.. role:: premium

.. 2006 Apr
.. ========

Exam 9 (64--70)
==========================

2006 Apr
--------

**Instructions**. Do as many problems as you can.  Complete solutions (except for minor flaws) to 5 problems would be considered a good performance.  Fewer than 5 complete solutions may still be passing, depending on the quality.

.. _2006Apr_p1:

.. proof:prob::

   Use only "epsilon-delta" techniques to prove the following discrete version of the Lebesgue :term:`dominated convergence theorem`: Suppose :math:`f_k: ℕ → ℝ` is a sequence of functions that converges pointwise to 0, and there is a function :math:`f: ℕ → [0, ∞)` with :math:`∑_{n∈ ℕ} f(n) < ∞` and :math:`|f_k(n)| ≤ f(n)` for all :math:`n` and for all :math:`k`.  Then :math:`\lim\limits_{k→∞}∑_{n∈ ℕ} f_k(n)=0`.

.. _2006Apr_p2:

.. proof:prob::

   For real numbers :math:`a` and :math:`b` with :math:`0<a<b`, evaluate the integral

   .. math:: ∫_0^∞ \frac{e^{-ax} - e^{-bx}}{x} \, dx := \lim\limits_{c→+∞}∫_{1/c}^c \frac{e^{-ax} - e^{-bx}}{x} \, dx.

   Justify (in writing) all crucial steps using well-known results.  **Hint**. Consider :math:`∫_a^b e^{-xy}\, dy`.

.. _2006Apr_p3:

.. proof:prob::

   Let :math:`f∈ L_2[0,1]` and define :math:`g(x) = ∫_0^x f(t)\, dt`, :math:`0≤ x ≤ 1`.  Show that :math:`\|g\|_2 ≤ \frac{1}{2}\|f\|_2`.

.. _2006Apr_p4:

.. proof:prob::

   a. Let :math:`\{f_n\}` be a sequence of :term:`measurable functions <measurable function>` on the :term:`measurable space` :math:`(X, 𝔐)`. Prove, from the definition of measurability, that :math:`\lim\sup_{n→∞} f_n` is measurable.

   b. Let :math:`\{f_n\}` be a sequence of measurable functions on the measurable space :math:`(X, 𝔐)`. Prove that the set of points where :math:`\lim_{n→∞} f_n` exists is measurable.

.. _2006Apr_p5:

.. proof:prob::

   Let :math:`f: ℝ→[0,∞)` be :term:`uniformly continuous` with :math:`∫_{-∞}^∞ f(x)\, dx < ∞`. Prove that :math:`f` is bounded.

.. _2006Apr_p6:

.. proof:prob::

   A function :math:`f` satisfies a :term:`Lipschitz condition` on an interval if there is a constant :math:`M` such that :math:`|f(x) - f(y)| ≤ M|x-y|` for all :math:`x`and :math:`y` in the interval.

   a. Show that a function satisfying a Lipschitz condition is :term:`absolutely continuous <absolutely continuous function>`.

   b. Show that an absolutely continuous function on an interval :math:`J` satisfyies a Lipschitz condition if and only if :math:`f' ∈ L_∞(J)`.

   .. index:: Radon-Nikodym derivative, Radon-Nikodym theorem

.. _2006Apr_p7:

.. proof:prob::

   Let :math:`μ` and :math:`ν` be :term:`finite <finite measure>` :term:`positive measures <positive measure>` on the :term:`measurable space` :math:`(X, 𝔐)` such that :math:`ν ≪ μ ≪ ν`, and let :math:`\frac{dν}{d(μ+ν)}` represent the :term:`Radon-Nikodym derivative` of :math:`ν` with respect to :math:`μ+ν`.  Show that

   .. math:: 0< \frac{dν}{d(μ+ν)} < 1 \quad μ\text{-a.e.}

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

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       **Solution** to :numref:`Problem {number} <2006Apr_p1>`

    It suffices to show that for all :math:`ε>0` there exists :math:`K>0` such that :math:`k > K` implies :math:`\bigl|∑_{n∈ ℕ}f_k(n)\bigr| < ε`.
    
