Exam 2 (10–16)¶

1994 Nov¶

Instructions Masters students: Do any 5 problems. Ph.D. students: Do any 6 problems.

Problem 10

Let \(E\) be a normed linear space. Show that \(E\) is complete if and only if, whenever \(∑_1^∞ \|x_n \| < ∞\), then \(∑_1^∞ x_n\) converges to an \(s ∈ E\).

Problem 11

Let \(f_n\) be a sequence of real-valued continuous functions on a compact Hausdorff space \(X\). Show that if \(f_1 ≥ f_2 ≥ f_3 ≥ \cdots\), and \(f_n(x) → 0\) for all \(x ∈ X\), then \(f_n → 0\) uniformly.

Problem 12

Let \(f\) be integrable on the real line with respect to Lebesgue measure. Evaluate \(\lim\limits_{n→∞} ∫_{-∞}^∞ f(x-n) \left(\frac{x}{1+|x|}\right)\, dx\). Justify all steps.

Problem 13

Let \(X\) and \(Y\) be metric spaces and let \(f\) be a continuous mapping from \(X\) onto \(Y\).

Show that if \(X\) is compact, then \(Y\) is compact.
Show that if \(X\) is compact, and \(f\) is one-one, then \(f\) is a homeomorphism.
Is the conclusion of (b) true if \(Y\) is compact, but \(X\) is not?

Problem 14

Show \(L_p[0,1]⊆ L_q[0,1]\) if \(p > q\). Is this true for \(L_p[0,∞)\) and \(L_q[0,∞)\)?
Clearly

\[⋂_{p≥ 1} L_p[0,1]⊇ L_∞[0,1].\]

Does equality hold?

Problem 15

Let \(λ\) be Lebesgue measure on \(ℬ(ℝ^2)\), the Borel sets of \(ℝ^2\). Define a Borel measure \(μ\) as follows: for each \(A ∈ ℬ(ℝ^2)\),

\[μ A = ∬_A \frac{dλ(x,y)}{1+x^2y^2}.\]

Show that \(λ\) is absolutely continuous with respect to \(μ\) and find \(\frac{dλ}{dμ}\), the Radon-Nikodym derivative.

Problem 16

Let \(\mathcal P\) be the set of continuous functions on \([0,3]\) whose graphs consist of a finite number of straight line segments. Prove that \(\mathcal P\) is dense in \(L_2[0,3]\).

(⇒) Suppose \(E\) is complete. Let \(\{x_n\} ⊆ E\) be absolutely convergent; i.e., \(∑_n\|x_n\| < ∞\).

We must show

\[∑_{n=1}^∞ x_n := \lim_{N→∞}∑_{n=1}^N x_n = s∈ E.\]

Let \(S_N = \sum_{n=1}^Nx_n\). Then, for any \(j\in \mathbb{N}\),

\[\|S_{N+j}-S_N\| = \left\|∑_{n=N+1}^{N+j} x_n \right\| ≤ ∑_{n=N+1}^{N+j}\|x_n\| → 0\]

as \(N → ∞\), since \(∑_n\|x_n\| < ∞\).

Therefore, \(\{S_N\}\) is a Cauchy sequence.

Since \(E\) is complete, there is an \(s ∈ E\) such that \(∑_{n=1}^∞ x_n = \lim\limits_{N→∞} S_N = s.\)

(⇐) Suppose whenever \(∑_1^∞ \|x_n \| < ∞\), then \(∑_1^∞ x_n\) converges to an \(s ∈ E\).

Let \(\{y_n\} ⊆ E\) be a Cauchy sequence. That is, \(\|y_n - y_m\| → 0\) as \(n, m → ∞\).

Let \(n_1 < n_2 < \cdots\) be a subsequence such that

\[n, m ≥ n_j \; ⟹ \; \|y_n - y_m\| < 2^{-j}.\]

Observe, for \(k>1\),

\[y_{n_k} = y_{n_1} + (y_{n_2} - y_{n_1}) + (y_{n_3}- y_{n_2}) + \cdots + (y_{n_k}- y_{n_{k-1}}) = y_{n_1}+∑_{j=1}^{k-1} (y_{n_{j+1}}- y_{n_j}),\]

and

\[∑_{j=1}^∞ \|y_{n_{j+1}}- y_{n_j}\| < ∑_{j=1}^∞ 2^{-j} = 1\]

By hypothesis, this implies that

\[y_{n_k} -y_{n_1} = ∑_{j=1}^{k-1} (y_{n_{j+1}}- y_{n_j}) → s ∈ E, \text{ as } k → \infty.\]

We have thus found a subsequence \(\{y_{n_k}\} ⊆ \{y_n\}\) having a limit in \(E\).

