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.

Problem 50

Let \(\{f_n\}\) be a sequence of Lebesgue measurable functions on a set \(E\subset \mathbb R\), where \(E\) is of finite Lebesgue measure. Suppose that there is \(M>0\) such that \(|f_n(x)|\leq M\) for \(n\geq 1\) and for all \(x \in E\), and suppose that \(\lim_{n \rightarrow \infty} f_n(x) = f(x)\) for each \(x\in E\). Use Egoroff’s theorem to prove that

\[\int_E f(x)\, dx = \lim_{n \to \infty} \int_E f_n(x)\, dx.\]
Problem 51

Let \(f(x)\) be a real-valued Lebesgue integrable function on \([0,1]\).

  1. Prove that if \(f>0\) on a set \(F\subset [0,1]\) of positive measure, then

    \[\int_F f(x)\, dx > 0.\]
  2. Prove that if

    \[\int_0^x f(x)\, dx =0, \quad \text{ for each } x\in [0,1],\]

    then \(f(x)=0\) for almost all \(x\in [0,1]\).

Problem 52

State each of the following:

  1. The Stone-Weierstrass theorem

  2. The Lebesgue (dominated) convergence theorem

  3. Hölder’s inequality

  4. The Riesz representation theorem for \(L_p\)

  5. The Hahn-Banach theorem.

Problem 53
  1. State the Baire category theorem.

  2. Prove the following special case of the uniform boundedness theorem: Let \(X\) be a (nonempty) complete metric space and let \(F\subseteq C(X)\). Suppose that for each \(x\in X\) there is a nonnegative constant \(M_x\) such that

    \[|f(x)| \leq M_x \quad \text{ for all } \quad f\in F.\]

    Prove that there is a nonempty open set \(G\subseteq X\) and a constant \(M>0\) such that

    \[|f(x)| \leq M \quad \text{ holds for all } x\in G \text{ and for all } f\in F.\]
Problem 54

Prove or disprove:

  1. \(L_2\) convergence implies pointwise convergence.

  2. \[\lim_{n \to \infty} \int_0^\infty \frac{\sin(x^n)}{x^n} \, dx = 0.\]
  3. Let \(\{f_n\}\) be a sequence of measurable functions defined on \([0,\infty)\). If \(f_n\rightarrow 0\) uniformly on \([0,\infty)\), as \(n\rightarrow \infty\), then

    \[\varliminf \int_{[0,\infty)} f_n(x) \, dx = \int_{[0,\infty)} \varliminf f_n(x) \, dx.\]
Problem 55

Let \(f:H\rightarrow H\) be a bounded linear functional on a separable Hilbert space \(H\) (with inner product denoted by \(\langle \cdot, \cdot \rangle\)). Prove that there is a unique element \(y\in H\) such that

\[f(x) = \langle x, y \rangle \quad \text{for all} \quad x\in H \quad \text{and} \quad \|f\| = \|y\|.\]

Hint. You may use the following facts: A separable Hilbert space, \(H\), contains a complete orthonormal sequence, \(\{\phi_k\}_{k=1}^\infty\), satisfying the following properties:

  1. If \(x,y\in H\) and if \(\langle x,\phi_k \rangle = \langle y,\phi_k \rangle\) for all \(k\), then \(x=y\).

  2. Parseval’s equality holds; that is, for all \(x\in H\), \(\langle x, x \rangle = \sum_{k=1}^\infty a_k^2\), where \(a_k = \langle x,\phi_k \rangle\).

Problem 56

Let \(X\) be a normed linear space and let \(Y\) be a Banach space. Let

\[B(X,Y) = \{A \mid A:X\to Y \text{ is a bounded linear operator}\}.\]

Then with the norm \(\|A\| = \sup_{\|x\|\leq 1} \|Ax\|\), \(B(X,Y)\) is a normed linear space (you need not show this). Prove that \(B(X,Y)\) is a Banach space; that is, prove that \(B(X,Y)\) is complete.


