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Math 55b Lecture Notes
Evan Chen Spring 2015
This is Harvard College’s famous Math 55b, instructed by Dennis Gaitsgory. The formal name for this class is “Honors Real and Complex Analysis” but it generally goes by simply “Math 55b”. It is an accelerated one-semester class covering the basics of analysis, primarily real but also some complex analysis. The permanent URL is http://web.evanchen.cc/~evanchen/coursework. html, along with all my other course notes. Special thanks to W Mackey for providing notes on the several day that I slept through class.
Contents
1 January 29, 2015 5 1.1 Definition and Examples............................ 5 1.2 Continuous Maps................................ 6 1.3 Forgetting the Metric – Topological Spaces................. 7 1.4 Intermediate Value Theorem......................... 8
2 February 3, 2015 10 2.1 Sequential Continuity............................. 10 2.2 Cauchy Completeness............................. 10 2.3 R is Complete................................. 11 2.4 1-countable and Hausdorff........................... 11
3 February 5, 2015 12 3.1 Completion................................... 12 3.2 Why no compactness.............................. 13 3.3 Sequential Compactness............................ 13 3.4 Compactness.................................. 14
4 February 12, 2015 16 4.1 Warm-Up.................................... 16 4.2 “Applied” Math................................ 16 4.3 Series...................................... 17 4.4 Convergent Things............................... 18 4.5 Convergence Tests............................... 18 4.6 The Series n−a^................................. 19 4.7 Exponential Function............................. 20
5 February 17, 2015 22 5.1 L^1 norm..................................... 22
Evan Chen (Spring 2015) Contents
Evan Chen (Spring 2015) Contents
22.3 Proof of Hurwitz’s Theorem.......................... 98 22.4 A Theorem We Tried to Use But Couldn’t................. 99
Evan Chen (Spring 2015) 1 January 29, 2015
§1 January 29, 2015
Grading system for 55B with ' 55A. We’ll be more or less following Baby Rudin.
§1.1 Definition and Examples
This subsection is copied from my Napkin project.
Definition 1.1. A metric space is a pair (M, d) consisting of a set of points M and a metric d : M × M → R≥ 0. The distance function must obey the following axioms.
- For any x, y ∈ M , we have d(x, y) = d(y, x); i.e. d is symmetric.
- The function d must be positive definite which means that d(x, y) ≥ 0 with equality if and only if x = y.
- The function d should satisfy the triangle inequality: for all x, y, z ∈ M ,
d(x, z) + d(z, y) ≥ d(x, y).
Abuse of Notation 1.2. Just like with groups, we will abbreviate (M, d) as just M.
Example 1.3 (Metric Spaces of R) (a) The real line R is a metric space under the metric d(x, y) = |x − y|.
(b) The interval [0, 1] is also a metric space with the same distance function.
(c) In fact, any subset S of R can be made into a metric space in this way.
Example 1.4 (Metric Spaces of R^2 ) (a) We can make R^2 into a metric space by imposing the Euclidean distance function d ((x 1 , y 1 ), (x 2 , y 2 )) =
(x 1 − x 2 )^2 + (y 1 − y 2 )^2.
(b) Just like with the first example, any subset of R^2 also becomes a metric space after we inherit it. The unit disk, unit circle, and the unit square [0, 1]^2 are special cases.
Example 1.5 (Taxicab on R^2 ) It is also possible to place the following taxicab distance on R^2 :
d ((x 1 , y 1 ), (x 2 , y 2 )) = |x 1 − x 2 | + |y 1 − y 2 |.
For now, we will use the more natural Euclidean metric. (One can also use max instead of a sum.)