Finally, since \(\{y_n\}\) is Cauchy, it is easy to verify that \(\{y_n\}\) must converge to the same limit.

This proves that every Cauchy sequence in \(E\) converges to a point in \(E\), as desired.

☐

Fix \(ε > 0\). We will show there exists \(N>0\) such that \(n ≥ N\) implies \(\|f_n\|_∞ < ε\).

For each \(x_α ∈ X\), let \(N_α\) be such that \(n ≥ N_α\) implies \(|f_n(x_α)| < ε/2\).

Let \(V_α\) be a neighborhood of \(x_α\) such that \(x ∈ V_α\) implies \(|f_{N_α}(x)-f_{N_α}(x_α)| < ε/2\).

Then, \(|f_{N_α}(x)| ≤ |f_{N_α}(x)-f_{N_α}(x_α)| + |f_{N_α}(x_α)| < ε/2\), and so \(f_1 ≥ f_2 ≥ f_3 ≥ \cdots\) implies that for all \(n≥ N_α\) and \(x ∈ V_α\) we have \(|f_n(x)| < ε\).

The collection \(\{V_α\}_{α∈𝔄}\) of these neighborhoods covers \(X\). Since \(X\) is compact, there exists a finite subcover \(\{V_{α_1}, \dots, V_{α_M}\}\), so that \(X = ⋃_{i=1}^M V_{α_i}\).

Let \(N = \max \{N_{α_i} : 1≤ i ≤ M\}\). Then for every \(x ∈ X\), we have \(x ∈ V_{α_i}\) for some \(i ∈ \{1, \dots, M\}\), and \(|f_n(x)|<ε\) whenever \(n ≥ N\). Therefore, \(\|f_n\|_∞ < ε\).

☐

Fix \(n>0\). Consider the change of variables, \(y = x-n\). Then \(dy = dx\) and \(x = y+n\), so

(7)¶\[\begin{split}\int_{-\infty}^\infty f(x-n) \frac{x}{1+|x|}\, dx &=\int_{-\infty}^\infty f(y) \frac{y+n}{1+|y+n|}\, dy\\[4pt] &=\int_{-n}^\infty f(y) \frac{y+n}{1+y+n}\, dy +\int_{-\infty}^{-n} f(y) \frac{y+n}{1-(y+n)}\, dy.\end{split}\]

Note that, when \(y \geq -n\), \(\frac{y+n}{1+y+n} \in [0,1)\), and increases to \(1\) as \(n\) tends to infinity. Thus,

\[0\leq |f(y)|\,\frac{y+n}{1+y+n} \leq |f(y)|,\]

for all \(y \geq -n\). Define the function 1

\[g_n(y) = f(y)\, \frac{y+n}{1+y+n} \,\mathbf{1}_{[-n,\infty)}(y).\]

Then \(|g_n|\leq |f|\) and \(\lim\limits_{n\rightarrow \infty} g_n = f\). Therefore, by the dominated convergence theorem,

\[\lim_{n\rightarrow \infty}\int_{-n}^\infty f(y) \frac{y+n}{1+y+n}\, dy = \lim_{n\rightarrow \infty} \int_{-\infty}^{\infty} g_n(y) \, dy =\int_{-\infty}^{\infty} f(y) \, dy.\]

Next, consider the second term in (7). Define the function

\[h_n(y) = f(y)\, \frac{y+n}{1-(y+n)} \,\mathbf{1}_{(-\infty,-n]}(y).\]

It is not hard to check that

\[\frac{|y+n|}{|1-(y+n)|} \,\mathbf{1}_{(-\infty,-n]}(y)\in [0,1),\]

from which it follows that \(|h_n| \leq |f|\). Also, it is clear that, for all \(y\),

\[\lim_{n\rightarrow \infty} h_n(y) = f(y)\,\lim_{n\rightarrow \infty}\frac{y+n}{1-(y+n)} \,\mathbf{1}_{(-\infty,-n]}(y) = 0.\]