Solutions

Solution to Problem 50

First note that \(|f(x)|\leq M\) for all \(x\in E.\) To see this, suppose it’s false for some \(x_0\in E\), so that \(|f(x_0)|>M\). Then there is some \(\epsilon>0\) such that \(|f(x_0)|=M+\epsilon\). By the triangle inequality, then, for all \(n\in \mathbb{N}\),

\[|f(x_0) - f_n(x_0)| \geq ||f(x_0)| - |f_n(x_0)|| = |M+\epsilon - |f_n(x_0)|| \geq \epsilon,\]

which contradicts \(f_n(x_0) \rightarrow f(x_0)\). Thus, \(|f(x)|\leq M\) for all \(x\in E\).

Next, fix \(\epsilon >0\). By Egoroff’s theorem, there is a \(G\subset E\) such that \(\mu(E\setminus G) < \epsilon\) and \(f_n \rightarrow f\) uniformly on \(G\). Furthermore, since \(|f_n| \leq M\) and \(|f|\leq M\) and \(\mu(E) < \infty\), it’s clear that \(\{f_n\}\subset L_1\) and \(f\in L_1\), so the following inequalities make sense (here we’re using the notation \(\|f\|_G = \sup\{|f(x)|: x\in G\}\)):

\[\begin{split}\left| \int_E f\, d\mu-\int_E f_n\, d\mu\right| &\leq \int_E |f- f_n|\, d\mu = \int_{E\setminus G} |f- f_n|\, d\mu + \int_G |f- f_n|\, d\mu\\ &\leq \int_{E\setminus G} |f|\, d\mu + \int_{E\setminus G} |f_n|\, d\mu + \|f(x)- f_n(x)\|_G \mu(G)\\ &\leq 2M \mu(E\setminus G) + \|f(x)- f_n(x)\|_G \mu(G)\\ &< \epsilon + \|f(x)- f_n(x)\|_G \mu(G).\end{split}\]

Finally, \(\mu(G) \leq \mu(E) < \infty\) and \(\|f(x)- f_n(x)\|_G \rightarrow 0\), which proves that \(\int_E f_n\, d\mu \rightarrow \int_E f\, d\mu\).

Solution to Problem 51

  1. Define \(F_n = \{x\in F: f(x) > 1/n\}\). Then

    \[F_1 \subseteq F_2 \subseteq \cdots \uparrow \bigcup_n F_n = F,\]

    and \(m(F)>0\) implies

    \[0< m(F) = m(\cup_n F_n) \leq \sum_n m(F_n).\]

    Therefore, \(m(F_k) >0\) for some \(k\in \mathbb N\), and then it follows from the definition of \(F_k\) that

    \[0< \frac{1}{k} m(F_k) \leq \int_{F_k} f \, dm \leq \int_F f\, dm.\]
  2. Suppose there is a subset \(E\subset [0,1]\) of positive measure such that \(f>0\) on \(E\). Then part (a) implies \(\int_E f\, dm >0\). Let \(F\subset E\) be a closed subset of positive measure. (That such a closed subset exists follows from Proposition 3.15 of [Roy88].) Then, again by (a), \(\int_F f\, dm >0\). Now consider the set \(G = [0,1]\setminus F\), which is open in \([0,1]\), and hence 1 is a countable union of disjoint open intervals, \(G = \bigcup_n (a_n, b_n)\). Therefore,

    \[0 = \int_{[0,1]} f \, dm = \sum_n \int_{(a_n,b_n)} f \, dm + \int_F f \, dm,\]

    so \(\int_F f\, dm >0\) implies

    \[\sum_n \int_{(a_n,b_n)} f \, dm < 0.\]

    Thus, \(\int_{(a_k,b_k)} f \, dm < 0\) for some \((a_k,b_k) \subset [0,1]\). On the other hand,

    \[\int_{(a_k,b_k)} f \, dm = \int_0^{b_k}f(x) \, dm(x) - \int_0^{a_k} f(x) \, dm(x).\]

    By the initial hypothesis, both terms on the right are zero, which gives the desired contradiction.