Evan Chen (Spring 2015) 1 January 29, 2015
In calculus you were also told (or have at least heard) of what it means for a function to be continuous. Probably something like A function f : R → R is continuous at a point p ∈ R if for every ε > 0 there exists a δ > 0 such that |x − p| < δ =⇒ |f x − f p| < ε. Question 1.11. Can you guess what the corresponding definition for metric spaces is? All we have do is replace the absolute values with the more general distance functions: this gives us a definition of continuity for any function M → N. Definition 1.12. Let M = (M, dM ) and N = (N, dN ) be metric spaces. A function f : M → N is continuous at a point p ∈ M if for every ε > 0 there exists a δ > 0 such that dM (x, p) < δ =⇒ dN (f x, f p) < ε. Moreover, the entire function f is continuous if it is continuous at every point p ∈ M. Notice that, just like in our definition of an isomorphism of a group, we use both the metric of M for one condition and the metric of N on the other condition.
Example 1. Let M be any metric space and D a discrete space. When is a map f : D → M continuous? Any map D → M is continuous.
Proof. Take an open ball of radius 12.
Example 1. The map R → R by x 7 → x^2 is continuous. So is x 7 → x^3.
Proof. Homework.
Example 1. Let X = Rn^ with one product metric and let Y = Rn^ with another product metric. Then id : X → Y is continuous. Thus we will generally not be pedantic about the choice of metric.
§1.3 Forgetting the Metric – Topological Spaces
Again excerpted from the Napkin project.
Definition 1.16. A topological space is a pair (X, T ), where X is a set of points, and T is the topology, which consists of several subsets of X, called the open sets of X. The topology must obey the following four axioms.
- ∅ and X are both in T.
- Finite intersections of open sets are also in T.
- Arbitrary unions (possibly infinite) of open sets are also in T. Abuse of Notation 1.17. We refer to the space (X, T ) by just X. (Do you see a pattern here?)
Evan Chen (Spring 2015) 1 January 29, 2015
Example 1. We can declare all sets are open: a discrete space is a topological space in which every set is open.
Example 1. Declare U open if and only if ∀x ∈ U , the ball of radius r centered at x is contained in U.
Proof. Check all the axioms. Blah.
Proposition 1. Let M be a metric space. Then for all x ∈ M and r > 0, the ball
B(x, r) = {y | dM (x, y) < r}
is open.
Proof. Pick y in the ball. Let t = d(x, y). Then t < r, so pick ε with t + ε < r. You can check using the triangle inequality that B(y, ε) ⊆ B(x, r).
Example 1. [0, 1] is not open since no ball at 0 is contained inside it.
Definition 1.22. A subset Y is closed iff X − Y is open.
Definition 1.23. A function f : X → Y of topological spaces is continuous at p ∈ X if the pre-image of any neighborhood of f p is also a neighborhood of p.
With some effort, we can show this is the same definition of continuity as with metric spaces.
§1.4 Intermediate Value Theorem
Theorem 1.24 (IVT) Let f : [0, 1] → R be continuous such that f (0) < 0 and f (1) > 0. Then ∃a ∈ [0, 1] such that f (a) = 0.
This theorem is not cheap, and requires the following theorem.
Theorem 1. Let A ⊆ R be bounded above. Then there exists a least upper bound y ∈ R.
Proof of IVT. Long and boring. Just draw a picture. The main point is to take y to be the least upper bound of the a for which f (a) ≤ 0.
Evan Chen (Spring 2015) 2 February 3, 2015
§2 February 3, 2015
Didn’t attend class.
§2.1 Sequential Continuity
Continuous functions send convergent sequences to convergent sequences, limits sent to limits. The converse is true in metric spaces.
§2.2 Cauchy Completeness
So far we can only talk about sequences converging if they have a limit. But consider the sequence x 1 = 1, x 2 = 1.4, x 3 = 1.41, x 4 = 1.414,.... It converges to
2 in R, of course. But it fails to converge in Q. And so somehow, if we didn’t know about the existence of R, we would have no idea that the sequence (xn) is “approaching” something. That seems to be a shame. Let’s set up a new definition to describe these sequences whose terms eventually get close to each other, but don’t necessarily converge to a point.
