2018 Spring Comprehensive in Analysis

Proof.

Suppose, toward a contradiction, that no such exists. Then for every , at least one of

is countable.

For , define

and

By the assumption, . Also , because if , then

which would make countable.

Both and are nonempty. Indeed, if , then , so for every rational , the set is countable. Since every point of lies in for some rational , we get

a countable union of countable sets, contradiction. Similarly, .

The set is downward closed in , and is upward closed in . Moreover, every element of is smaller than every element of ; otherwise, if with and , then by downward closure of , contradicting .

Let

Choose rational sequences and such that

Then

so is countable. Similarly,

so is countable. Therefore

is countable, a contradiction.

Hence there exists such that both and are uncountable.

Proof.

(a) Since , continuity at means

Along the sequence , this requires

Thus is necessary.

Conversely, assume . Let . Choose such that for all . If , then either is not of the form , in which case , or with , in which case . Therefore is continuous at .

So is continuous at if and only if

(b) Since , differentiability at means that

exists. For not of the form , the quotient is . For , the quotient is

Therefore differentiability at requires

Conversely, if , then for , the quotient is either or equals for some large . Hence

Thus is differentiable at , and .

So is differentiable at if and only if

(c) Since is assumed bounded, is Riemann integrable on for every such bounded sequence . Thus there is no additional condition beyond boundedness.

We prove this directly. Let

If , then and there is nothing to prove. Assume .

Let . Choose so large that

All points with lie in . The contribution of this interval to any upper oscillation estimate is at most

The remaining exceptional points

are finitely many. Around them choose finitely many open intervals whose total length is less than

Refine a partition so that these intervals and are unions of subintervals of the partition. On the complement of these intervals, , so the oscillation is . On the chosen intervals, the oscillation is at most . Hence for this partition ,

Therefore is Riemann integrable.

Moreover, the same construction gives upper sums arbitrarily close to and lower sums arbitrarily close to , so the integral is

Proof.

Let

be a partition of . The polygonal length of the graph along this partition is

Using

we get

The first sum is . It remains to bound the second sum.

For a convex function, the secant slopes over consecutive intervals are nondecreasing. Indeed, if , convexity implies

Hence the increments can change sign at most once: first they may be nonpositive, and afterwards they may be nonnegative.

Therefore, for this partition, the total variation along the partition is at most the amount by which decreases plus the amount by which increases. Since , this gives

Thus every polygonal approximation to the graph has length at most

Taking the supremum over all partitions, the arclength of the graph of is at most .

Proof.

(a) Since , the derivative is continuous on the compact interval . Hence is bounded. Let

If , then is constant and is absolutely continuous. Assume .

Let and choose

If , , are disjoint subintervals with

then by the mean value theorem,

Therefore

Thus is absolutely continuous.

(b) Assume is absolutely continuous. Apply the definition with . Then there exists such that for every finite disjoint family of subintervals with total length less than , the corresponding sum of absolute changes of is less than .

Choose an integer so large that

Divide into subintervals

with

for each .

Now let

be any partition of . Refine by adding the points . Refining a partition can only increase the sum

so it is enough to bound the variation on the refined partition.

Inside each interval , the subintervals of the refined partition are disjoint and have total length . Hence the sum of the corresponding absolute changes of is less than . Summing over , we get

Thus is of bounded variation, with variation at most .

Proof.

Let be an open cover of in . Then

For each , the preimage is open in because is continuous. Also,

Thus is an open cover of .

Since is compact, there exist finitely many indices such that

Applying , we get

Therefore every open cover of has a finite subcover. Hence is compact.

Proof.

(a) We prove the stronger estimate

for all .

Fix . For every , the triangle inequality gives

Taking the infimum over gives

Thus

Interchanging and gives

Hence

Therefore is Lipschitz with constant , and hence uniformly continuous on .

(b) First suppose . Then every ball centered at meets . Hence for every , there exists such that

Therefore

Since this holds for every , we get .

Conversely, suppose . Then for every , by the definition of infimum, there exists such that

Thus every ball centered at meets , so .

Therefore

Proof.

Let

Since is compact and is continuous on , the function is uniformly continuous on .

Let . If , then , so is continuous. Assume . By uniform continuity, there exists such that whenever

and

we have