Download Solution to Homework 8

Denker
Spring 2010
312 Analysis - Assignment 8
Due Date: Monday, 03/22/2010 in class.
Problem 1: Prove that the following functions are continuous at x0 = 0:
√
a) f (x) = x
b) f (x) = x sin x1 for x 6= 0 and f (0) = 0.
Solution:
(a) Suppose f is not continuous at 0. Then there exists a sequence xn converging to
0 such that the sequence f (xn ) does not converge. Since for 0 ≤ x ≤ 1, |f (x)| ≤ 1, the
sequence is bounded and therefore we have at least two subsequential limits x 6= y of
the sequence f (xn ). But f (xn )2 = xn converges to 0, so x2 = y 2 = 0, hence x = y, a
contradiction.
(b) Since |f (x)| ≤ |x| it follows that for xn converging to 0,
0 ≤ |f (xn ) − f (0)| ≤ |xn |.
The continuity follows by squeezing.
Problem 2: Prove that the following functions are discontinuous at x0 = 0:
a) f (x) = 0 for x < 0 and f (x) = 2 for x ≥ 0.
b) f (x) = cos x1 .
Solution:
(a) Let xn = (−1)n n1 . Then lim xn = 0, but the sequence f (xn ) has two subsequential
limit points, 0 and 2.
1
. Then lim xn = 0, f (xn ) = 1 for n even and f (x) = −1 for n odd.
(b) Take xn = πn
Hence f (xn ) has again two different subsequential limit points and canot be continuous
at 0.
Problem 3: For each rational number x, write x as pq where p, q are integers with no
common factors and q > 0. Define f (x) = 1q . Also, define f (x) = 0 for x ∈ R \ Q. Show
that f is continuous at every irrational number x ∈ R \ Q.
Solution
Let x be an irrational number. Let ǫ > 0. Choose first q0 such that 1 < q0 ǫ. Then the
set of rational points p/q with q < q0 and p ∈ Z, where we assume that p and q have no
common divisors, has inter-point distances
p p′
1
| − ′| ≥ 2,
q q
q0
and x is not a point of this set. So x has some positive distance from that set:
Let a = sup{ pq :
p
q
≤ x; q < q0 ; p ∈ Z} and b = inf{ pq :
p
q
≥ x; q < q0 ; p ∈ Z}. Then
a<x<b
and we can take δ = min{b − x, x − a} > 0. Then for y ∈ R with |x − y| < δ, either
f (y) = 0 (if y is irrational) or f (y) = 1q ≤ q10 for some q ≥ q0 . It follows that
|f (x) − f (y)| ≤
1
< ǫ.
q0
Problem 4: Prove that two continuous functions f and g with domain [a, b] agree if they
agree on each rational number in [a, b]. Hint: Use the fact that each number is the limit
of rational numbers.
Solution
Let x ∈ [a, b] and chhose a sequence of rational numbers xn converging to x. Since f
and g are continuous at x and since f (xn ) = g(xn ) we obtain
f (x) = lim f (xn ) = lim g(xn ) = g(x).
Since x was any point in the interval the claim is proved.
Problem 5: Let f be continuous on [0, 2] and f (0) = f (2). Prove that there are x, y ∈
[0, 2] such that |x − y| = 1 and f (x) = f (y).
Solution
Define the function h : [0, 1] → R by h(x) = f (x) − f (x + 1). Then h is continuous on
[0, 1] and h(0) = f (0) − f (1) and h(1) = f (1) − f (2) = −h(0).
If h(0) = 0, then f (0) = f (1) and we are done. If h(0) 6= 0 then h(0) and h(1) have
values < 0 and > 0 (not both are positive or negative), hence by the intermediate value
theorem there is some x with h(x) = 0, equivalently f (x) = f (x + 1).
Problem 6: Prove that each polynomial function
f (x) = a0 + a1 x + a2 x2 + ... + an xn
is continuous. Moreover, if n is odd, show that there exists x0 with f (x0 ) = 0.
Solution
We prove the first statement by induction over n ∈ N for the statements
Pn : A polynomial of degree n − 1 is continuous.
For n = 1 the polynomial is constant so it is continuous.
Induction step: Assume that any polynomial of the form a0 +a1 x+a2 x2 +...+an−1 xn−1
is continuous.
Let a0 + a1 x + a2 x2 + ... + an xn be any polynomial of degree n. Then define f (x) =
a0 + a1 x + a2 x2 + ... + an−1 xn−1 and g(x) = an x2 . We know that g is continuous, hence the
sum f + g is continuous, proving that Pn+1 is true. The claim follows by mathematical
induction.
For the second statement observe that for n odd, lim xn = +∞ for x diverging to +∞
and lim xn = −∞ for x diverging to −∞. Pick x and y real numbers where the polynomial
is neagtive and positive. Then by the intermediate value theorem there is a point z where
the polynomial vanishes.
Problem 7: Let f be a continuous function on a closed interval [a, b]. Show that the
following statement is true:
∀ǫ > 0∃δ > 0 such that ∀x, y ∈ [a, b] with |x − y| < δ =⇒ |f (x) − f (y)| < ǫ.
(Hint: see Theorem 19.2.)
Solution
The proof is in the book, proof of Theorem 19.2.