Definition 2.1. Let x 1 , x 2 ,... be a sequence which lives in a metric space M = (M, dM ). We say it is Cauchy if for any ε > 0, we have
dM (xm, xn) < ε
for all sufficiently large m and n.
Note that, unlike the rest of this chapter, this is a notion which applies only to metric spaces. In a general topological space there is not a good enough notion of “distance” to make suremake sure make this definition work.
Question 2.2. Show that a sequence which converges is automatically Cauchy. (Draw a picture.)
Now we can define the following.
Definition 2.3. A metric space M is complete if every Cauchy sequence converges.
Most metric spaces aren’t complete, like Q. But it turns out that every metric space can be completed by “filling in the gaps” somehow, resulting in a space called the completion of the metric space. The construction is left as an (in my opinion) fun problem. It’s a theorem that R is complete. To prove this I’d have to define R rigorously, which I won’t do here (yet). In fact, there are some competing definitions of R. It is sometimes defined as the completion of the space Q. Other times it is defined using something called Dedekind cuts. For now, let’s just accept that R behaves as we expect and is complete.
Example 2.4 (Examples of Complete Sets) (a) R is complete.
(b) The discrete space is complete, as the only Cauchy sequences are eventually constant.
(c) The closed interval [0, 1] is complete.
(d) Rn^ is in fact complete as well. (You are welcome to prove this by induction on n.)
Evan Chen (Spring 2015) 2 February 3, 2015
Example 2.5 (Non-Examples of Complete Sets) (a) The rationals Q are not complete.
(b) The open interval (0, 1) is not complete, as the sequence xn = (^) n^1 is Cauchy but does not converge.
§2.3 R is Complete
Theorem 2.6 (Bolzano-Weirestraß) Any sequence in [0, 1] has a convergent subsequence.
§2.4 1 -countable and Hausdorff
Most of the “nice” properties of metric spaces carry over to 1-countable Hausdorff general topological subspaces.
Evan Chen (Spring 2015) 3 February 5, 2015
§3.2 Why no compactness
Lemma 3. A function f : [0, 1] → R is bounded above.
Proof. Otherwise we get a sequence of points x 1 , x 2 ,... such that f (xm) > m for all m. Then we can find a convergent subsequence using Bolzano-Weirestraß. This breaks sequential continuity.
Theorem 3. Any function f : [0, 1] → R has a global maximum.
Proof. By the lemma, f “[0, 1] is bounded above and we can take the least upper bound y. We claim this is actually in the limit. If not, then we can construct a sequence x 1 , x 2 ,... such that f (xm) ≥ y − (^) m^1. Take a convergent subsequence by Bolzano-Weirestraß. Then (xm) converges to some x ∈ [0, 1], but then we must have f (x) = y.
This sucks. Compactness is better. Rawr.
Second Proof, from Jane. As before, there is a least upper bound y. Assume for contra- diction that the bound is never obtained. Then the function [0, 1] → R by
x 7 →
y − f (x)
is an unbounded continuous function on [0, 1], impossible.
§3.3 Sequential Compactness
Definition 3.6. A topological space is sequentially compact if every sequence has a convergent subsequence.
Proposition 3. Let f : X → Y. If X is sequentially compact then f (X) is sequentially compact.
Proof. Trivial. If (f xn) is a sequence, take a converge subsequence of (xn) ∈ X.
Theorem 3.8 (Heine-Borel) A ⊆ R is sequentially compact if and only if it closed and bounded.
We can now kill the original maximum theorem. Given f : [0, 1] → R, its image in R is bounded and closed and we can easily use this to show that f has a maximum. To prove Heine-Borel, we first prove one direction in greater generality.
Evan Chen (Spring 2015) 3 February 5, 2015
Lemma 3. Let Y ′^ ⊆ Y , where Y ′^ is sequentially compact and Y is 1-countable and Hausdorff. Then Y ′^ is closed in Y.
