.. 2000 Nov 17
.. ===========

Exam 6 (43--49)
========================

2000 Nov
--------

**Instructions** Do as many problems as you can. Complete solutions to five problems would be considered a good performance.

.. index:: inverse function theorem, Euclidean norm, invertible linear map

.. _2000Nov_p1:

.. proof:prob::

   a. State the :term:`inverse function theorem`. [1]_ 

   b. Suppose :math:`L: ℝ^3 → ℝ^3` is an invertible linear map and that :math:`g: ℝ^3 → ℝ^3` has continuous first order partial derivatives and satisfies :math:`\|g(x)\| ≤ C\|x\|^2` for some constant :math:`C` and all :math:`x ∈ ℝ^3`. Here :math:`\|x\|` denotes the usual :term:`Euclidean norm` on :math:`ℝ^3`. Prove that :math:`f(x) = L(x) + g(x)` is locally invertible near 0.

.. _2000Nov_p2:

.. proof:prob::

   Let :math:`f` be a differentiable real valued function on the interval :math:`(0,1)`, and suppose the derivative of :math:`f` is bounded on this interval.

   Prove the existence of the limit :math:`L = \lim_{x → 0^+} f(x)`.

   .. index:: pair: Fubini's theorem; Tonelli's theorem

.. _2000Nov_p3:

.. proof:prob::

   Let :math:`f` and :math:`g` be :term:`Lebesgue integrable` functions on :math:`[0,1]`, and let :math:`F` and :math:`G` be the integrals

   .. math:: F(x) = ∫_0^x f(t) \, dt, \quad G(x) = ∫_0^x g(t) \, dt.

   Use :term:`Fubini's theorem` and/or :term:`Tonelli's theorem` to prove that

   .. math:: ∫_0^1 F(x)g(x) \, dx = F(1) G(1) - ∫_0^1 f(x)G(x) \, dx.

   Other approaches to this problem are possible, but credit will be given only to solutions based on these theorems.

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

.. _2000Nov_p4:

.. proof:prob::

   Let :math:`(X, 𝔐, μ)` be a finite :term:`measure space` and suppose :math:`ν` is a :term:`finite measure` on :math:`(X, 𝔐)` that is :term:`absolutely continuous <absolutely continuous measure>` with respect to :math:`μ`. Prove that the norm of the :term:`Radon-Nikodym derivative` :math:`f = \left[\frac{dν}{dμ}\right]` is the same in :math:`L_∞ (μ)` as it is in :math:`L_∞(ν)`.

.. _2000Nov_p5:

.. proof:prob::

   Suppose that :math:`\{f_n\}` is a sequence of (Lebesgue) :term:`measurable functions <measurable function>` on :math:`[0,1]` such that :math:`\lim\limits_{n→∞} ∫_0^1 |f_n| \, dx = 0` and there is an :term:`integrable` function :math:`g` on :math:`[0,1]` such that :math:`|f_n|^2 ≤ g`, for each :math:`n`. Prove that :math:`\lim\limits_{n→∞}∫_0^1 |f_n|^2 \, dx =0`.

   .. index:: polynomials, density (of continuous functions)

.. _2000Nov_p6:

.. proof:prob::

   Denote by :math:`\mathcal{P}_e` the family of all even polynomials. Thus a polynomial :math:`p` belongs to :math:`\mathcal{P}_e` if and only if :math:`p(x) = \frac{p(x) + p(-x)}{2}` for all :math:`x`. Determine, with proof, the :term:`closure` of :math:`\mathcal{P}_e` in :math:`L_1[-1,1]`. You may use without proof the fact that :term:`continuous functions <continuous function>` on :math:`[-1,1]` are :term:`dense <dense set>` in :math:`L_1[-1,1]`.

.. _2000Nov_p7:

.. proof:prob::

   Suppose that :math:`f` is real valued and integrable with respect to Lebesgue measure :math:`m` on and that there are real numbers :math:`a<b` such that

   .. math:: a ⋅ m(U) ≤ ∫_U f \, dm ≤ b ⋅ m(U),

   for all :term:`open sets <open set>` :math:`U` in :math:`ℝ`. Prove that :math:`a ≤ f(x) ≤ b` a.e.