Solution to Problem 53

  1. See the statements of the Baire category theorem and its corollaries.

  2. Define \(A_m = \{x∈ X: |f(x)|≤ m, \, ∀ f∈ F\}\). Then \(X = \bigcup_{m=1}^\infty A_m\), since for every \(x\) there is a finite number \(M_x\) such that \(|f(x)| \leq M_x\) for all \(f\in F\). Now note that \(A_m = \bigcap_{f\in F} \{x\in X: |f(x)|\leq m\}\), and, since \(f\) and \(a\mapsto |a|\) are continuous functions, each \(\{x\in X: |f(x)|\leq m\}\) is closed, so \(A_m\) is closed. Therefore, Corollary 2 of the Baire category theorem implies that there must be some \(m\in \mathbb N\) such that \(A_m^\circ \neq \emptyset\), so the set \(G = A_m^\circ\) and the number \(M=m\) satisfy the given criteria.

Solution to Problem 54

  1. This is false, as the following example demonstrates: For each \(k\in \mathbb{N}\), define \(f_{k,j} = \chi_{[\frac{j-1}{k}, \frac{j}{k})}\) for \(j = 1, \ldots, k\), and let \(\{g_n\}\) be the sequence defined by

    \[\begin{split}g_1 &= f_{1,1}, \\ g_2 &= f_{2,1}, \; g_3 = f_{2,2}, \\ g_4 &= f_{3,1}, \; g_5 = f_{3,2}, \; g_6 = f_{3,3}, \\ g_7 &= f_{4,1}, \ldots\end{split}\]

    Then \(\int |f_{k,j}|^2 \, d\mu = 1/k\) for each \(j = 1, \ldots, k\), so \(\|f_{k,j}\|_2 = 1/\sqrt{k} \rightarrow 0\), as \(k\rightarrow \infty\). Therefore \(\|g_n\|_2 \rightarrow 0\) as \(n\rightarrow \infty\). However, \(\{g_n\}\) does not converge pointwise since, for every \(x\in [0,1]\) and every \(N \in \mathbb N\), we can always find some \(k\in \mathbb N\) and \(j\in \{1,\ldots, k\}\) such that \(g_n(x) = f_{k,j}(x) = 1\) with \(n\geq N\), and we can also find a \(k'\in \mathbb N\) and \(j'\in \{1, \ldots, k'\}\) such that \(g_{n'}(x) = f_{k',j'}(x) = 0\) with \(n'\geq N\).

  2. For any fixed \(0<x<1\), \(\lim_{n\to \infty}x^n = 0\). Also, \(\frac{\sin t}{t} \rightarrow 1\), as \(t\rightarrow 0\), which can be proved by L’Hopital’s rule. Together, these two facts yield

    \[\lim_{n\to \infty}\frac{\sin x^n}{x^n} = 1.\]

    Now, recall that \(|\sin \theta| \leq |\theta|\) for all real \(\theta\). Indeed, since \(\sin \theta = \int_0^\theta \cos x \, dx\), we have, for \(\theta \geq 0\),

    \[|\sin \theta| \leq \int_0^\theta |\cos x| \, dx \leq \int_0^\theta 1 \, dx = \theta,\]

    and, for \(\theta < 0\),

    \[|\sin \theta| = |\sin (-\theta)| \leq |-\theta| = |\theta|.\]

    In particular, for any \(0<x<1\), \(\frac{|\sin x^n|}{|x^n|} \leq 1\). Therefore, we can apply the dominated convergence theorem to the function \(\frac{\sin x^n}{x^n}\), to obtain

    (30)\[\lim_{n\to \infty}\int_0^1 \frac{\sin x^n}{x^n} \, dx = \int_0^1 1 \, dx = 1.\]

    Next consider the part of the integral over \(1\leq x < N\), for any real \(N>1\). Fix \(n\geq 2\). The change of variables \(u = x^n\) results in \(du = n x^{n-1} dx\), and, since \(u^{1/n} = x\), we have \(x^{n-1} = u^{1-\frac{1}{n}}\). Therefore,

    \[\int_1^N \frac{\sin x^n}{x^n} \, dx = \int_1^{N^n} \frac{\sin u}{u} \, \frac{du}{n u^{1-\frac{1}{n}}} = \frac{1}{n} \int_1^{N^n} \frac{\sin u}{u^{2-\frac{1}{n}}} \, du.\]

