Good handwriting. Complete the assignments on time. Put as much effort as you can

Greg Kuperberg

MAT 127C: Real Analysis

Homework 6

This problem set is due Friday, May 8, by 9pm Pacific time in Gradescope.

Brief reminders: Please write each solution on separate pages. Please list all main study partners

on the first page of your homework. You should have justification in complete sentences for all

questions, even those that are mainly calculation.

Note: Homework problems in this class are not to be posted online until after the end of the quarter.

6.1. In the proof of the inverse function theorem, I had a sequence (~xk ) in Rn where I also gave

names to the differences between consecutive terms:

~xk+1 =~xk + ∆k~x

I also assumed that

||∆k+1~x|| < c||∆k~x||
for some constant c < 1. Prove (as I needed) that (~xk ) converges under these conditions.
(Hint: Show that the sequence is Cauchy.)
6.2. Error estimates related to Newton’s method.
(a) Prove that if f : Rn → R is C2 , then
f (~x + ∆~x) = f (~x) + D f (~x) · ∆~x + O(||∆~x||2 ).
(b) Give an example of a C1 function f : Rn → R and a point ~x ∈ Rn such that
f (~x + ∆~x) 6= f (~x) + D f (~x) · ∆~x + O(||∆~x||α )
for any α > 1. (Hint: You are allowed to choose both f and n, so why not try n = 1.

Also see problem 4.1 for possible ideas.)

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6.3. Let U 2 ⊆ R2 be the upper half plane of points (x, y) with y > 0, and also consider the coordinates

p

def

(x, r) = ~g(x, y) = (x, x2 + y2 )

on the same region U 2 .

(a) Draw a picture of the coordinate system (x, r), analogous to (but of course not the same

as) the usual diagram of polar coordinates (r, θ ).

(b) Let:

z = f (x, r) = f (~g(x, y))

For some C1 function f . Show that the partial derivative ∂ z/∂ x defined when the

coordinates are x and r does not have to equal ∂ z/∂ x defined when the coordinates are

x and y. How is this possible? (Hint: I warned you in an earlier lecture that this can

happen.)

6.4. Define a function f : R2≥0 → R by

(

(sin πx)(sin πy) n ≤ x ≤ n + 2 and n ≤ y ≤ n + 1 for some n ∈ Z≥0

f (x, y) =

.

0

otherwise

(a) Prove that f is continuous and bounded. (Hint: To argue that f is continuous, it may help

to draw a heat map of f as I have sometimes done in lecture. See also the continuity

theorem stated at the end of the April 10 lecture.)

(b) Evaluate the two integrals

Z ∞Z ∞

Z ∞Z ∞

f (x, y) dx dy

0

0

f (x, y) dy dx

0

0

and observe that they are not equal.

(c) Explain why this example does not contradict Fubini’s theorem. Explain why it still

wouldn’t contradict Fubini’s theorem if you made the integrals from 0 to π/2 using the

substitutions x = tan u and y = tan v. (Specifically, do the substitutions following the

rules of calculus so that the values of the integrals don’t change.)

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Craig A. Tracy

MAT 135B, Spring 2020

HW#5, Due Friday, May 8

1. Consider simple random walk in continuous time on the two-dimensional square lattice Z2 ; that is, the walker

jumps at random times (exponential distribution with parameter λ) to one of the four neighbors with equal

probability. Formulate and solve the Kolmogorov forward equation

dP

=P ·Q

dt

2. Recall from the lectures that ASEP on the integer lattice Z is a continuous time Markov process following the

rules: (1) A particle at x waits an exponential time with parameter 1, and then chooses y with probability

p(x, y). (2) If y is vacant at that time it moves to y, while if y is occupied it remains at x. The adjective

“simple” refers to the fact that the allowed jumps are only one step to the right, p(x, x + 1) = p, or one step

to the left, p(x, x − 1) = q = 1 − p.

For 2-particle ASEP the states are labelled (x1 , x2 ) ∈ Z2 with x1 < x2 and we showed that if P(y1 ,y2 ) (x1 , x2 ; t)
denotes the transition probability (y1 , y2 ) → (x1 , x2 ) at time t, then P satisfies the forward equations (write
Y = (y1 , y2 ))
p PY (x1 − 1, x2 ; t) + q PY (x1 + 1, x2 ; t) + p PY (x1 , x2 − 1; t)+
d
PY (x1 , x2 ; t) =
q PY (x1 , x2 + 1; t) − 2PY (x1 , x2 ; t),
if x2 > x1 + 1,

dt

p PY (x1 − 1, x2 ; t) + q PY (x1 , x2 + 1; t) − PY (x1 , x2 ; t),

if x2 = x1 + 1

subject to the initial condition PY (x1 , x2 ; 0) = δx1 ,y1 δx2 ,y2 .

Consider now 3-particle ASEP on the integer lattice Z. States are labeled by (x1 , x2 , x3 ) ∈ Z3 with x1 < x2 <
x3 . Let PY (x1 , x2 , x3 ; t) denote the transition probability Y = (y1 , y2 , y3 ) → (x1 , x2 , x3 ) at time t subject to
the initial condition PY (x1 , x2 , x3 ; 0) = δx1 ,y1 δx2 ,y2 δx3 ,y3 .
Problem: Explicitly write the forward equations for PY (x1 , x2 , x3 ; t). (You are not asked to solve these
equations.) Hint: There are three generic cases: (1) None of the particles have adjacent neighboring particles,
(2) two particles are neighbors but the third is not a neighbor of the other two, and (3) all three particles have
neighbors; that is the particles are located at (x, x + 1, x + 2) for x ∈ Z.
Remarks: Clearly there is a formulation for N -particle ASEP and one can seek the transition function PY (X; t)
where states are labelled by (x1 , x2 , . . . , xN ) ∈ ZN with x1 < x2 < · · · < xN . See
https://arxiv.org/abs/0704.2633
Start with the initial condition YN = (1, 2, . . . , N ) and let Xm (t) denote the position of the mth particle from
the left. Let q > p, then an important quantity is

PZ (Xm (t) = x) = lim PYN (Xm (t) = x).

N →∞

See

https://arxiv.org/abs/0807.1713

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3. Let {Xn } be a sequence of independent and identically distributed random variables with E(Xj ) = µ. Let

Sn := X1 + X2 + · · · + Xn and set

Yn := Sn − nµ.

Show that {Yn } is a martingale. Does Yn satisfy the hypothesis of the Martingale Convergence Theorem? Why

or why not.

4. An urn initially contains r red balls and g green balls. At each time n we draw a ball out, then replace it, and

add c more balls of the color drawn. Let Xn be the fraction of green balls after the nth draw. Show that {Xn }

is a martingale.

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