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

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       **Solution** to :numref:`Problem {number} <2000Nov_p1>`

    a. A statement of the :term:`inverse function theorem` appears in the :ref:`appendix of theorems <appendix-theorems>`.

    b. First note that :math:`L` and :math:`g` both have continuous first order partial derivatives; i.e., :math:`L, g ∈ C^1(ℝ^3)`. Therefore, the derivative of :math:`f = L + g`,

       .. math:: f'(x) \triangleq J_f(x) \triangleq \left(\frac{\partial f_i}{\partial x_j}\right)_{i,j=1}^3

       exists. Furthermore, :math:`J_f(x)` is continuous in a neighborhood of the zero vector, because this is true of the partials of :math:`g(x)`, and the partials of :math:`L(x)` are the constant matrix :math:`L`. Therefore, :math:`f ∈ C^1(ℝ^3)`. By the IFT, then, we need only show that :math:`f'(0)` is invertible. Since :math:`f'(x) = L + g'(x)`, we must show :math:`f'(0) = L + g'(0)` is invertible. Consider the matrix :math:`g'(0) = J_g(0)`. We claim, :math:`J_g(0)=0`. Indeed, if :math:`x_1, x_2, x_3` are the elementary unit vectors (also known as **i, j, k**), then the elements of :math:`J_g(0)` are

       .. math:: \frac{∂ g_i}{∂ x_j}(0) = \lim_{h→0} \frac{g_i(0 + h x_j) - g_i(0)}{h} = \lim_{h→0} \frac{g_i(h x_j)}{h}.
          :label: 2000-1

       The second equality follows by the hypothesis that :math:`g` is continuous and satisfies :math:`\|g(x)\| ≤ C\|x\|^2`, which implies that :math:`g(0) =0`. Finally, to show that :eq:`2000-1` is zero, consider

       .. math:: |g_i(h x_j)| ≤ \|g(h x_j)\| ≤ C \|h x_j\|^2 = C|h|^2,

       which implies

       .. math:: \frac{|g_i(h x_j)|}{|h|} ≤ C \frac{|h x_j|^2}{|h|} = C|h| → 0, \text{ as } h→0.

       This proves that :math:`f'(0) = L`, which is invertible by assumption, so the :term:`inverse function theorem` implies that :math:`f(x)` is locally invertible near :math:`0`.

       ☐

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       **Solution** to :numref:`Problem {number} <2001Nov_p2>`

    First consider the sequence :math:`\{x_n\} := \{\frac{1}{n}\}` for :math:`n = 2, 3, 4, \dots`.

    **Claim 1**. :math:`\{f(x_n)\}` is a :term:`Cauchy sequence`.

    *Proof*. Observe

      .. math::
      
          \left| f\bigl( \frac{1}{n+j} \bigr)- f\bigl( \frac{1}{n} \bigr) \right| = \left| \frac{ f( \frac{1}{n+j} ) - f( \frac{1}{n} ) } { \frac{1}{n+j} - \frac{1}{n} } \right| \left| \frac{1}{n+j} - \frac{1}{n} \right|,

      and 

      .. math:: \frac{1}{n+j} - \frac{1}{n} = \frac{n - (n + j)}{n(n+j)}  = \frac{-j}{n+j}.

      By the :term:`mean value theorem`, there exists :math:`c ∈ \left[\frac{1}{n+j},\frac{1}{n}\right]` such that 

      .. math:: \left| f'(c)\right| = \frac{\left| f(\frac{1}{n+j})- f(\frac{1}{n})\right|}{\left| \frac{1}{n+j}- \frac{1}{n} \right|}.

      Therefore,
      
      .. math:: \left| f\left(\frac{1}{n+j}\right)- f\left(\frac{1}{n}\right)\right| ≤ M \left| \frac{1}{n+j}- \frac{1}{n} \right| ≤ M/n,

      so given :math:`ε > 0` we have :math:`|f(1/m)- f(1/n)| < ε` for all :math:`m,n ≥ N > M / ε`.

      This proves that :math:`\{f(\frac{1}{n})\}` is a Cauchy sequence, and it follows that :math:`\lim_{n→ ∞} f(\frac{1}{n}) = ℓ` exists in :math:`ℝ`, since :math:`ℝ` is a :term:`complete metric space`.

      ☐

    **Claim 2**. :math:`\lim\limits_{x↘0} f(x) = ℓ`.

    *Proof*. Fix :math:`ε>0`. We show :math:`∃ δ>0` such that :math:`0 < x < δ` implies :math:`|f(x) - ℓ| < ε`.
    