Proof. Because Y is 1-countable, we can use the sequence definition of closed in the tricky direction: it suffices to prove that if y 1 , y 2 , · · · → y with yi ∈ Y ′^ then y ∈ Y ′. But yi has a convergent subsequence to some y′. It had also better converge to y. By Hausdorff, y = y′.
Lemma 3. Suppose Y ′^ ⊆ Y for some topological spaces, with Y a Hausdorff space. If Y ′^ is closed and Y is sequentially compact then Y ′^ is sequentially compact.
Proof. Triviality: given yn in Y ′^ use sequential compactness of Y to force some subse- quence to converge to y ∈ Y. Since Y ′^ is closed, y ∈ Y ′.
Proof of Heine-Borel. First, suppose A is bounded and closed; by bounded-ness it lives in some A ⊆ [a, b]. By Bolzano-Weirestraß, the set [a, b] is sequentially compact. By our lemma, A ⊆ [a, b] is sequentially compact. For the converse, suppose A is sequentially compact. We did a lemma that show A is closed. Hence A is bounded.
§3.4 Compactness
Apparently Gaitsgory does not like sequential compactness QQ.
Definition 3.11. A topological space is compact if every open cover has a finite subcover.
Proposition 3. If f : X → Y is continuous and Y is compact then f (X) is compact.
Proof. Tautological using the definition of continuity.
We can mirror this for sequential compactness now.
Lemma 3. Let Y be a Hausdorff space and Y ′^ a compact subspace. Then Y ′^ is closed.
Proof. We will show that for any y ∈ Y \ Y ′, there is a neighborhood U ⊆ Y \ Y ′ containing y. For every y′^ ∈ Y ′, we can use the Hausdorff condition to find neighborhoods Uy′^3 y′^ and Vy′^3 y. By definition, Y ′^ is covered by Uy′^ , and we can find a finite subcover
Y ′^ ⊆
⋃^ n
i=
Uyi
Now take (^) n ⋂
i=
Vyi 3 y.
Evan Chen (Spring 2015) 4 February 12, 2015
§4 February 12, 2015
“Today we will be doing applied math” – Gaitsgory “This can’t be happening” “Applied by my standards”.
§4.1 Warm-Up
Lemma 4. Let ai, bi be sequences converging to a and b. Then ai + bi → a + b, aibi → ab.
Proof. The maps +, × : R^2 → R are continuous.
§4.2 “Applied” Math
Also known as: cooking up stupid bounds. Also known as: suppose an ≤ xn ≤ bn for all n, where an and bn are the locations of two cops and xn is the location of a drunkard. Now a surprisingly nontrivial statement.
Example 4. Let |a| < 1. Then the sequence (an)n converges to 0.
Proof. WLOG a > 0 (the other case is easy). We bound the sequence an^ between two guys. Put the estimate (^) ( 1 a
)n ≥ n
a
From this we deduce 0 ≤ an^ ≤
1 a −^1
n
Example 4. For any real number a > 0, the sequence ( n
a)n converges to 1.
The existence of n
a follows from the Intermediate Value Theorem.
Proof. It suffices to consider a > 1, since otherwise we can consider 1/a. Let xn be such that xnn = a for each n. Then 1 ≤ a <
a n
)n
so 1 ≤ x ≤ 1 + (^) na.
Example 4. The sequence ( n
n)n converges to 1.
Proof. 1 ≤ n ≤ (1 + (^1) n )n^ by Bernoulli’s inequality.
Evan Chen (Spring 2015) 4 February 12, 2015
§4.3 Series
We’ll put curly brackets around series for emphasis.
Definition 4.5. A series {an} converges if the sequence bn = a 1 + · · · + an converges.
Here is a better notion.
Definition 4.6. The series {an} converges absolutely if bn = |a 1 |+· · ·+|an| converges.
Example 4. For |a| < 1 the series {an}n converges absolutely.
