.. 2001 Nov 26 .. =========== Exam 7 (50--56) =============== 2001 Nov -------- **Instructions** Masters students do any **4** problems Ph.D. students do any **5** problems. Use a separate sheet of paper for each new problem. .. index:: Egoroff's theorem .. _2001Nov_p1: .. proof:prob:: Let :math:`\{f_n\}` be a sequence of Lebesgue measurable functions on a set :math:`E\subset \mathbb R`, where :math:`E` is of finite Lebesgue measure. Suppose that there is :math:`M>0` such that :math:`|f_n(x)|\leq M` for :math:`n\geq 1` and for all :math:`x \in E`, and suppose that :math:`\lim_{n \rightarrow \infty} f_n(x) = f(x)` for each :math:`x\in E`. Use Egoroff's theorem to prove that .. math:: \int_E f(x)\, dx = \lim_{n \to \infty} \int_E f_n(x)\, dx. .. _2001Nov_p2: .. proof:prob:: Let :math:`f(x)` be a real-valued Lebesgue integrable function on :math:`[0,1]`. a. Prove that if :math:`f>0` on a set :math:`F\subset [0,1]` of positive measure, then .. math:: \int_F f(x)\, dx > 0. b. Prove that if .. math:: \int_0^x f(x)\, dx =0, \quad \text{ for each } x\in [0,1], then :math:`f(x)=0` for almost all :math:`x\in [0,1]`. .. index:: Riesz representation theorem, Stone-Weierstrass theorem, Hölder's inequality, Hahn-Banach theorem .. _2001Nov_p3: .. proof:prob:: State each of the following: (a) The Stone-Weierstrass theorem (b) The Lebesgue (dominated) convergence theorem (c) Hölder's inequality (d) The Riesz representation theorem for :math:`L_p` (e) The Hahn-Banach theorem. .. index:: Baire category theorem, uniform boundedness theorem .. _2001Nov_p4: .. proof:prob:: a. State the :term:`Baire category theorem`. b. Prove the following special case of the uniform boundedness theorem: Let :math:`X` be a (nonempty) complete metric space and let :math:`F\subseteq C(X)`. Suppose that for each :math:`x\in X` there is a nonnegative constant :math:`M_x` such that .. math:: |f(x)| \leq M_x \quad \text{ for all } \quad f\in F. Prove that there is a nonempty *open* set :math:`G\subseteq X` and a constant :math:`M>0` such that .. math:: |f(x)| \leq M \quad \text{ holds for all } x\in G \text{ and for all } f\in F. .. _2001Nov_p5: .. proof:prob:: Prove or disprove: a. :math:`L_2` convergence implies pointwise convergence. b. .. math:: \lim_{n \to \infty} \int_0^\infty \frac{\sin(x^n)}{x^n} \, dx = 0. c. Let :math:`\{f_n\}` be a sequence of measurable functions defined on :math:`[0,\infty)`. If :math:`f_n\rightarrow 0` uniformly on :math:`[0,\infty)`, as :math:`n\rightarrow \infty`, then .. math:: \varliminf \int_{[0,\infty)} f_n(x) \, dx = \int_{[0,\infty)} \varliminf f_n(x) \, dx. .. index:: separable Hilbert space, Hilbert space, Parseval's equality .. _2001Nov_p6: .. proof:prob:: Let :math:`f:H\rightarrow H` be a bounded linear functional on a separable Hilbert space :math:`H` (with inner product denoted by :math:`\langle \cdot, \cdot \rangle`). Prove that there is a unique element :math:`y\in H` such that .. math:: f(x) = \langle x, y \rangle \quad \text{for all} \quad x\in H \quad \text{and} \quad \|f\| = \|y\|. *Hint.