      Given :math:`0<a<b<1` there exists :math:`c∈ [0,1]` such that :math:`f(b) - f(a) = f'(c) (b-a)`, and :math:`|f'(c)|≤ M` by assumption.
      
      Therefore, :math:`|f(b) - f(a)| ≤ M |b-a|` for all :math:`0<a<b<1`.
    
      Let :math:`N> 0` be chosen so large that :math:`N > ε/(2M)` and :math:`n≥ N \, ⟹ \, |f(1/n) - ℓ| < ε/2`.

      Then for all :math:`0< x < 1/n` we have 

      .. math:: |f(x) - ℓ| ≤ |f(x) - f(1/n)| + |f(1/n) - f(ℓ)| ≤ M/n + ε/2 < M\frac{ε}{2M} + \frac{ε}{2} = ε.

      ☐

    ☐

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       **Solution** to :numref:`Problem {number} <2001Nov_p3>`

    The left-hand side is 

    .. math:: ∫_0^1\left( ∫_0^x f(t) \, dt\right) g(x) \, dx = ∫_0^1∫_0^x f(t) g(x) \, dt\, dx.

    Consider :math:`f̃(t,x):= f(t)` and :math:`g̃(t,x):= g(x)`.  Then,

    .. math:: f̃^{-1}(G) = f^{-1}(G) × [0,1] \; \text{ and } \; g̃^{-1}(G) = [0,1]× g^{-1}(G)

    are measurable subsets of :math:`[0,1] × [0,1]:= T × X`, so :math:`f̃` and :math:`g̃` are measurable on :math:`T× X`.

    Therefore, :math:`f̃ g̃ = f g` is measurable and we can apply :term:`Fubini's theorem` which asserts the following:

    .. math:: \text{ if } \; ∬ |f̃ g̃ |\, dt\, dx < ∞ \; \text{ or } \; ∬ |f̃ g̃ |\, dx\, dt < ∞,
    
    then :math:`f̃ g̃ ∈ L_1([0,1] × [0,1])` and

    .. math:: ∬ f̃  g̃ \, dt\, dx = ∬ f̃  g̃ \, dx\, dt.

    So we calculate,
    
    .. math:: ∫_0^1 ∫_0^x |f̃ g̃ |\, dt\, dx &= ∫_0^1 ∫_0^x |f(t)||g(x)|\, dt\, dx\\
                   & ≤  ∫_0^1 ∫_0^1 |f||g| = \left(∫ |f|\right) \left(∫ |g|\right) < ∞, 

    since :math:`f, g ∈ L_1` by assumption.  Therefore, the hypotheses of Fubini's theorem are satisfied and we have

    .. math:: ∫_0^1 ∫_0^x f(t)g(x)\, dt\, dx &= ∫_0^1 ∫_t^1 f(t)g(x)\, dx\, dt \\
                     &= ∫_0^1 \left(∫_t^1 g(x)\, dx\right) f(t) \, dt \\
                     &= ∫_0^1 [G(1) - G(t)] f(t)\, dt = F(1)G(1) - ∫_0^1 f(t)G(t) \, dt,

    as desired.

    ☐

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       **Solution** to :numref:`Problem {number} <2001Nov_p4>`

    It seems an implicit assumption here is that the measures are :term:`positive <positive measure>`, so we make that assumption in what follows.

    Define (as usual) the :term:`∞-norm` relative to :math:`μ` of a (real- or complex-valued) function :math:`f` on :math:`X` as follows:

    .. math:: \|f\|_∞ := \inf \{a∈ ℝ^∗ ∣ μ\{x : |f(x)| > a\} = 0\},
       :label: norm

    where :math:`ℝ^∗ = ℝ ∪ \{-∞, ∞\}` and :math:`\inf ∅ = ∞`.

    Since there are two measures in context, let us denote the norm defined in :eq:`norm` by :math:`\|f\|_μ`, and let :math:`\|f\|_ν` denote the :term:`∞-norm` relative to :math:`ν`.

    The claim to prove is :math:`\left\|\frac{dν}{dμ}\right\|_μ = \left\|\frac{dν}{dμ}\right\|_ν`.

    Here is the proof.
    