Proof. 1 + · · · + an^ = 1 −a n+ 1 −a →^
1 1 −a.
Lemma 4. If the series {an} converges then the sequence (an) converges to 0.
Remark 4.9. The converse is false; for example, take the harmonic series.
Proof. The partial sums Sn converge as sequences to some b. Then (Sn+1 − Sn)n is a convergent sequence to b − b = 0.
Lemma 4. The series {an} converges if and only if ∀ε > 0 there is a natural N such that ∣∣ ∣∣ ∣
∑^ n^2
i=n 1
ai
< ε
for all n 1 , n 2 > N.
Proof. The condition just says that the partial sums are a Cauchy sequence. Since R is complete, that’s equivalent to partial sums converging.
Lemma 4. A series converging absolutely also converges.
Proof. Use the triangle inequality in the previous lemma in the form
ε >
|ai|
ai
Evan Chen (Spring 2015) 4 February 12, 2015
For part (b), note that infinitely many terms are actually greater than 1 which is impossible.
No conclusion when the lim sup equals 1.
Proposition 4.17 (Ratio Test) Let {an} be a series and assume that
lim sup
∣∣^ an+ an
Then {an} converges absolutely.
Proof. Let r be the lim sup and let ε > 0 such that r + ε < 1. For sufficiently large n we have |an+1| < (r + ε) |an| and bound the sequence above.
Example 4. The series {an} given by an = x
n n! converges absolutely.
Proof. Direct application of ratio test.
§4.6 The Series n−a
In homework, we will define what xc^ means for real numbers x and c. In fact I can tell you:
xc^ def = limq→c q∈Q
xq.
Now let’s consider the sequence an = n−c, where c > 0 is a real number. The Ratio and Root tests both fail. Here is the answer.
Theorem 4.19 (Zeta Series) The sequence {an} given by an = n−c^ converges for c > 1 and diverges for c ≤ 1. In particular the harmonic series 1 1
is divergent.
The idea is the “powers of two” estimate, which we prove in full generality here.
Lemma 4. Let {an} be a series of positive real numbers which is monotically decreasing, and let
bn = 2n^ · a 2 n.
Then {bn} converges if and only if {an} converges.
Evan Chen (Spring 2015) 4 February 12, 2015
The lemma immediately implies the theorem on zeta series.
§4.7 Exponential Function
We define
exp(x) def =
∑^ ∞
n=
xn n!
We showed earlier that this always converges.
Lemma 4.21 (Physicist’s Lemma) Let {an} and {bn} be absolutely convergent series. Define
cn =
∑^ n
i=
aibn−i.
Then {cn} converges to the product of
an ·
bn.
Proof. We want to show that there for an ε > 0 there exists an N such that when n ≥ N
∑^2 n
i=
ci − AB < ε
where A and B are sums of {an} and {bn}. For sufficiently large N , ∣∣ ∣∣ ∣
( (^) n ∑
i=
an
) ( (^) n ∑
i=
bn
− AB
ε.
So we want to estimate the value of ∣∣ ∣ ∣∣
∑^2 n
i=
ci −
( (^) n ∑
i=
an
) ( (^) n ∑
i=
bn
Expanding, we find that the quantity is actually ∣∣ ∣∣ ∣∣ ∣∣
n<i≤ 2 n j≤n
aibj +
n<j≤ 2 n i≤n
aibj
1 ≤j≤n
bj
n<i≤ 2 n
ai
1 ≤i≤n
bi
n<j≤ 2 n
aj
So it suffices to prove that for large enough n we have ∣ ∣ ∣∣ ∣∣
1 ≤j≤n
bj
n<i≤ 2 n
ai
∣∣ <^
ε.
Using absolute convergency, (^) ∑
1 ≤j≤n
|bj |
n<i≤ 2 n
|ai|
The left term is at most B, while the right term can be made to be at most ε/^1000 B , as desired.
You can actually weaken the condition to just one series being absolutely convergent.