* You may use the following facts: A separable Hilbert space, :math:`H`, contains a complete orthonormal sequence, :math:`\{\phi_k\}_{k=1}^\infty`, satisfying the following properties: (1) If :math:`x,y\in H` and if :math:`\langle x,\phi_k \rangle = \langle y,\phi_k \rangle` for all :math:`k`, then :math:`x=y`. (2) *Parseval's equality* holds; that is, for all :math:`x\in H`, :math:`\langle x, x \rangle = \sum_{k=1}^\infty a_k^2`, where :math:`a_k = \langle x,\phi_k \rangle`. .. index:: Banach space, normed linear space, complete Banach space .. _2001Nov_p7: .. proof:prob:: Let :math:`X` be a normed linear space and let :math:`Y` be a Banach space. Let .. math:: B(X,Y) = \{A \mid A:X\to Y \text{ is a bounded linear operator}\}. Then with the norm :math:`\|A\| = \sup_{\|x\|\leq 1} \|Ax\|`, :math:`B(X,Y)` is a normed linear space (you need not show this). Prove that :math:`B(X,Y)` is a Banach space; that is, prove that :math:`B(X,Y)` is complete. ---------------------------------------- Solutions --------- .. container:: toggle .. container:: header **Solution** to :numref:`Problem {number} <2001Nov_p1>` First note that :math:`|f(x)|\leq M` for all :math:`x\in E.` To see this, suppose it’s false for some :math:`x_0\in E`, so that :math:`|f(x_0)|>M`. Then there is some :math:`\epsilon>0` such that :math:`|f(x_0)|=M+\epsilon`. By the triangle inequality, then, for all :math:`n\in \mathbb{N}`, .. math:: |f(x_0) - f_n(x_0)| \geq ||f(x_0)| - |f_n(x_0)|| = |M+\epsilon - |f_n(x_0)|| \geq \epsilon, which contradicts :math:`f_n(x_0) \rightarrow f(x_0)`. Thus, :math:`|f(x)|\leq M` for all :math:`x\in E`. Next, fix :math:`\epsilon >0`. By :term:`Egoroff's theorem`, there is a :math:`G\subset E` such that :math:`\mu(E\setminus G) < \epsilon` and :math:`f_n \rightarrow f` uniformly on :math:`G`. Furthermore, since :math:`|f_n| \leq M` and :math:`|f|\leq M` and :math:`\mu(E) < \infty`, it’s clear that :math:`\{f_n\}\subset L_1` and :math:`f\in L_1`, so the following inequalities make sense (here we’re using the notation :math:`\|f\|_G = \sup\{|f(x)|: x\in G\}`): .. math:: \left| \int_E f\, d\mu-\int_E f_n\, d\mu\right| &\leq \int_E |f- f_n|\, d\mu = \int_{E\setminus G} |f- f_n|\, d\mu + \int_G |f- f_n|\, d\mu\\ &\leq \int_{E\setminus G} |f|\, d\mu + \int_{E\setminus G} |f_n|\, d\mu + \|f(x)- f_n(x)\|_G \mu(G)\\ &\leq 2M \mu(E\setminus G) + \|f(x)- f_n(x)\|_G \mu(G)\\ &< \epsilon + \|f(x)- f_n(x)\|_G \mu(G). Finally, :math:`\mu(G) \leq \mu(E) < \infty` and :math:`\|f(x)- f_n(x)\|_G \rightarrow 0`, which proves that :math:`\int_E f_n\, d\mu \rightarrow \int_E f\, d\mu`. .. container:: toggle .. container:: header **Solution** to :numref:`Problem {number} <2001Nov_p2>` (a) Define :math:`F_n = \{x\in F: f(x) > 1/n\}`. Then .. math:: F_1 \subseteq F_2 \subseteq \cdots \uparrow \bigcup_n F_n = F, and :math:`m(F)>0` implies .. math:: 0< m(F) = m(\cup_n F_n) \leq \sum_n m(F_n). Therefore, :math:`m(F_k) >0` for some :math:`k\in \mathbb N`, and then it follows from the definition of :math:`F_k` that .. math:: 0< \frac{1}{k} m(F_k) \leq \int_{F_k} f \, dm \leq \int_F f\, dm. (b) Suppose there is a subset :math:`E\subset [0,1]` of positive measure such that :math:`f>0` on :math:`E`. Then part (a) implies :math:`\int_E f\, dm >0`. Let :math:`F\subset E` be a closed subset of positive measure. (That such a closed subset exists follows from Proposition 3.15 of :cite:`Royden:1988`.) Then, again by (a), :math:`\int_F f\, dm >0`. Now consider the set :math:`G = [0,1]\setminus F`, which is open in :math:`[0,1]`, and hence [1]_ is a countable union of disjoint open intervals, :math:`G = \bigcup_n (a_n, b_n)`. Therefore, .. math:: 0 = \int_{[0,1]} f \, dm = \sum_n \int_{(a_n,b_n)} f \, dm + \int_F f \, dm, so :math:`\int_F f\, dm >0` implies .. math:: \sum_n \int_{(a_n,b_n)} f \, dm < 0. Thus, :math:`\int_{(a_k,b_k)} f \, dm < 0` for some :math:`(a_k,b_k) \subset [0,1]`. On the other hand, .. math:: \int_{(a_k,b_k)} f \, dm = \int_0^{b_k}f(x) \, dm(x) - \int_0^{a_k} f(x) \, dm(x). By the initial hypothesis, both terms on the right are zero, which gives the desired contradiction. .. container:: toggle .. container:: header **Solution** to :numref:`Problem {number} <2001Nov_p3>` a. See the :term:`Stone-Weierstrass theorem`. b. See the :term:`dominated convergence theorem`. c. See :term:`Hölder's inequality`. d. See the :term:`Riesz representation theorem`. e. See the :term:`Hahn-Banach theorem` .. container:: toggle .. container:: header **Solution** to :numref:`Problem {number} <2001Nov_p4>` a. See the statements of the :term:`Baire category theorem` and its :ref:`corollaries `. b. Define :math:`A_m = \{x∈ X: |f(x)|≤ m, \, ∀ f∈ F\}`. Then :math:`X = \bigcup_{m=1}^\infty A_m`, since for every :math:`x` there is a finite number :math:`M_x` such that :math:`|f(x)| \leq M_x` for all :math:`f\in F`. Now note that :math:`A_m = \bigcap_{f\in F} \{x\in X: |f(x)|\leq m\}`, and, since :math:`f` and :math:`a\mapsto |a|` are continuous functions, each :math:`\{x\in X: |f(x)|\leq m\}` is closed, so :math:`A_m` is closed. Therefore, :ref:`Corollary 2 of the Baire category theorem ` implies that there must be some :math:`m\in \mathbb N` such that :math:`A_m^\circ \neq \emptyset`, so the set :math:`G = A_m^\circ` and the number :math:`M=m` satisfy the given criteria. .. container:: toggle .. container:: header **Solution** to :numref:`Problem {number} <2001Nov_p5>` a. This is false, as the following example demonstrates: For each :math:`k\in \mathbb{N}`, define :math:`f_{k,j} = \chi_{[\frac{j-1}{k}, \frac{j}{k})}` for :math:`j = 1, \ldots, k`, and let :math:`\{g_n\}` be the sequence defined by .. math:: g_1 &= f_{1,1}, \\ g_2 &= f_{2,1}, \; g_3 = f_{2,2}, \\ g_4 &= f_{3,1}, \; g_5 = f_{3,2}, \; g_6 = f_{3,3}, \\ g_7 &= f_{4,1}, \ldots Then :math:`\int |f_{k,j}|^2 \, d\mu = 1/k` for each :math:`j = 1, \ldots, k`, so :math:`\|f_{k,j}\|_2 = 1/\sqrt{k} \rightarrow 0`, as :math:`k\rightarrow \infty`. Therefore :math:`\|g_n\|_2 \rightarrow 0` as :math:`n\rightarrow \infty`. However, :math:`\{g_n\}` does not converge pointwise since, for every :math:`x\in [0,1]` and every :math:`N \in \mathbb N`, we can always find some :math:`k\in \mathbb N` and :math:`j\in \{1,\ldots, k\}` such that :math:`g_n(x) = f_{k,j}(x) = 1` with :math:`n\geq N`, and we can also find a :math:`k'\in \mathbb N` and :math:`j'\in \{1, \ldots, k'\}` such that :math:`g_{n'}(x) = f_{k',j'}(x) = 0` with :math:`n'\geq N`.   