Therefore, the dominated convergence theorem implies that

\[\lim_{n\rightarrow \infty}\int_{-\infty}^{-n} f(y) \frac{y+n}{1-(y+n)}\, dy = 0.\]

Combining the two results above, we see that \(\lim\limits_{n\rightarrow \infty} \int_{-\infty}^\infty f(x-n) \left(\frac{x}{1+|x|}\right)\, dx = \int_{-\infty}^{\infty}f(x) \, dx\).

☐

Remark Intuitively, this is the result we expect because the translation \(f(x-n) = T_n f(x)\) is merely shifting the support of \(f\) to the right tail of the measure \(d\mu := \frac{x}{1+|x|}\, dx\), and in the tail this measure looks like \(dx\).

Let \(\{V_α\}_{α∈𝔄} \) of \(Y\).

Since \(Y\) is a metric space, this cover can be reduced to a countable subcover \(\{V_{α_i}\}_{i=1}^∞\).

Define \(U_{α_i} = f^{-1}(V_{α_i})\). Then \(X = ⋃_{i=1}^∞ U_{α_i}\).

Since \(X\) is compact, there is a finite subcover \(\{U_{α_{i_1}}, \dots, U_{α_{i_M}}\}\) so that \(X = ⋃_{j=1}^M U_{α_{i_j}}\).

Now, \(f(U_{α_{i_j}}) = f f^{-1}(V_{α_{i_j}}) ⊆ V_{α_{i_j}}.\) Therefore,

\[Y = f(X) = f(⋃_{j=1}^M U_{α_{i_j}}) = ⋃_{j=1}^M f(U_{α_{i_j}}) ⊆ ⋃_{j=1}^M V_{α_{i_j}}.\]

Thus, we have reduced an arbitrary cover of \(Y\) to a finite subcover, thereby showing that \(Y\) is compact.

☐
Suppose \(X\) is compact and \(f\) is a bijection.

We must show that \(f^{-1}\) exists and is continuous.

That \(f^{-1}\) exists is obvious since \(f\) is bijective.

To show \(f^{-1}\) is continuous, we show that for each open set \(U ⊆ X\) the set \([f^{-1}]^{-1} (U) = f(U)\) is open in \(Y\).

Let \(U\) be an open subset of \(X\). Let \(U'\) denote the complement of \(U\) in \(X\), which is closed, thus compact (since \(X\) is compact).

Since \(f ∈ C(X, Y)\), the argument in Part a implies \(f(U')\) is a compact subset of \(Y\). Therefore, \(f(U')\) is closed (and bounded), so \([f(U')]'\) is open in \(Y\).

Now, it’s not hard to show that if \(f\) is one-one, then \(f(U) = f(X)- f(U')\), from which \(f(U) = [f(U')]'\) is open in \(Y\), as desired.

☐
Finally, suppose \(f: X → Y\) is a continuous bijection and \(Y\) is compact.

If \(f\) is a homeomorphism, then \(f^{-1}\) is a continuous mapping from \(Y\) onto \(X\) and, by a, \(X\) is compact. Thus, if \(X\) is not compact, then \(f\) cannot be a homeomorphism.

☐

Fix \(f\in L_p[0,1]\). Let \(\alpha = p/q\) (which is greater than \(1\), since \(p>q\)).

By Hölder’s inequality, with \(\frac{1}{\alpha}+\frac{1}{\beta} = 1\),

(8)¶\[\int |f|^q \, d\mu = \int |f|^q \chi_{[0,1]}\, d\mu \leq \| |f|^q\|_\alpha \|\chi_{[0,1]}\|_\beta = \| |f|^q \|_\alpha.\]

Since \(f \in L_p[0,1]\),

(9)¶\[\| |f|^q \|_\alpha = \left(\int |f|^{q\alpha} \, d\mu\right)^{1/\alpha} = \left(\int |f|^p \, d\mu\right)^{1/\alpha} = \|f\|_p^{p/\alpha}= \|f\|_p^{q} < \infty,\]

By (8) and (9), \(\|f\|_q^{q} \leq \|f\|_p^{q} < \infty\), so \(f \in L_q[0,1]\), as desired.

