,

     .. math:: \| |f|^q \|_\alpha = \left(\int |f|^{q\alpha} \, d\mu\right)^{1/\alpha} =  \left(\int |f|^p \, d\mu\right)^{1/\alpha} = \|f\|_p^{p/\alpha}= \|f\|_p^{q} < \infty,
        :label: 5.2

     By :eq:`5.1` and :eq:`5.2`, :math:`\|f\|_q^{q} \leq \|f\|_p^{q} < \infty`, so :math:`f \in L_q[0,1]`, as desired.

     The same result does *not* hold for :math:`L_p[0,\infty)` and :math:`L_q[0,\infty)`.  Here is a counterexample.

     For some :math:`\alpha` (to be determined), let

     .. math:: f(x) = \begin{cases}x^{-\alpha},& x\in [1, \infty)\\ 0, & x\in [0,1).\end{cases}

     Then,

     .. math:: \|f\|_p^p = \int_0^\infty |f(x)|^p \, dx = \int_1^\infty x^{-\alpha p} \, dx = \left. \frac{x^{1-\alpha p}}{1-\alpha p} \right|_{x=1}^{x\to \infty}.
        :label: 5.3

     Similarly,

     .. math:: \|f\|^q_q = \left. \frac{x^{1-\alpha q}}{1-\alpha q} \right|_{x=1}^{x\to \infty}.
        :label: 5.4

     If :math:`p>q`, then we can choose :math:`\alpha` so that :math:`1-\alpha p < 0 < 1-\alpha q` (for example, let :math:`\alpha = \frac{2}{p+q}`).

     Therefore, by :eq:`5.3` and :eq:`5.4`, :math:`\|f\|_p < \infty` while :math:`\|f\|_q = \infty`.

     ☐

  b. As asserted in the problem statement, :math:`\bigcap_{p\geq 1} L_p[0,1]\supseteq L_\infty[0,1].`

     (Indeed, if :math:`\exists M >0` such that :math:`|f|\leq M` a.e., then for all :math:`p\geq 1`, :math:`\|f\|_p \leq \left(M^p\mu[0,1]\right)^{1/p} = M < \infty`.)

     Equality does *not* hold. That is, there are functions belonging to :math:`L_p[0,1]` for every finite :math:`p\geq 1` that do not belong to  :math:`L_{\infty}[0,1]`. Here is an example.

     Define :math:`f : [0,1] \to ℝ` as follows:

     .. math:: f = \sum_{n=1}^\infty c_n \chi_{[\frac{1}{n+1}, \frac{1}{n})}.

     Then

     .. math:: \left(\int |f|^p \, d\mu\right)^{1/p} = \left(\sum_{n=1}^\infty c_n^p \mu[\frac{1}{n+1}, \frac{1}{n}]\right)^{1/p}= \left(\sum_{n=1}^\infty \frac{c_n^p}{n(n+1)}\right)^{1/p}.
         :label: 5.5

     Let :math:`c_n = \log n`. Then, for :math:`n` large enough, :math:`(\log n)^p < \sqrt{n}`.  Therefore, there exists :math:`N>0` such that

     .. math:: \sum_{n=N+1}^\infty \frac{c_n^p}{n(n+1)} <  \sum_{n=N+1}^\infty \frac{\sqrt{n}}{n(n+1)} < \sum_{n=N+1}^\infty \left(\frac{1}{n}\right)^{3/2} < \infty.

     This proves that for every finite :math:`p\geq 1` there is an :math:`N>0` such that the tail of the series in :eq:`5.5` converges, so the whole series converges. This proves that :math:`f\in L_p[0,1]` for every finite :math:`p\geq 1`.

     On the other hand, consider the growth of :math:`f` near zero, say, :math:`\frac{1}{n+1} \leq x < \frac{1}{n}`. Over that interval, :math:`f(x) = \log(n)`. So as :math:`x` approaches zero, :math:`f` grows without bound.  Therefore, :math:`f\notin L_\infty[0,1]`.

     ☐

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     **Solution** to :numref:`Problem {number} <1994Nov_p6>`

  Recall the definition of an :term:`absolutely continuous measure`, where we used :math:`λ ≪ μ` to denote the fact that :math:`λ` is :term:`absolutely continuous <absolutely continuous measure>` (AC) with respect to :math:`μ`.

  In the present case, we clearly have :math:`μ ≪ λ` and :math:`dμ/dλ = (1 + x^2y^2)^{-1}`.

  Suppose :math:`A ∈ ℬ(ℝ^2)` (:term:`Borel sets <Borel set>` of :math:`ℝ^2`) and :math:`λ A > 0`.  We will show :math:`μ A>0`.  Since both measure are positive, this will suffice to prove that :math:`μ A=0` implies :math:`λ A = 0`; that is, :math:`λ ≪ μ`.

  Partition :math:`ℝ^2` into a grid :math:`ℝ^2 = ⋃_{m,n} [m, m+1) × [n, n+1)`.