b. For any fixed :math:`01`. Fix :math:`n\geq 2`. The change of variables :math:`u = x^n` results in :math:`du = n x^{n-1} dx`, and, since :math:`u^{1/n} = x`, we have :math:`x^{n-1} = u^{1-\frac{1}{n}}`. Therefore, .. math:: \int_1^N \frac{\sin x^n}{x^n} \, dx = \int_1^{N^n} \frac{\sin u}{u} \, \frac{du}{n u^{1-\frac{1}{n}}} = \frac{1}{n} \int_1^{N^n} \frac{\sin u}{u^{2-\frac{1}{n}}} \, du. Now, .. math:: \lim_{N \to \infty} \frac{1}{n} \left|\int_1^{N^n} \frac{\sin u}{u^{2-\frac{1}{n}}} \, du\right| \leq \lim_{N \to \infty} \frac{1}{n} \int_1^{N^n} u^{\frac{1}{n}-2} \,du = \lim_{N \to \infty} \frac{1}{n} \left.\frac{u^{\frac{1}{n}-1}}{\frac{1}{n}-1} \right|_1^{N^n} = \frac{1}{n-1}. Therefore, .. math:: \left|\int_1^\infty \frac{\sin x^n}{x^n} \, dx\right| \leq \frac{1}{n-1}, and so, .. math:: \lim_{n\to \infty}\int_1^\infty \frac{\sin x^n}{x^n} \, dx = 0. :label: 2001-b Combining results :eq:`2001-b-1` and :eq:`2001-b` yields .. math:: \lim_{n\to \infty}\int_0^\infty \frac{\sin x^n}{x^n} \, dx = 1. c. This is false, as the following example demonstrates: Let :math:`f_n = \frac{1}{n}\chi_{[0,n)}.` Then :math:`f_n \rightarrow 0` uniformly and so :math:`\int \varliminf f_n = 0`. On the other hand, :math:`\int f_n = 1` for all :math:`n\in \mathbb N`. Therefore, :math:`\varliminf \int f_n = 1 \neq 0 = \int \varliminf f_n`. .. container:: toggle .. container:: header **Solution** to :numref:`Problem {number} <2001Nov_p6>` Define :math:`y = \sum_{k=1}^\infty f(\phi_k) \phi_k`, and check that this :math:`y\in H` has the desired properties. First observe that, by properties (1) and (2) given the hint, any :math:`x\in H` can be written as :math:`x = \sum_{k=1}^\infty a_k \phi_k`, where :math:`a_k = \langle x, \phi_k \rangle`, for each :math:`k\in \mathbb N`. Therefore, by linearity of :math:`f`, .. math:: f(x) = f(\sum_k a_k \phi_k) = \sum_k a_k f(\phi_k). :label: 2001-11-6.1 Now .. math:: \langle x, y \rangle = \langle \sum_k a_k \phi_k, \, y \rangle = \sum_k a_k \langle \phi_k, y \rangle, :label: 2001-11-6.2 and, by definition of :math:`y`, .. math:: \langle \phi_k, y \rangle = \langle \phi_k, \, \sum_j f(\phi_j) \phi_j \rangle = \sum_j f(\phi_j) \langle \phi_k, \phi_j \rangle = f(\phi_k). :label: 2001-11-6.3 The last equality holds by orthonormality; i.e., :math:`\langle \phi_k, \phi_j \rangle` is 1 when :math:`j=k` and 0 otherwise. Putting it all together, we see that, for every :math:`x\in H`, .. math:: f(x) &= \sum_k a_k f(\phi_k)\\ &= \sum_k a_k \langle \phi_k, y \rangle\\ &= \langle x,y \rangle The first equation holds by :eq:`2001-11-6.1`, the second, by :eq:`2001-11-6.3`, and the third, by :eq:`2001-11-6.2`. Moreover, this :math:`y` is unique. For, suppose there is another :math:`y'\in H` such that :math:`f(x) = \langle x,y' \rangle` for all :math:`x\in X`. Then :math:`\langle x, y \rangle = f(x) = \langle x, y' \rangle` for all :math:`x\in X`. In particular, :math:`\langle \phi_k, y \rangle = f(\phi_k) = \langle \phi_k, y' \rangle` for each :math:`k\in N`, which, by property (1) of the hint, proves that :math:`y=y'`. Finally, we must show :math:`\|f\| = \|y\|`. Observe, .. math:: \|f\| = \sup_{x\in X}\{|f(x)|: \|x\|\leq 1\} = \sup_{x\in X} \frac{|f(x)|}{\|x\|} = \sup_{x\in X}\frac{|(x,y)|}{\|x\|}, and recall that :math:`|(x,y)| \leq \|x\| \|y\|` holds for all :math:`x, y\in X`. Whence, .. math:: \|f\| = \sup_{x\in X}\frac{|(x,y)|}{\|x\|} \leq \|y\|. :label: 2001-11-6.4 On the other