Exercises in Real Analysis II

This post is a sequel to Exercises in Real Analysis.

4. Let T be the triangle \{(x, y) \in {\mathbb{R}}^2: 0 \leq |x| \leq y \leq 1\}, and \mu be the restriction of the planar Lebesgue measure on T. Suppose that f \in L^2(T, \mu). Prove that

\liminf_{y \rightarrow 0^+} \int_{-y}^y |f(x, y)| dx = 0.

Solution. Assume on the contrary that

L = \liminf_{y \rightarrow 0^+} \int_{-y}^y |f(x, y)| dx > 0.

Then for some \delta > 0, we have

\int_{-y}^y |f(x, y)| dx > \frac{L}{2}      (1)

for all 0 < y < \delta.

To link L^1 to L^2, we use the Cauchy-Schwarz inequality:

\int_{-y}^y |f(x, y)| dx \leq \left( \int_{-y}^y |f(x, y)|^2 dx \right)^{\frac{1}{2}} (2y)^{\frac{1}{2}}.

Hence from (1), we get

\int_{-y}^y |f(x, y)|^2 dx \geq \frac{L^2}{8} \frac{1}{y},    0 < y < \delta.

Integrating over y (Fubini’s theorem), we get

\|f\|^2_{L^2(T, \mu)} \geq \int_0^{\delta} \int_{-y}^y |f(x, y)|^2 dx dy \geq \int_0^{\delta} \frac{L^2}{8} \frac{1}{y} dy = \infty.

Contradiction. \Box

**********

5. Let X be a compact metric space, and \mu be a finite positive Borel measure on X. Suppose that \mu(\{x\}) = 0 for every x \in X. Prove that for every \epsilon > 0 there is a \delta > 0 such that, if E is any Borel subset of X having diameter less than \delta, then \mu(E) < \epsilon.

Solution. Since \mu(X) < \infty, by continuity of measure we have

\lim_{r \downarrow 0} \mu(B(x, r)) = 0

for any x \in X. Thus, given \epsilon > 0, there exists r_x > 0 such that

\mu(B(x, r_x)) < \epsilon.

Clearly \{B(x, r_x)\}_{x \in x} is an open cover of X, and by compactness there is a finite subcover, say

X \subset B(x_1, r_{x_1}) \cup \cdots \cup B(x_n, r_{x_n}).

Recall that the finite cover has some Lebesgue number \delta > 0, i.e., every subset E of X with diameter less than \delta is contained entirely in some B(x_i, r_{x_i}). Hence (assuming B is Borel)

\mu(E) \leq \mu(B(x_i, r_{x_i})) < \epsilon.   \Box

**********

6. Let X and Y be compact metric spaces, and let C(X) and C(Y) denote the Banach spaces of continuous, real-valued functions on X and Y respectively (with sup norm). Suppose that \phi: X \rightarrow Y is a continuous surjective map. Let

D = \{f \in C(X): f(x) = f(x') \text{ whenever } \phi(x) = \phi(x')\}.

(a) Show that D is a closed subspace of C(X), and that

D = \{g \circ \phi: g \in C(Y)\}.

(b) Let \nu be a finite positive Borel measure on Y. Prove that there is a finite positive Borel measure \mu on X such that \mu(\phi^{-1}(F)) = \nu(F) for all Borel subsets F of Y.

Solution. (a) It is easy to show that D is a closed subspace.

It is also clear that \{g \circ \phi: g \in C(Y)\} \subset D. To show the reverse inclusion, consider any f \in D. Since \phi is surjective, for each y \in Y there is some x \in X such that \phi(x) = y. We define a function g on Y by setting g(y) = f(x). The function g is well defined, for if \phi(x) = \phi(x') = y, then f(x) = f(x') since f \in D.

It remains to show that g is continuous. Fix y \in Y and suppose y_n \in Y \rightarrow y. Fix x_n \in X such that \phi(x_n) = y_n. Then

g(y_n) = f(x_n).

Since X is compact, some subsequence x_{n_k} converges to some x \in X. By continuity of \phi,

\phi (x) = \lim_n \phi (x_n) = \lim_n y_n = y.

Finally, by continuity of f, we have

g(y) = f(x) = \lim_n f(x_n) = \lim_n g(y_n).

Hence f \in \{g \circ \phi: g \in C(Y)\}.