    Let :math:`f := \frac{dν}{dμ}` so that for each :math:`E ∈ 𝔐` we have :math:`ν E = ∫_E f\, dμ`.

    Observe that :math:`f ≥ 0`, since we assumed :math:`μ, ν` positive.

    For each :math:`a ∈ ℝ^∗`, let :math:`E_a := \{x ∈ X ∣ f(x) > a\}`.
    
    Then :math:`μ E_a = 0` only if :math:`ν E_a = 0`, since :math:`ν ≪ μ`.
    
    Therefore, :math:`\{a ∈ ℝ^∗ ∣ μ E_a = 0\}  ⊆ \{a ∈ ℝ^∗ ∣ ν E_a = 0\}`, so

    .. math:: \|f\|_ν ≤ \|f\|_μ.
       :label: norm ineq

    Suppose :math:`\|f\|_ν < \|f\|_μ`. Then there exists :math:`a ∈ ℝ` such that :math:`ν E_a = 0` and :math:`μ E_a > 0`, and 

    .. math:: 0 = ν E_a = ∫_{\{f > a\}} f\, dμ ≥ a μ E_a > 0,

    a contradiction.  Therefore, equality must hold in :eq:`norm ineq`.

    ☐

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       **Solution** to :numref:`Problem {number} <2001Nov_p5>`

    Fix an arbitrarily small :math:`ε > 0` and let :math:`m` denote :term:`Lebesgue measure` on :math:`ℝ`.
    
    Since :math:`g ∈ L_1[0,1]`, there exists :math:`A ⊆ [0,1]` such that :math:`g` is bounded on :math:`A` and :math:`∫_{[0,1]-A} g \, dm < ε`.
    
    This and the assumption :math:`g ≥ |f_n|^2 ≥ 0` imply that there exists :math:`M < ∞` such that :math:`0≤ g(x) ≤ M` for all :math:`x ∈ A`.
    
    Define,

    .. math:: A_n^+ := {A ∩ \{|f_n|>1\}} \quad \text{ and } \quad A_n^- := {A ∩ \{|f_n|≤1\}}.
    
    and observe that :math:`A = A_n^+ ∪ A_n^-`, a disjoint union. Thus, for every :math:`n∈ ℕ`,
    
    .. math:: ∫_{[0,1]} |f_n|^2 \, dm &= ∫_A |f_n|^2 \, dm + ∫_{[0,1]-A} |f_n|^2 \, dm\\
                                  &= ∫_{A_n^+} |f_n|^2 \, dm + ∫_{A_n^-} |f_n|^2 \, dm + ∫_{[0,1]-A} |f_n|^2 \, dm.

    On :math:`A_n^+ := A ∩ \{|f_n|>1\}` we have :math:`|f_n| ≤ |f_n|^2 ≤ g ≤ M`, so 

    .. math:: ∫_{A_n^+} |f_n|^2 \, dm ≤ M ∫_{A_n^+} |f_n|\, dm ≤ M ∫_{[0,1]} |f_n|\, dm.
    
    On :math:`A_n^-:= A ∩ \{|f_n|≤1\}` we have :math:`|f_n|^2 ≤ |f_n|`, so 

    .. math:: ∫_{A_n^-} |f_n|^2 \, dm ≤ ∫_{A_n^-} |f_n|\, dm ≤ ∫_{[0,1]} |f_n|\, dm.

    Thus, for every :math:`n∈ ℕ`,
    
    .. math:: ∫_{[0,1]} |f_n|^2 \, dm &= ∫_{A_n^+} |f_n|^2 \, dm + ∫_{A_n^-} |f_n|^2 \, dm + ∫_{[0,1]-A} |f_n|^2 \, dm\\
                                      &≤ ∫_{[0,1]}|f_n| \, dm + M ∫_{[0,1]}|f_n| \, dm + ∫_{[0,1] - A}g \, dm.

    By assumption :math:`\lim\limits_{n→∞}∫_{[0,1]}|f_n| \, dm= 0`, so the right-hand side approaches :math:`∫_{[0,1] - A}g \, dm` as :math:`n→∞` and we conclude that
    
    .. math:: \lim\limits_{n→∞}∫_0^1 |f_n|^2 \, dx ≤ ∫_{[0,1] - A}g \, dm < ε.

    Since :math:`ε>0` was arbitrary, this completes the proof.