    Fix :math:`ε>0`.  Since :math:`∑_n f(n) < ∞`, there exists :math:`N>0` such that :math:`∑_{n≥N} f(n) < ε/2`.

    For each :math:`0≤ j < N`, there exists :math:`n_j` such that

    .. math:: k > n_j \quad ⟹ \quad |f_k(j)| < \frac{ε}{2N}.

    Define :math:`K := \max \{n_j ∣ 0≤ j< N\}`.

    To complete the proof, we show that :math:`k ≥ K` implies :math:`\bigl|∑_{n∈ ℕ}f_k(n)\bigr| < ε`.

    Indeed, if :math:`k≥ K`, then

    .. math::  \bigl|∑_{n ∈ ℕ} f_k(n)\bigr| & ≤ ∑_{n ∈ ℕ} |f_k(n)| &\\
                                            & = ∑_{n=0}^{N-1} |f_k(n)| + ∑_{n=N}^∞ |f_k(n)| &\\
                                            & = N \frac{ε}{2N} + ∑_{n=N}^∞ f(n)&(∵ k > K)\\
                                  & = \frac{ε}{2} + \frac{ε}{2} = ε &(\text{by definition of } N).

    ☐

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       **Solution** to :numref:`Problem {number} <2006Apr_p2>`

    Following the hint, consider
    
    .. math:: ∫_a^b e^{-xy}\, dy = \left.\frac{-e^{-xy}}{x}\right|_{y=a}^{y=b} = \frac{-e^{-xb}+e^{-xa}}{x} = \frac{e^{-xa}-e^{-xb}}{x}.

    Therefore, 

    .. math:: ∫_0^∞ \frac{e^{-ax} - e^{-bx}}{x} \, dx = 
             ∫_0^∞ \bigl(∫_a^b e^{-xy}\, dy\bigr)\, dx = \lim\limits_{c→+∞}∫_{1/c}^c\bigl( ∫_a^b e^{-xy}\, dy \bigr)\, dx.

    By :term:`Tonelli's theorem` we have 

    .. math:: ∫_Y \left(∫_X g(x,y)\, dx\right)\, dy = ∬_{X×Y}g(x,y)\, dm(x×y) = ∫_X \left(∫_Y g(x,y)\, dy\right)\, dy,
       :label: tone

    whenever :math:`g: X × Y → [0,∞]` is measurable with respect to product measure on :math:`X × Y`.

    In the present case, :math:`g(x,y) := e^{-xy}`, which is continuous, hence measurable, and :math:`g≥0`.

    Therefore, we can compute the integral as either of the iterated integrals in :eq:`tone`.  We choose to integrate with respect to :math:`x` first.

    .. math:: ∫_a^b \bigl(\lim\limits_{c→+∞}∫_{1/c}^c e^{-xy}\, dx \bigr)\, dy 
               &= ∫_a^b \left[\frac{-e^{-xy}}{y}\right]_{x=0}^{x → ∞}\, dy \\ 
               &= ∫_a^b \lim\limits_{c→∞}\left(\frac{-e^{-cy} + e^{-y/c}}{y}\right)\, dy \\ 
               &= ∫_a^b \frac{1}{y}\, dy = \left. \log y \right|_a^b = \log b - \log a = \log (b/a).

    ☐

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       **Solution** to :numref:`Problem {number} <2006Apr_p3>`

    This is false as the following example shows:
    
    Let :math:`f(x) = 1-x`.  Then

    .. math:: \|f\|_2^2 &= ∫_0^1 |f(x)|^2\, dx=∫_0^1 (1-x)(1-x)\, dx= ∫_0^1 1-2x +x^2\, dx \\&= \left[x-x^2 +\frac{x^3}{3}\right]_0^1 = 1-1 +\frac{1}{3} = \frac{1}{3}.

