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 \([0,1]\).

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

\[\int_F f(x)\, dx > 0.\]
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:

The Stone-Weierstrass theorem
The Lebesgue (dominated) convergence theorem
Hölder’s inequality
The Riesz representation theorem for \(L_p\)
The Hahn-Banach theorem.

Problem 53

State the Baire category theorem.
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:

\(L_2\) convergence implies pointwise convergence.
\[\lim_{n \to \infty} \int_0^\infty \frac{\sin(x^n)}{x^n} \, dx = 0.\]
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:

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

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¶

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 \(\int_E f_n\, d\mu \rightarrow \int_E f\, d\mu\).

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

\[0< \frac{1}{k} m(F_k) \leq \int_{F_k} f \, dm \leq \int_F f\, dm.\]
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.

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

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 \(n'\geq N\).
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

\[\lim_{n\to \infty}\int_0^\infty \frac{\sin x^n}{x^n} \, dx = 1.\]
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, \(\varliminf \int f_n = 1 \neq 0 = \int \varliminf f_n\).

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.

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 \(n\geq N\).