    Now,

    \[\lim_{N \to \infty} \frac{1}{n} \left|\int_1^{N^n} \frac{\sin u}{u^{2-\frac{1}{n}}} \, du\right| \leq \lim_{N \to \infty} \frac{1}{n} \int_1^{N^n} u^{\frac{1}{n}-2} \,du = \lim_{N \to \infty} \frac{1}{n} \left.\frac{u^{\frac{1}{n}-1}}{\frac{1}{n}-1} \right|_1^{N^n} = \frac{1}{n-1}.\]

    Therefore,

    \[\left|\int_1^\infty \frac{\sin x^n}{x^n} \, dx\right| \leq \frac{1}{n-1},\]

    and so,

    (31)\[\lim_{n\to \infty}\int_1^\infty \frac{\sin x^n}{x^n} \, dx = 0.\]

    Combining results (30) and (31) yields

    \[\lim_{n\to \infty}\int_0^\infty \frac{\sin x^n}{x^n} \, dx = 1.\]
  3. This is false, as the following example demonstrates: Let \(f_n = \frac{1}{n}\chi_{[0,n)}.\) Then \(f_n \rightarrow 0\) uniformly and so \(\int \varliminf f_n = 0\). On the other hand, \(\int f_n = 1\) for all \(n\in \mathbb N\). Therefore, \(\varliminf \int f_n = 1 \neq 0 = \int \varliminf f_n\).

Solution to Problem 55

Define \(y = \sum_{k=1}^\infty f(\phi_k) \phi_k\), and check that this \(y\in H\) has the desired properties.

First observe that, by properties (1) and (2) given the hint, any \(x\in H\) can be written as \(x = \sum_{k=1}^\infty a_k \phi_k\), where \(a_k = \langle x, \phi_k \rangle\), for each \(k\in \mathbb N\). Therefore, by linearity of \(f\),

(32)\[f(x) = f(\sum_k a_k \phi_k) = \sum_k a_k f(\phi_k).\]

Now

(33)\[\langle x, y \rangle = \langle \sum_k a_k \phi_k, \, y \rangle = \sum_k a_k \langle \phi_k, y \rangle,\]

and, by definition of \(y\),

(34)\[\langle \phi_k, y \rangle = \langle \phi_k, \, \sum_j f(\phi_j) \phi_j \rangle = \sum_j f(\phi_j) \langle \phi_k, \phi_j \rangle = f(\phi_k).\]

The last equality holds by orthonormality; i.e., \(\langle \phi_k, \phi_j \rangle\) is 1 when \(j=k\) and 0 otherwise. Putting it all together, we see that, for every \(x\in H\),

\[\begin{split}f(x) &= \sum_k a_k f(\phi_k)\\ &= \sum_k a_k \langle \phi_k, y \rangle\\ &= \langle x,y \rangle\end{split}\]

The first equation holds by (32), the second, by (34), and the third, by (33).

Moreover, this \(y\) is unique. For, suppose there is another \(y'\in H\) such that \(f(x) = \langle x,y' \rangle\) for all \(x\in X\). Then \(\langle x, y \rangle = f(x) = \langle x, y' \rangle\) for all \(x\in X\). In particular, \(\langle \phi_k, y \rangle = f(\phi_k) = \langle \phi_k, y' \rangle\) for each \(k\in N\), which, by property (1) of the hint, proves that \(y=y'\).

Finally, we must show \(\|f\| = \|y\|\). Observe,

\[\|f\| = \sup_{x\in X}\{|f(x)|: \|x\|\leq 1\} = \sup_{x\in X} \frac{|f(x)|}{\|x\|} = \sup_{x\in X}\frac{|(x,y)|}{\|x\|},\]

and recall that \(|(x,y)| \leq \|x\| \|y\|\) holds for all \(x, y\in X\). Whence,

(35)\[\|f\| = \sup_{x\in X}\frac{|(x,y)|}{\|x\|} \leq \|y\|.\]

On the other hand,

(36)\[\|f\| = \sup_{x\in X}\frac{|(x,y)|}{\|x\|} \geq \frac{|(y,y)|}{\|y\|} = \frac{\|y\|^2}{\|y\|} = \|y\|.\]

Together, (35) and (36) give \(\|f\| = \|y\|\), as desired.

