MA2108S, Tutorial 10, let’s go. 16th April, 23:34. All the files are on my home server but it’s not like it’s public anyway, so maybe I’ll upload them to Mediafire some time. I lost focus somewhere around the last part of question 6, so the times are kinda meaningless.

### 1

Claim: is continuous on .

Let and . Then since and is continuous over that interval, is continuous at . Therefore is continuous on .

Claim: is not necessarily continuous on .

Consider over , where is clearly discontinuous at .

(23:38)

### 2

For any , the truncation of the (canonical) binary expansion of (i.e. ) is a rational sequence converging to also bounded by , and since is stated to be continuous,

(23:44)

### 3

Um.

Since the set of continuous functions is closed under addition, scalar multiplication, and taking the absolute value, the set is also closed under and .

(23:53)

### 4

Assumed above, but I guess this actually needs to be proven? (If we assume 3, we can use , but circular arguments are circular.)

Let be some element of ‘s domain and assume that it is a cluster point too. (If it’s not a cluster point, continuity is trivial.)

(00:00)

### 5(a)

Consider . If , then . If , then . Otherwise, , so .

is continuous over and discontinuous elsewhere.

### 5(b)

Continuous over the irrationals, discontinuous over the rationals. Gone through in class etc.

### 5(c)

Continuous at 0, discontinuous elsewhere. Given any nonzero , for any , consider the -neighbourhood of . There exists both a rational and an irrational in this neighbourhood; let these be respectively. (WLOG assume are all positive.)

Therefore is discontinuous at every nonzero value.

### 5(d)

This is clearly discontinuous at the reciprocals of the square roots of nonsquare integers, and continuous between the reciprocals of the square roots of integers. This leaves the cases and for some nonzero .

In the former case, is constant and hence continuous at .

In the latter case, suppose WLOG that . We have that while , so is not continuous at such .

To summarise, is discontinuous on and continuous elsewhere.

(00:45) (I got distracted, okay.)

### 6(a)

### 6(b)

### 6(c)

### 6(d)

Apparently, for this kind of questions you’re supposed to use cheating methods like Taylor series to get a guess of what the behaviour is, then rigorously prove it without Taylor series! Seems kinda backwards.

Hmm. Not quite the right approach.

### 6(e)

### 6(f)

Second derivative?

(01:57)

### 7

Let and consider behaviour around .

(02:25)

Actually, I might as well do tutorial 11 too.

### 1

Follows from intermediate value theorem and induction; in this case it would be easier to prove the more general statement for any convex combination, not just the mean.

### 2

Again, intermediate value theorem. For sufficiently large and sufficiently small , , and since , IVT gives us the existence of at least two distinct roots.

### 3

Bi-Lipschitz blah blah. is well defined (cf. tutorial 9 question 1), so all that remains is to show that is continuous.

is Lipschitz continuous and therefore also continuous. (A special case of a more general theorem on how a small Lipschitz perturbation of the identity function gives a bi-Lipschitz result.)

### 4

If is either nonincreasing or nondecreasing, then it is a convergent sequence, where we may let and .

Otherwise, there exists some for which .

But also

where, by IVT, there exists some value of between and such that .

### 5

This is a homework question.

### 6

Suppose for contradiction that is continuous. Then it attains a minimum value at two points , and likewise a maximum value for two other points . WLOG let and .

If , let ; if , let . Let . If , we have a contradiction. Suppose . By IVT, at least three of these statements are true (with the fourth possibly failing due to asserting the existence of an element in an empty set):

As the open intervals are all disjoint, this implies there are at least three distinct values of such that , a contradiction.

Therefore and . By the same argument, and . However, this means that ; in other words, is a constant function. This is also a contradiction.

Consequently, the initial assumption that is continuous is untenable.

### 7

“A continuous endomorphism has a fixed point.”

If or we are done. Suppose otherwise, and let .

and , so by IVT there exists some such that .

### 8

This is exactly what I said for question 1.

### 9

Definition quoting time!

Suppose that is uniformly continuous and that . For any , there exists some such that , and there exists some such that for all , . Putting two and two together, .

In the other direction, suppose that is not uniformly continuous. Then there exists some such that for all there exist some such that and . Then construct sequences and , where and .

### 10

Oh god it’s 4 am I can’t think any more.