    ☐

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       **Solution** to :numref:`Problem {number} <2001Nov_p6>`
    
    Define the operators :math:`E, O: L_1[-1,1] → L_1[-1,1]` as follows: for :math:`f∈ L_1[-1,1]`, 

    .. math:: E f(x) = \frac{f(x) + f(-x)}{2} \quad \text{ and } \quad O f(x) = \frac{f(x) - f(-x)}{2}.

    Denote by :math:`Λ` the collection of a.e.-even functions in :math:`L_1[-1,1]`; that is, :math:`g ∈ Λ` iff :math:`g ∈ L_1[-1,1]` and :math:`g(x) = (g(x) + g(-x))/2` holds for almost every :math:`x∈ [-1,1]`.

    **Claim 1**.  :math:`E` is continuous (i.e., bounded).

    *Proof*. Evidently,

      .. math:: \|E f\|_1 &= ∫_{-1}^1 |E f(x)|\, dx = ∫_{-1}^1 \left| \frac{f(x) + f(-x)}{2} \right|\, dx \\ 
                          &≤ \frac{1}{2} \left( ∫_{-1}^1 |f(x)|\, dx + ∫_{-1}^1 |f(-x)|\, dx\right) = \|f\|_1.

      ☐                        

    **Claim 2**. If :math:`p_n ∈ \mathcal{P}_e`, then :math:`E p_n = p_n`. (obvious)

    Since :term:`continuous functions on [-1,1] are dense in L₁[-1,1]`, for each :math:`g∈ L_1[-1,1]` there is a sequence :math:`\{f_n\}⊆ \overline{C[-1,1]}` such that :math:`f_n → g` in :math:`L_1[-1,1]`.

    Fix :math:`ε>0` and let :math:`N` be a natural number large enough so that :math:`\|f_N - g\|_1 < ε/2`.
    
    There exists :math:`\{p_n\}⊆ \mathcal P` such that :math:`\|p_n - f\|_∞ → 0` as :math:`n→∞`, so

    .. math:: \|p_n - g\|_1 &= \|p_n - f_N + f_N - g\|_1\\
                            &≤  \|p_n - f_N\|_1 + \|f_N - g\|_1\\
                            &≤ \|p_n - f_N\|_∞ \, μ[-1,1] + ε/2 < ε,

    for large enough :math:`n ∈ ℕ`.

    Since :math:`E` is continuous, :math:`E p_n → E g`.

    **Claim 3**. :math:`E p_n ∈ \mathcal P_e`. (obvious)

    **Claim 4**. If :math:`g ∈ L_1[-1,1]` and :math:`g` is even, then :math:`E g = g`. (obvious)

    **Claim 5**. If :math:`g ∈ L_1[-1,1]` and :math:`g` is even, then :math:`O g = 0` a.e. (why?)

    **Claim 6**. :math:`O(\bar{\mathcal P}_e) = 0`. (why?)
    
    **Claim 7**. :math:`\bar{\mathcal P}_e = \{g ∈ L_1[-1,1]: g \text{ even a.e.}\} = Λ`.

    *Proof*. The inclusion :math:`⊆` follows from Claims 5 and 6.

      To prove :math:`⊇`, fix :math:`g∈ Λ` and observe that there exists :math:`\{p_n\} ⊆ \mathcal P_e` such that :math:`p_n \xrightarrow{L_1} g`. Therefore, :math:`E p_n → E g` by continuity of :math:`E`, and :math:`E g = g` a.e. since :math:`g` is even a.e.

      Since :math:`E p_n` is an even polynomial, we have :math:`g = \lim\limits_{n→ ∞} E p_n`, so :math:`g ∈ \bar{\mathcal P}_e`.

    ☐

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       **Solution** to :numref:`Problem {number} <2001Nov_p7>`
     
    Suppose on the contrary that there exists a measurable set :math:`E` of positive measure such that :math:`f(x) > b` for all :math:`x ∈ E`.

    Define :math:`E_n := \{x∈ ℝ ∣ f(x) > b + \frac{1}{n}\}`. Then :math:`E = ⋃_{n=1}^∞ E_n` and there exists :math:`N ∈ ℕ` such that :math:`m (E_N) > 0`.

    Since :math:`f ∈ L_1(m)`, for all :math:`ε>0` there exists :math:`δ>0` such that

    .. math:: m(S) < δ \; ⟹ \; ∫_S |f| \, dm < ε.

