In this post I type up solutions of some interesting questions in the real analysis qualifying examinations. Most questions will be from UW. This post will be updated continuously. Of course, you are encouraged to think about the question before looking at the solution. And it may happen that the solution is not correct. Some motivations / comments are also included in the solutions.

Actually many questions in such an exam are not really difficult. Usually they do not involve advanced theorems. But you need to think about the relevant tools and use them in some natural way.

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1. Let be a continuous function on . For and , define .

(a) Show that for all , and that is continuous.

(b) Show that if is a bounded sequence in , the sequence contains a uniformly convergent subseqeunce.

(c) Assume that is one-to-one. Show that does not map onto .

*Solution*. General thoughts: We are studying an *integral operator* on , and the kernel is continuous. Functional analysis!

(a) This shows that is a bounded linear operator from to equipped with the sup norm. The solution is standard. For any ,

.

Since we need the -norm of in the estimate, we use the *Cauchy-Schwarz* in equality. We get

.

Continuity is also easy to check. For ,

.

Since is continuous on , it is uniformly continuous. Thus, given , we may find such that

whenever . Again using Cauchy-Schwarz inequality, we have

. (*)

So is continous.

(b) We need to prove that is a *compact* linear operator. Thus we need a certain compactness result in . This is provided by the *Arzelà–Ascoli theorem*: has a uniformly convergent subsequence if is uniformly bounded and equicontinuous.

By (a), . Since is bounded in , we see that is uniformly bounded. To check equicontinuity, note that by (*),

whenever . This is enough since depends only on but not on the .

(c) We need to show that if is one-one, then it is not onto. In such an abstract setting, it is hopeless to construct some function outside the range of . We shall proceed by *contradiction*.

Hence, suppose that is one-one and onto. Hence it is bijective and is well defined. In such a situation, you should be able to recall the *Open Mapping Theorem*, which implies that is also bounded. Thus, there exists a constant such that

for all . (**)

Now we have used the assumption that is one-one and we also applied a big theorem. So we should be almost there!

It takes a little thought to derive a contradiction from (**). A hint is given by (b), so perhaps we need to construct a sequence of continuous functions which does not has a uniformly convergent subsequence. For example, consider defined by on , on , and is linear in between. As is well known, this sequence converges pointwise to a discontinuous function. (So no subseqeuence converges uniformly.)

Since is assumed to be onto, there exist functions such that . Now by (**), we have

for all . Hence is bounded in . And so by the compactness result in (b), we can extract a uniformly convergent subsequence from . This is clearly a contradiction.

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2. Let be the family of all non-empty compact subsets of . For , define

.

(a) Prove that is a metric space.

(b) Prove that is separable.

*Solution*. First, let us understand the definition of the proposed metric by drawing a picture (here I draw it for the plane):

The distance is the sum of the length of the two lines.

(a) The only difficulty is to prove the triangle inequality. Let and be non-empty compact set. Consider

.

We do not expect that there are special tricks involved. The inequality should follow from some application of the triangle inequality of . To make something in appear, let and insert them inside:

.

We use two independent elements in to ensure more flexibility when handling and . Simplifying the above expression, we get

.

Note that

.

So these two terms are alright. So, now we have

for all . It might appear initially things are not fine because we cannot interchange and . But don’t worry. By compactness and continuity, we may replace all by . Now we may formulate and prove the following result:

**Lemma.** Suppose that is a real continuous function of where and belong respectively to some compact sets. If for all , then .

*Proof*: Exercise.

Hence, we get

.

(b) Now we need to prove that the metric space is separable. That is, we need to find countable dense set. Note that we are working on the real line; there are not many reasonable choices for the countable set.

Let be the *dyadic rationals*. Given and , let

be the corresponding *dyadic interval*. Let be the set of all finite unions of intervals with endpoints in . Clearly, is countable and is a subset of . We claim that is dense in .

To show this, let be any compact set on . Define

.

That is, we consider covers of by small dyadic intervals. Clearly, for all . Moreover, for every , there exists such that . Now we consider

.

Since , the first term is . And by the second property of mentioned above, for all we have

.

Hence as .

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3. Let be a positive, continuously differentiable function on satisfying for all . Suppose that for some constant ,

.

Show that

.

(Hint: Use the fact that if , then .)

*Solution.* We need to show that , i.e., the derivative cannot be too large. However, it is impossible to obtain an estimate of in terms of (draw a few pictures to see!), and this is where analysis (rather than mere calculus) comes into play.

Suppose on the contrary that . To use the hint, let be arbitrary. Then we may find such that

.

The idea is that we may use this to obtain an estimate of the *distribution function* of . We use to denote the Lebesgue measure. By the *Markov inequality*, for any , we have

.

(This method is also called the *first moment estimate*.) Rearranging, we get

for . This shows that cannot be too small in a measure-theoretic sense. Consider

.

(Fundamental theorem of calculus.) For ,

.

Use calculus to maximize the last function. We get

for . Since is arbitrary, we conclude that the estimate cannot hold. This contradiction establishes that .

*Remark:* We also observe that the bound is sharp. If , then . KKK may want to think about how this exercise relates to isoperimetric inequalities.

I think there are some typos in the chain of inequalities after (Fundamental Theorem of Calculus) in Q3.

Corrected! Common typos when typing indefinite integrals. Thank you.

I think the third term in this chain of inequality is a typo. More precisely I think the correct one should be:

And also, a very minor mistake: after maximizing the function (in ) , we should have . This of course doesn’t affect your conclusion.

I have thought a more elementary proof of question 3, see if I am correct.

Fix any . For , by Cauchy-Schwarz inequality,

Since , f is strictly increasing and for , and so the above inequality rewrites as

Now since for , for sufficently large , and

Now take , we have

Now take , we have

The result then follows from the hint.

[Nice proof. Hope you don’t mind me to do some typesetting in your comment -KKK]Sorry for the typo, the last formula is

Q1 : the Theorem should hold true if .

This is called the Hilbert-Schmidt integral operator.

Q2: I didn’t think seriously, is it the Hausdorff metric?

Paullai: for Q2, this metric is equivalent to the Hausdorff distance:

by the elementary inequality for .

I haven’t read your solution of Q2(b) in details (and I don’t know dyadic intervals), but can we do it this way?

Let be the collection of all subsets of which consist of only finitely many rational numbers. Clearly is countable. Given a compact and , we then cover by finitely many balls of radius , whose centers are in . We choose in each ball a rational point and let be the collection of all such points, then it is easy to see that .