The same result does not hold for \(L_p[0,\infty)\) and \(L_q[0,\infty)\). Here is a counterexample.

For some \(\alpha\) (to be determined), let

\[\begin{split}f(x) = \begin{cases}x^{-\alpha},& x\in [1, \infty)\\ 0, & x\in [0,1).\end{cases}\end{split}\]

Then,

(10)¶\[\|f\|_p^p = \int_0^\infty |f(x)|^p \, dx = \int_1^\infty x^{-\alpha p} \, dx = \left. \frac{x^{1-\alpha p}}{1-\alpha p} \right|_{x=1}^{x\to \infty}.\]

Similarly,

(11)¶\[\|f\|^q_q = \left. \frac{x^{1-\alpha q}}{1-\alpha q} \right|_{x=1}^{x\to \infty}.\]

If \(p>q\), then we can choose \(\alpha\) so that \(1-\alpha p < 0 < 1-\alpha q\) (for example, let \(\alpha = \frac{2}{p+q}\)).

Therefore, by (10) and (11), \(\|f\|_p < \infty\) while \(\|f\|_q = \infty\).

☐
As asserted in the problem statement, \(\bigcap_{p\geq 1} L_p[0,1]\supseteq L_\infty[0,1].\)

(Indeed, if \(\exists M >0\) such that \(|f|\leq M\) a.e., then for all \(p\geq 1\), \(\|f\|_p \leq \left(M^p\mu[0,1]\right)^{1/p} = M < \infty\).)

Equality does not hold. That is, there are functions belonging to \(L_p[0,1]\) for every finite \(p\geq 1\) that do not belong to \(L_{\infty}[0,1]\). Here is an example.

Define \(f : [0,1] \to ℝ\) as follows:

\[f = \sum_{n=1}^\infty c_n \chi_{[\frac{1}{n+1}, \frac{1}{n})}.\]

Then

(12)¶\[\left(\int |f|^p \, d\mu\right)^{1/p} = \left(\sum_{n=1}^\infty c_n^p \mu[\frac{1}{n+1}, \frac{1}{n}]\right)^{1/p}= \left(\sum_{n=1}^\infty \frac{c_n^p}{n(n+1)}\right)^{1/p}.\]

Let \(c_n = \log n\). Then, for \(n\) large enough, \((\log n)^p < \sqrt{n}\). Therefore, there exists \(N>0\) such that

\[\sum_{n=N+1}^\infty \frac{c_n^p}{n(n+1)} < \sum_{n=N+1}^\infty \frac{\sqrt{n}}{n(n+1)} < \sum_{n=N+1}^\infty \left(\frac{1}{n}\right)^{3/2} < \infty.\]

This proves that for every finite \(p\geq 1\) there is an \(N>0\) such that the tail of the series in (12) converges, so the whole series converges. This proves that \(f\in L_p[0,1]\) for every finite \(p\geq 1\).

On the other hand, consider the growth of \(f\) near zero, say, \(\frac{1}{n+1} \leq x < \frac{1}{n}\). Over that interval, \(f(x) = \log(n)\). So as \(x\) approaches zero, \(f\) grows without bound. Therefore, \(f\notin L_\infty[0,1]\).

☐

Recall the definition of an absolutely continuous measure, where we used \(λ ≪ μ\) to denote the fact that \(λ\) is absolutely continuous (AC) with respect to \(μ\).

In the present case, we clearly have \(μ ≪ λ\) and \(dμ/dλ = (1 + x^2y^2)^{-1}\).

Suppose \(A ∈ ℬ(ℝ^2)\) (Borel sets of \(ℝ^2\)) and \(λ A > 0\). We will show \(μ A>0\). Since both measure are positive, this will suffice to prove that \(μ A=0\) implies \(λ A = 0\); that is, \(λ ≪ μ\).

Partition \(ℝ^2\) into a grid \(ℝ^2 = ⋃_{m,n} [m, m+1) × [n, n+1)\).