    .. math:: \|g\|_2^2 &= ∫_0^1 |g(x)|^2\, dx =∫_0^1 \left|∫_0^x f(t)\, dt\right|^2 \, dx= ∫_0^1 \left|x-\frac{x^2}{2}\right|^2 \, dx\\
                        &= ∫_0^1 (x^2-x^3 + \frac{x^4}{4}) \, dx= \left[\frac{x^3}{3} - \frac{x^4}{4}+ \frac{x^5}{20}\right]_0^1\\
                        &=\frac{1}{3} - \frac{1}{4}+ \frac{1}{20} = \frac{1}{12} + \frac{1}{20} = \frac{32}{240} = \frac{4}{30}.

    Then, 

    .. math:: \|g\|_2 = \sqrt{\frac{4}{30}} = \frac{2}{\sqrt{2}\sqrt{3}\sqrt{5}} = \frac{\sqrt{2}}{\sqrt{5} \sqrt{3}} \quad \text{ and } \quad \|f\|_2 = \frac{1}{\sqrt{3}}.

    **Claim**. :math:`\frac{2\sqrt{2}}{\sqrt{5} \sqrt{3}} ≰ \frac{1}{2\sqrt{3}}`.

    .. math:: \frac{\sqrt{2}}{\sqrt{5} \sqrt{3}} &≤ \frac{1}{2\sqrt{3}}\\
              \frac{2\sqrt{3}\sqrt{2}}{\sqrt{5} \sqrt{3}} &≤1\\
              \frac{2\sqrt{2}}{\sqrt{5}} &≤ 1\\
               1.264911064 &≰ 1\\

    
    ☐


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       **Solution** to :numref:`Problem {number} <2006Apr_p4>`

    a. Let :math:`(X, 𝔐)` be the :term:`measurable space` under consideration, and define 
    
       .. math:: g_k(x) := (\sup_{n≥k} f_n)(x).
    
       We first show that this supremum of a sequence of measurable functions is measurable.  This requires we prove for all :math:`α∈ ℝ` and all :math:`k∈ ℕ` that :math:`g_k^{-1}(α, ∞] ∈ 𝔐`.
    
       Fix :math:`α∈ ℝ` and :math:`k∈ ℕ`.  Then,
    
       .. math:: g_k^{-1}(α, ∞] &= \{x∈ X ∣ g_k(x) > α\}\\
                                & = ⋃_{n≥ k} \{x ∈ X ∣ f_n(x) > α\}\\
                                & = ⋃_{n≥ k} f_n^{-1}(α, ∞].
          :label: g
    
       Now, since :math:`f_n^{-1}(α, ∞] ∈ 𝔐` by measurability of :math:`f_n`, and since :math:`𝔐` is a :term:`σ-algebra`, we have :math:`g_k^{-1}(α, ∞] ∈ 𝔐` by :eq:`g`.

       Next observe
    
       .. math:: \lim\limits_{k→∞} g_k = \limsup\limits_{k→∞} f_n = \inf\limits_k \sup\limits_{n≥k} f_n = \inf\limits_k g_k.

       To complete the proof then, we must show that :math:`\inf\limits_k g_k` is measurable whenever :math:`\{g_k\}` is a sequence of measurable functions.

       Observe that :math:`\inf\limits_{k≥ 0} g_k = -\sup\limits_{k≥ 0} (-g_k)`, and :math:`-g_k` is measurable since each :math:`g_k` is measurable.
       
       We have already proved that the supremum of a sequence of measurable functions is measurable, so the proof for Part a is complete.

       ☐

    b. We must show that the set :math:`E:= \{x∈ X ∣ \lim\limits_{n→∞} f_n(x) \text{ exists }\}` is measurable.

       Fix :math:`x∈ X` and recall that :math:`\lim\limits_{n→∞} f_n(x)` exists if and only if :math:`\liminf\limits_{n→∞} f_n(x) = \limsup\limits_{n→∞} f_n(x)`.  Thus,

       .. math:: E= \{x∈ X ∣ \liminf\limits_{n→∞} f_n(x) = \limsup\limits_{n→∞} f_n(x)\}.