Solution to Problem 56

Let \(\{T_n\} \subset B(X,Y)\) be a Cauchy sequence; i.e.,\(\|T_n - T_m\| \rightarrow 0\) as \(m, n \rightarrow \infty\). Fix \(x\in X\). Then,

\[\|T_n x - T_m x\|_Y \leq \|T_n - T_m\|\|x\|_X \rightarrow 0, \; \text{ as } \; n, m\rightarrow \infty.\]

Therefore, the sequence \(\{T_n x\} \subset Y\) is a Cauchy sequence in \((Y, \|\cdot \|_Y)\). Since the latter is complete, the limit \(\lim_{n\to \infty}T_n x = y \in Y\) exists. Define \(T:X\rightarrow Y\) by \(Tx = \lim_{n\to \infty}T_n x\), for each \(x\in X\). To complete the proof, we must check that \(T\) is linear, bounded, and satisfies \(\lim_{n\to \infty}\|T_n - T\| = 0\).

  • \(T\) is linear:

    For \(x_1, x_2\in X\),

    \[\begin{split}T(x_1+x_2) &= \lim_{n\to \infty} T_n (x_1+x_2) & & \\ &= \lim_{n\to \infty} (T_n x_1+ T_n x_2) & & (\because T_n \text{ is linear})\\ &= \lim_{n\to \infty} T_n x_1 + \lim_{n\to \infty}T_n x_2 & & (\because \text{ both limits exist})\\ &= T x_1+ T x_2 & &.\end{split}\]
  • \(T\) is bounded:

    First, note that \(\{\|T_n\|\}\) is a Cauchy sequence of real numbers, since \(\left|\, \|T_n\| - \|T_m\|\, \right| \leq \|T_n - T_m\| \rightarrow 0\), as \(n, m \rightarrow \infty\). Therefore, there is a \(c\in \mathbb R\) such that \(\|T_n\|\rightarrow c\), as \(n\rightarrow \infty\). For some \(N\in \mathbb N\), then, \(\|T_n\| \leq c+1\) for all \(n\geq N\). Thus,

    (37)\[\|T_n x\|_Y \leq \|T_n\| \|x\|_X \leq (c+1)\|x\|_X \quad (\forall x\in X).\]

    Now, by definition, \(T_n x \rightarrow Tx\), for all \(x\in X\) and, since the norm \(\|\cdot \|_Y\) is uniformly continuous,

    (38)\[\|T_n x \|_Y\rightarrow \|T x \|_Y \quad (\forall x\in X).\]

    Taken together, (37) and (38) imply \(\|T x\|_Y \leq (c+1)\|x\|_X\), for all \(x\in X\). Therefore, \(T\) is bounded.

  • \(\lim_{n\to \infty}\|T_n - T\| = 0\):

    Fix \(\epsilon >0\) and choose \(N\in \mathbb N\) such that \(n,m\geq N\) implies \(\|T_n - T_m\| < \epsilon\). Then,

    \[\|T_n x - T_m x\| \leq \|T_n - T_m\| \|x\|_X < \epsilon \|x\|_X\]

    holds for all \(n, m\geq N\), and \(x\in X\). Letting \(m\) go to infinity, then,

    \[\|T_n x - T x\| = \lim_{m \to \infty} \|T_n x - T_m x\| \leq \epsilon \|x\|_X.\]

    That is, \(\|T_n x - T x\| \leq \epsilon \|x\|_X\), for all \(n\geq N\) and \(x\in X\). Whence, \(\|T_n - T\| \leq \epsilon\) for all \(n\geq N\).


Footnotes

1

Every open set of real numbers is the union of a countable collection of disjoint open intervals (Royden [Roy88], Proposition 8, page 42).


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