    If :math:`ε_0 = 1/(2N)`, and if :math:`δ_0>0` is chosen appropriately, then

    .. math::  \left|∫_S f \, dm\right| ≤ ∫_S |f| \, dm < ε_0 = \frac{1}{2N},
       :label: bound

    whenever and :math:`m(S) < δ_0`.

    Now, for all :math:`δ_1> 0` there exists an open set :math:`U` containing :math:`E_N` such that,
    
    .. math:: m(U-E_N) < δ_0 \quad \text{ and } \quad m(E_N) > m(U)-δ_1.
       :label: two ineqs
    
    From the first of these inequalities along with :eq:`bound` it follows that
    
    .. math:: \left|∫_{U-E_N} f\, dm\right| < \frac{1}{2N}.
       :label: ineq0

    From the second inequality in :eq:`two ineqs` follows the second inequality in the sequence of calculations below.

    .. math:: ∫_U f \, dm &= ∫_{U-E_N} f \, dm + ∫_{E_N} f \, dm \\
                          &≥ ∫_{U-E_N} f \, dm + \bigl(b + \frac{1}{N}\bigr) ⋅ m(E_N)\\
                          &> ∫_{U-E_N} f \, dm + \bigl(b + \frac{1}{N}\bigr) ⋅ (m(U) - δ_1).
       :label: ineq1

    Combining :eq:`ineq0` and :eq:`ineq1` we have, for arbitrary :math:`δ_1>0`,

    .. math:: ∫_U f \, dm > \bigl(b + \frac{1}{N}\bigr) ⋅ (m(U) - δ_1) - \frac{1}{2N}.
       :label: ineq2

    We will show that an appropriate choice of :math:`δ_1>0` yields
    
    .. math:: \bigl(b + \frac{1}{N}\bigr)⋅(m(U) - δ_1) > b ⋅ m(U) + \frac{1}{2N}.
       :label: goal

    From this and :eq:`ineq2` will follow,

    .. math:: ∫_U f \, dm > b ⋅ m(U),

    contradicting the hypothesis that :math:`∫_U f \, dm ≤ b⋅m(U)` holds for every :term:`open set` :math:`U` in :math:`ℝ`.

    Calculate,
    
    .. math:: \bigl(b + \frac{1}{N}\bigr)⋅(m(U) - δ_1)  &= \bigl(b + \frac{1}{N}\bigr)⋅m(U) - \bigl(b + \frac{1}{N}\bigr)⋅δ_1  \\
                                                      &= b⋅m(U) + \frac{m(U)}{N} - b ⋅ δ_1  - \frac{δ_1}{N}.

    We want,
    
    .. math:: b⋅m(U) + \frac{m(U)}{N} - b ⋅ δ_1  - \frac{δ_1}{N} > b ⋅ m(U) + \frac{1}{2N},

    equivalently, :math:`m(U)/N - (b - 1)⋅ δ_1/N > 1/(2N)`.

    Equivalently, we must prove any one (hence all) of the following (equivalent) conditions:

    .. math:: \begin{array}{rrcl}
               & m(U) - (b ⋅ N  + 1)⋅ δ_1 &>& \frac{1}{N}\\
               ⟺ & m(U) - \frac{1}{N} &>& (b ⋅ N  + 1)⋅ δ_1\\
               ⟺ & \bigl(m(U) - \frac{1}{N}\bigr) ⋅ (|b| ⋅ N  + 1)^{-1} &>& δ_1.
               \end{array}

    Thus if we choose :math:`δ_1` so that it satisfies

    .. math::    δ_1 <  \frac{m(E_N) - 1/N}{|b| ⋅ N  + 1} <  \frac{m(U) - 1/N}{|b| ⋅ N  + 1}, 

    then our goal :eq:`goal` will be satisfied and the desired contradiction will obtain.

    A similar argument can be used to show that the existence of a nonnegligible set :math:`E` on which :math:`f < a` will lead to a contradiction. Therefore, :math:`a ≤ f(x) ≤ b` holds for almost every :math:`x ∈ ℝ`.

    ☐


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.. rubric:: Footnotes

.. [1]
   Historically speaking, questions involving the :term:`inverse function theorem` do not seem very popular. This is one of the few exams on which it appears.

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