       It follows from Part a above that if :math:`\{f_n\}` is a sequence of measurable functions on :math:`(X, 𝔐)`, then :math:`\liminf\limits_{n→∞} f_n(x)` is measurable. Indeed (dropping :math:`(x)` for brevity),
       
       .. math:: \liminf\limits_{n→∞} f_n = \sup\limits_{k≥ 0}\inf\limits_{n≥k}f_n = \sup\limits_{k≥ 0}(-\sup\limits_{n≥k}(-f_n)) = -\inf\limits_{k≥ 0}\sup\limits_{n≥k}(-f_n).

       The following lemma completes the proof.

       **Lemma**. If :math:`g` and :math:`h` are :term:`measurable functions <measurable function>`, then the set :math:`\{x∈ X ∣ g(x) = h(x)\}` is measurable.

       *Proof*. Let :math:`\{r_n ∣ n ∈ ℕ\}` be an enumeration of the rationals.  Then,

         .. math:: E_1 &:= \{x ∈ X ∣ g(x) < h(x)\}\\
                       &= ⋃_{n=0}^∞\left(\{x∈ X ∣ g(x) < r_n\} ∩ \{x∈ X ∣ r_n < h(x)\}\right),

         is measurable since each set involved in the union is measurable.  For the same reason, the set :math:`E_2 := \{x ∈ X ∣ g(x) > h(x)\}` is also measurable.

         Therefore, :math:`\{x∈ X ∣ g(x) = h(x)\} = X - E_1 - E_2` is measurable.

       ☐

.. We take ":math:`\lim_{n→∞} f_n(x)` exists" to mean that there is a number :math:`ℓ(x) ∈ ℝ^∗ := ℝ∪ \{-∞, ∞\}` such that :math:`f_n(x) → ℓ(x)` as :math:`n→∞`.

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       **Solution** to :numref:`Problem {number} <2006Apr_p5>`

    We must show there exists :math:`M>0` such that :math:`∀ x∈ ℝ`, :math:`f(x) < M`.

    Define :math:`I := ∫_{-∞}^∞ f(x) \, dx` and keep in mind that :math:`I<∞`.

    Suppose on the contrary that :math:`f` is unbounded.

    By uniform continuity, :math:`∀ ε>0`, :math:`∃ δ>0`, :math:`∀ x, y ∈ ℝ`,
    
    .. math:: |x-y| < δ \; ⟹ \; |f(x) - f(y)|<ε.
       :label: uc

    Consider :math:`M := I/δ`.  Suppose there exists :math:`x∈ ℝ` such that :math:`f(x) > M + ε`.

    By :eq:`uc`, for all :math:`y∈ B(x;δ)` we have :math:`|f(x) - f(y)|<ε`, so :math:`f(y) > M` for all :math:`y∈ B(x;δ)`. Therefore,

    .. math:: I := ∫_{-∞}^∞ f(x) \, dx ≥ ∫_{B(x;δ)} f(x) \, dx > M ⋅ m(B(x;δ)) = M ⋅ 2δ = 2I, 

    a contradiction.
    
    We have thus proved that :math:`ε + I/δ` is an upper bound on :math:`f`.  By assumption, :math:`f` is bounded below by 0, so the proof is complete.
    
    ☐

..  so :math:`0≤ f(x) ≤ ε + I/δ` for all :math:`x∈ ℝ`, so :math:`f` is bounded.

.. there exists :math:`δ>0` such that :math:`|f(x) - f(y)|<ε` for all :math:`|x-y| < δ`.

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       **Solution** to :numref:`Problem {number} <2006Apr_p6>`

    a. Assume :math:`f` is :term:`Lipschitz continuous` in :math:`[a,b]`.  We must prove :math:`f ∈ AC[a,b]`.

       That is, we must show for all :math:`ε>0` there exists :math:`δ>0` such that for every finite collection :math:`\{(a_i,b_i) ∣ i=1,2,\dots, n\}` of open intervals in :math:`(a,b)`, we have

       .. math:: ∑_{i=1}^n (b_i - a_i) < δ \; ⟹ \; ∑_{i=1}^n |f(b_i)- f(a_i)| < ε.

       Fix :math:`ε>0`. Let :math:`M` be the Lipschitz constant and define :math:`δ := ε/M`.
       