  For all :math:`m, n ∈ ℤ`, define :math:`A_{m,n} = A ∩ ([m, m+1) × [n, n+1))`. Then,

  .. math:: λ A = ∑_{m,n} λ A_{m,n} > 0

  implies that there exist :math:`n_0` and :math:`m_0` such that :math:`λ A_{m_0, n_0} > 0`. Thus,

  .. math:: ∬_{A_{m_0,n_0}} \frac{d λ (x,y)}{1 + x^2y^2} ≥ \inf \left\{\frac{1}{1 + x^2y^2} ∣ n_0 ≤ x < n_0+1, \, m_0 ≤ y < m_0+1\right\} ⋅ λ A_{m_0,n_0} > 0.

  Therefore,

  .. math:: μ A = ∬_A \frac{d λ(x,y)}{1 + x^2y^2} ≥ ∬_{A_{m_0,n_0}} \frac{d λ(x,y)}{1 + x^2y^2} > 0.

  We now have two mutually AC positive measures, :math:`λ ≪ μ ≪ λ`.

  We want a function :math:`f` which satisfies :math:`λ A = ∫_A f \, dμ`.

  Recall, for a measure :math:`ν`, the function :math:`φ A = ∫_A g \, dν` defines a measure and :math:`∫ h \, dφ = ∫ hg\, dν.`

  In the present case, we have

  .. math:: ∬_A (1 + x^2y^2) \, dμ = ∬_A \frac{1 + x^2y^2}{1 + x^2y^2} \, dλ = λ A,

  so the :term:`Radon-Nikodym derivative` of :math:`λ` with respect to :math:`μ` is :math:`\frac{dλ}{dμ} = 1 + x^2y^2`.

  ☐

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     **Solution** to :numref:`Problem {number} <1994Nov_p7>`

  We take for granted that :math:`C[0,3]` is dense in :math:`L_2[0,3]`, so it suffices to prove that :math:`\mathcal P` is dense in :math:`C[0,3]`.

  Fix :math:`g \in C[0,3]` and :math:`\epsilon > 0`.

  Since :math:`[0,3]` is closed and bounded, hence compact, :math:`g` is :ref:`uniformly continuous <uniformly-continuous>` in :math:`[0,3]` (by the :ref:`basic continuity theorem <basic-continuity-theorem>`).

  For :math:`n > 0` and :math:`0 < \frac{1}{n} < \delta`, partition :math:`[0,3]` as follows:

  Let :math:`E_k = [\frac{k-1}{n}, \frac{k}{n})` for :math:`k = 1, 2, \dots, 3n-1` and let :math:`E_{3n} = [\frac{3n-1}{n}, \frac{3n}{n}]`. Then,

  .. math:: [0,3] = \bigcup_{k=1}^{3n} E_k.

  Define :math:`p \in \mathcal P` as follows:

  .. math:: \text{Let } \quad p_k(x) = (k - nx)g\left(\frac{k-1}{n}\right) + (nx - (k-1))g\left(\frac{k}{n}\right), \quad x \in E_k, \quad \text{ and let}

  .. math:: p(x) = \sum_{k=1}^{3n} p_k(x) \chi_{E_k}(x).

  Then :math:`p` is piece-wise linear and for :math:`x \in E_k` (:math:`k = 1,2,\dots, 3n`),

  .. math:: |p(x) - g(x)| &= |p_k(x) - g(x)|\\
            &= |(k - nx)g\left(\frac{k-1}{n}\right) + (nx - (k-1))g\left(\frac{k}{n}\right) - g(x)|\\
            &= |(k - nx)[g\left(\frac{k-1}{n}\right) - g(x)] + (nx - (k-1))[g\left(\frac{k}{n}\right) - g(x)]|\\
            &\leq (k - nx)|g\left(\frac{k-1}{n}\right) - g(x)| + (nx - (k-1))|g\left(\frac{k}{n}\right) - g(x)|\\
            &< (k - nx)\epsilon' + (nx - (k-1))\epsilon' = \epsilon',

  where :math:`\epsilon'` is the maximum of :math:`|g\left(\frac{k-1}{n}\right) - g(x)|` and :math:`|g\left(\frac{k}{n}\right) - g(x)|` as :math:`x` is allowed to range over :math:`E_k`.

  By choosing :math:`n` large enough, we can ensure :math:`\epsilon' < \epsilon/\sqrt{3}`, so that

  .. math:: \|p - g\|_2 = \left(\int_0^3 |p(x) - g(x)|^2 \, dx\right)^{1/2} < (3 \epsilon'^2)^{1/2} = \sqrt{3}\epsilon' <\epsilon.

  That is, :math:`\|p - g\|_2 < \epsilon`, as desired.

  ☐

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.. rubric:: Footnotes

.. [1]
   Here :math:`\mathbf{1}_A(x)` denotes the indicator function of the set :math:`A`, which is 1 if :math:`x\in A` and 0 if :math:`x\notin A`.

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