Math 140B - Winter 2015 - Midterm II
Problem 1.
Consider fn : [0, 1] R given by
nx
.
1 + nx3
What is the pointwise limit of the sequence fn as n ? Does the sequence fn converge uniformly?
fn (x) =
Solution:
Keeping x xed, while making n , we have
nx
x
1
x
June 13, 2007 Math 140B Solutions to Sample Final 1. (20 pts.) Give an example of a sequence cfw_fn of continuous, realvalued functions on [0, 1] such that limn fn (x) = 0 for all x [0, 1], but
1 n
lim
fn (x) dx = 0.
0
Explain your example briey. Solutio
1.
(i) We apply LHopital rule:
f (t) f (t)
f (t) f (0) f (t) f (0)
f (t) + f (t) 2f (0)
= lim
= lim
+
.
2
t0
t0
t0
t
2t
2t
2t
lim
(0)
Each of the fractions above give the limit f 2 by using the denition of the derivative
(0)
(0)
applied to the function f
1. We use the supremum norm to test uniform convergence.
(i) We show that fn
f . For xed n, we have
n
x
= sup 1
x + n x[0,a]
x+n
x[0,a]
|fn f |[0,a] = sup
=1
n
a
=
.
a+n
a+n
Then, since
a
=0
n
n a + n
we conclude that the fn s converge uniformly to f .
(
Math 140B
Test 1
100 points
April 25, 2014
Directions: Justify all answers. No calculators. If you appeal to a theorem,
show that the hypotheses of that theorem are satised. As usual, R and C
denote the reals and the complexes, respectively. Each problem
140B HOMEWORK 4 SOLUTIONS
Problem 1
(i)
0 if x [0, 1)
xn1
= 1
n
n 1 + x
2 if x = 1
(a) The limit function is discontinuous.
(b) The limit function is discontinuous, and therefore not dierentiable.
(c) The limit function is integrable because it has a nit
Math 140B
Test 2
100 points
May 23, 2014
Directions: Justify all answers. No calculators. If you appeal to a theorem,
show that the hypotheses of that theorem are satised. The notation [x]
denotes the greatest integer x. Each problem is worth 25 points.
2
Math 140B
Final
155 points
June 12, 2014
Directions: Justify all answers. If you appeal to a theorem, show that the
hypotheses of that theorem are satised.
b
a
(1) If F is Riemann integrable on [a, b], prove that
(25 pts)
F (x)dx = F (b)F (a).
SOLUTION: S
140B HOMEWORK 5 SOLUTIONS
Problem 1
Fix > 0 and let fn (x) = f (xn ). Since f and xn are continuous functions, fn (x) is continuous,
hence integrable.
Moreover, since xn 0 as n for all x [0, 1/2], we have that fn (x) f (0) pointwise for all
x [0, 1/2].
We
140B SOLUTIONS
Problem 1
(i) Problem from Rudin, solution can be found online.
(ii) Let f [a, b] R be dened by f (x) = 0 if x (a, b] and f (a) = 1. It has only a single
discontinuity, so Theorem 6.10 implies f is Riemann integrable. Let P = cfw_x0 , x1 ,
140B HOMEWORK 6 SOLUTIONS
Problem 1
See online solutions.
Problem 2
For a polynomial p in P we can write
2015
p(x) = ck xk .
k=0
For the interval [a, b], let = maxcfw_ a , b . Since the coecients ck are in [1, 1] we have ck 1.
By the triangle inequality
2