       If :math:`∑_{i=1}^n (b_i - a_i) < δ`, then

       .. math:: ∑_{i=1}^n |f(b_i)- f(a_i)| ≤ ∑_{i=1}^n M ⋅ (b_i-a_i) =M ∑_{i=1}^n  (b_i-a_i) < M ⋅ δ = ε.

       ☐

    b. Assume :math:`f∈ AC(J)`.
    
       (⇒) Suppose :math:`f` is :term:`Lipschitz continuous` with :term:`Lipschitz constant` :math:`M`.

         For :math:`x` in the :term:`interior` of :math:`J`,

         .. math:: f'(x) = \lim\limits_{h→ 0} \frac{f(x+h) - f(x)}{h}.

         For :math:`h ≠ 0` and :math:`x + h ∈ J`,

         .. math:: \left|\frac{f(x+h) - f(x)}{h}\right| ≤ \frac{Mh}{h} = M.

         Thus if :math:`h ≠ 0` and :math:`x + h ∈ J`, then

         .. math:: -M ≤ \frac{f(x+h) - f(x)}{h} ≤ M,

         so :math:`-M ≤ f'(x) ≤ M` since limits preserve finite bounds.

         Therefore, :math:`f'(x)` is bounded for almost every :math:`x∈ J`.

         Measurability of :math:`f'` follows from Problem 4. Whence :math:`f'∈ L_∞(J)`.

       (⇐) Suppose :math:`f' ∈ L_∞(J)`, so :math:`|f'| ≤ M` a.e. for some :math:`M>0`.
       
         Suppose :math:`x ≤ y` in :math:`J`.  Then,

         .. math:: |f(x) - f(y)| &= \left|∫_x^y f'(t) \, dt\right|  \qquad (∵ f ∈ AC(J))\\
                                &≤ ∫_x^y |f'(t)| \, dt ≤ M(y-x) = M |x - y|.
         
         Therefore, :math:`f` satisfies a Lipschitz condition on :math:`J`.

       ☐

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       **Solution** to :numref:`Problem {number} <2006Apr_p7>`

    Clearly :math:`μ ≪ μ+ν` and :math:`ν ≪ μ+ν`, so the :term:`Radon-Nikodym theorem` gives unique nonnegative functions :math:`f,g ∈ L_1(μ+ν)` such that 

    .. math:: μ E = ∫_E f \, d(μ+ν) \quad \text{ and } \quad ν E = ∫_E g \, d(μ+ν),
       :label: 7

    for every :math:`E ∈ 𝔐`.  Here :math:`g = dν/d(μ+ν)`.

    We must show :math:`0<g(x)<1` holds for :math:`\mu`-almost every :math:`x ∈ X`.

    Since :math:`μ+ν` and :math:`ν` are both positive, it follows from :eq:`7` that :math:`g(x) ≥ 0` for all :math:`x∈ X`.

    Suppose there exists :math:`E ∈ 𝔐` such that :math:`g(x) = 0` for all :math:`x ∈ E`.

    We will show that :math:`μ E = 0`, as this will allow us to conclude :math:`g>0` :math:`μ`-a.e.

    If :math:`g(x) = 0` for all :math:`x∈ E`, then

    .. math:: ν E = ∫_E g \, d(μ+ν) = 0 \quad \text{ and } \quad μ ≪ ν \; ⟹ \; μ E = 0.

    So, to finish the proof, we must show :math:`g(x) < 1` for :math:`μ`-a.e. :math:`x∈X`.

    Suppose :math:`g(x) ≥ 1` for :math:`x ∈ E`. Then,

    .. math:: ν E = ∫_E g\, d(μ+ν) ≥ ∫_E 1 \, d(μ+ν) = μ E + ν E.

    Now, since :math:`ν E` is finite, we can subtract it from both sides of the last expression yielding :math:`0≥ μ E`.  On the other hand, :math:`μ ≥ 0`, so :math:`μ E = 0`.

    We have thus proved that :math:`g(x) < 1` for :math:`μ`-almost every :math:`x ∈ X`.

    ☐

       

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.. .. rubric:: Footnotes

.. .. [1]