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