(b) To construct a measure on a compact metric space, a good idea is to use the Riesz representation theorem.

Define a positive linear functional \Phi on the subspace D by setting

\Phi f = \int_Y g(y) d\nu(y)    if f = g \circ \phi \in D.

(You should verify that this definition makes sense.) The linear functional \Phi is bounded on D, since

|\Phi f| \leq \|g\|_{sup} \nu(X) = \|f\|_{sup} \nu(X) .

(You should verify the last inequality.) To use the Riesz representation theorem, we need a linear functional defined on the whole C(X). To get such a functional, we simply use the Hahn-Banach theorem to extend \Phi to a bounded linear functional on C(X). (Note that the question only requires us to construct a measure. Uniqueness is not guarenteed.)

Now, the Riesz representation theorem provides us with a positive Borel measure \mu on X such that

\Phi f = \int_X f(x) d\mu(x),    f \in C(X).

And when f = g \circ \phi \in D, we have

\Phi f = \int_X g \circ \phi (x) d\mu(x) = \int_X g(y) d\nu(y).   (2)

Finally, we want to show that (2) implies that

\mu(\phi^{-1}(F)) = \nu(F)   (3)

for all Borel subsets F of Y.

It suffices to prove (3) where F is open. Fix such a set F. Then, there exists a sequence g_n(x) \in C(X) such that g_n \uparrow \chi_F

\int_X g_n \circ \phi (x) d\mu(x) = \int_X g_n(y) d\nu(y).

By the monotone convergence theorem, we get

\mu(\phi^{-1}(F)) = \int_X \chi_F \circ \phi(x) d\mu(x) = \int_X \chi_F(y) d\nu(y) = \nu(F). \Box

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2 Responses to Exercises in Real Analysis II

  1. KKK says:

    Here is another solution for Real analysis prelim provided by Wongting.

    Question:
    (a) For a nonnegative {L^1} function {f} on {[0, \infty)}, show that there is a nonnegative strictly increasing function {a(x)} with {\displaystyle\lim_{x\rightarrow \infty}a(x)=\infty} and {af\in L^1}.
    (b) For a sequence {f_n} of nonnegative {L^1} functions on {[0, \infty)} with {|f_n|_{L^1}\rightarrow 0}, find a nonnegative strictly increasing function {b(x)} with {\displaystyle\lim_{x\rightarrow \infty}b(x)=\infty} and {|bf_n|_{L^1}\rightarrow0}.

    Solution:
    (a) Let {I=\int_0^\infty f}. We can choose {0=x_0<x_1<\cdots} with {x_i\rightarrow\infty} such that

    \displaystyle \int_{x_i}^\infty f \leq \frac{I}{2^i}.

    We then define {a(x_i)=i} and is linear between {x_i}‘s. Clearly we only have to check {af\in L^1}.

    To see this, let t_0=0 and {t_i= \inf\{t\geq 0: \int_0^t f=\frac{I}{2^1}+\cdots +\frac{I}{2^i}\}}, i\geq 1. Observe that {t_i\leq x_i}, and so {a(t_i)\leq i}. We then have

    \displaystyle \int_0^\infty af= \sum_{i=1}^\infty \int_{t_{i-1}}^{t_{i}}af\leq \sum _{i=1}^\infty \int_{t_{i-1}}^{t_{i}}i f= \sum _{i=1}^\infty \frac{iI}{2^i}=CI

    where {\displaystyle C=\sum _{i=1}^\infty \frac{i}{2^i}<\infty}. Thus {af\in L^1}.
    (b) We modify the above proof. Let {I_n:=\int _0^\infty f_n}. There is {x_1} such that

    \displaystyle \int_{x_1}^\infty f_1 \leq\frac{I_1}{2}.

    Inductively, we can find {0=x_0<x_1<\cdots} with {x_i\rightarrow \infty } such that for all {n\leq i},

    \displaystyle  \int_{x_i}^\infty f_n \leq\frac{I_n}{2^i}.

    We then choose {b(x)} exactly as before, i.e. {b(x_i)=i} and linear in between. The argument in (a) then shows that {\int bf_n\leq CI_n\rightarrow 0}, noting that {C} is independent of {n}.

  2. Leonard Wong says:

    Nice proof!

    In the last sentence, it should be \int bf_n \leq C I_n. [Yes, thanks! -KKK]

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