hand, .. math:: \|f\| = \sup_{x\in X}\frac{|(x,y)|}{\|x\|} \geq \frac{|(y,y)|}{\|y\|} = \frac{\|y\|^2}{\|y\|} = \|y\|. :label: 2001-11-6.5 Together, :eq:`2001-11-6.4` and :eq:`2001-11-6.5` give :math:`\|f\| = \|y\|`, as desired. .. container:: toggle .. container:: header **Solution** to :numref:`Problem {number} <2001Nov_p7>` Let :math:`\{T_n\} \subset B(X,Y)` be a Cauchy sequence; i.e.,\ :math:`\|T_n - T_m\| \rightarrow 0` as :math:`m, n \rightarrow \infty`. Fix :math:`x\in X`. Then, .. math:: \|T_n x - T_m x\|_Y \leq \|T_n - T_m\|\|x\|_X \rightarrow 0, \; \text{ as } \; n, m\rightarrow \infty. Therefore, the sequence :math:`\{T_n x\} \subset Y` is a Cauchy sequence in :math:`(Y, \|\cdot \|_Y)`. Since the latter is complete, the limit :math:`\lim_{n\to \infty}T_n x = y \in Y` exists. Define :math:`T:X\rightarrow Y` by :math:`Tx = \lim_{n\to \infty}T_n x`, for each :math:`x\in X`. To complete the proof, we must check that :math:`T` is linear, bounded, and satisfies :math:`\lim_{n\to \infty}\|T_n - T\| = 0`. - :math:`T` is linear: For :math:`x_1, x_2\in X`, .. math:: T(x_1+x_2) &= \lim_{n\to \infty} T_n (x_1+x_2) & & \\ &= \lim_{n\to \infty} (T_n x_1+ T_n x_2) & & (\because T_n \text{ is linear})\\ &= \lim_{n\to \infty} T_n x_1 + \lim_{n\to \infty}T_n x_2 & & (\because \text{ both limits exist})\\ &= T x_1+ T x_2 & &. - :math:`T` is bounded: First, note that :math:`\{\|T_n\|\}` is a Cauchy sequence of real numbers, since :math:`\left|\, \|T_n\| - \|T_m\|\, \right| \leq \|T_n - T_m\| \rightarrow 0`, as :math:`n, m \rightarrow \infty`. Therefore, there is a :math:`c\in \mathbb R` such that :math:`\|T_n\|\rightarrow c`, as :math:`n\rightarrow \infty`. For some :math:`N\in \mathbb N`, then, :math:`\|T_n\| \leq c+1` for all :math:`n\geq N`. Thus, .. math:: \|T_n x\|_Y \leq \|T_n\| \|x\|_X \leq (c+1)\|x\|_X \quad (\forall x\in X). :label: 2001-11-7.0 Now, by definition, :math:`T_n x \rightarrow Tx`, for all :math:`x\in X` and, since the norm :math:`\|\cdot \|_Y` is uniformly continuous, .. math:: \|T_n x \|_Y\rightarrow \|T x \|_Y \quad (\forall x\in X). :label: 2001-11-7.1 Taken together, :eq:`2001-11-7.0` and :eq:`2001-11-7.1` imply :math:`\|T x\|_Y \leq (c+1)\|x\|_X`, for all :math:`x\in X`. Therefore, :math:`T` is bounded. - :math:`\lim_{n\to \infty}\|T_n - T\| = 0`: Fix :math:`\epsilon >0` and choose :math:`N\in \mathbb N` such that :math:`n,m\geq N` implies :math:`\|T_n - T_m\| < \epsilon`. Then, .. math:: \|T_n x - T_m x\| \leq \|T_n - T_m\| \|x\|_X < \epsilon \|x\|_X holds for all :math:`n, m\geq N`, and :math:`x\in X`. Letting :math:`m` go to infinity, then, .. math:: \|T_n x - T x\| = \lim_{m \to \infty} \|T_n x - T_m x\| \leq \epsilon \|x\|_X. That is, :math:`\|T_n x - T x\| \leq \epsilon \|x\|_X`, for all :math:`n\geq N` and :math:`x\in X`. Whence, :math:`\|T_n - T\| \leq \epsilon` for all :math:`n\geq N`. -------------------------------------------- .. rubric:: Footnotes .. [1] Every open set of real numbers is the union of a countable collection of disjoint open intervals (Royden :cite:`Royden:1988`, Proposition 8, page 42). ------------------------------------------------ .. blank .. .. [2] *Proof* :math:`\left|\, \|a\|_Y - \|b\|_Y\right|\leq \|a - b\|_Y\; (\forall a, b\in Y)`.