For all \(m, n ∈ ℤ\), define \(A_{m,n} = A ∩ ([m, m+1) × [n, n+1))\). Then,

\[λ A = ∑_{m,n} λ A_{m,n} > 0\]

implies that there exist \(n_0\) and \(m_0\) such that \(λ A_{m_0, n_0} > 0\). Thus,

\[∬_{A_{m_0,n_0}} \frac{d λ (x,y)}{1 + x^2y^2} ≥ \inf \left\{\frac{1}{1 + x^2y^2} ∣ n_0 ≤ x < n_0+1, \, m_0 ≤ y < m_0+1\right\} ⋅ λ A_{m_0,n_0} > 0.\]

Therefore,

\[μ A = ∬_A \frac{d λ(x,y)}{1 + x^2y^2} ≥ ∬_{A_{m_0,n_0}} \frac{d λ(x,y)}{1 + x^2y^2} > 0.\]

We now have two mutually AC positive measures, \(λ ≪ μ ≪ λ\).

We want a function \(f\) which satisfies \(λ A = ∫_A f \, dμ\).

Recall, for a measure \(ν\), the function \(φ A = ∫_A g \, dν\) defines a measure and \(∫ h \, dφ = ∫ hg\, dν.\)

In the present case, we have

\[∬_A (1 + x^2y^2) \, dμ = ∬_A \frac{1 + x^2y^2}{1 + x^2y^2} \, dλ = λ A,\]

so the Radon-Nikodym derivative of \(λ\) with respect to \(μ\) is \(\frac{dλ}{dμ} = 1 + x^2y^2\).

☐

We take for granted that \(C[0,3]\) is dense in \(L_2[0,3]\), so it suffices to prove that \(\mathcal P\) is dense in \(C[0,3]\).

Fix \(g \in C[0,3]\) and \(\epsilon > 0\).

Since \([0,3]\) is closed and bounded, hence compact, \(g\) is uniformly continuous in \([0,3]\) (by the basic continuity theorem).

For \(n > 0\) and \(0 < \frac{1}{n} < \delta\), partition \([0,3]\) as follows:

Let \(E_k = [\frac{k-1}{n}, \frac{k}{n})\) for \(k = 1, 2, \dots, 3n-1\) and let \(E_{3n} = [\frac{3n-1}{n}, \frac{3n}{n}]\). Then,

\[[0,3] = \bigcup_{k=1}^{3n} E_k.\]

Define \(p \in \mathcal P\) as follows:

\[\text{Let } \quad p_k(x) = (k - nx)g\left(\frac{k-1}{n}\right) + (nx - (k-1))g\left(\frac{k}{n}\right), \quad x \in E_k, \quad \text{ and let}\]

\[p(x) = \sum_{k=1}^{3n} p_k(x) \chi_{E_k}(x).\]

Then \(p\) is piece-wise linear and for \(x \in E_k\) (\(k = 1,2,\dots, 3n\)),

\[\begin{split}|p(x) - g(x)| &= |p_k(x) - g(x)|\\ &= |(k - nx)g\left(\frac{k-1}{n}\right) + (nx - (k-1))g\left(\frac{k}{n}\right) - g(x)|\\ &= |(k - nx)[g\left(\frac{k-1}{n}\right) - g(x)] + (nx - (k-1))[g\left(\frac{k}{n}\right) - g(x)]|\\ &\leq (k - nx)|g\left(\frac{k-1}{n}\right) - g(x)| + (nx - (k-1))|g\left(\frac{k}{n}\right) - g(x)|\\ &< (k - nx)\epsilon' + (nx - (k-1))\epsilon' = \epsilon',\end{split}\]

where \(\epsilon'\) is the maximum of \(|g\left(\frac{k-1}{n}\right) - g(x)|\) and \(|g\left(\frac{k}{n}\right) - g(x)|\) as \(x\) is allowed to range over \(E_k\).

By choosing \(n\) large enough, we can ensure \(\epsilon' < \epsilon/\sqrt{3}\), so that

\[\|p - g\|_2 = \left(\int_0^3 |p(x) - g(x)|^2 \, dx\right)^{1/2} < (3 \epsilon'^2)^{1/2} = \sqrt{3}\epsilon' <\epsilon.\]

That is, \(\|p - g\|_2 < \epsilon\), as desired.

☐

Footnotes