SOLUTION OF HW7
MINGFENG ZHAO
April 10, 2013
1. [10 Points] Show that if f and g are bounded and satisfy a Lipschitz condition on an interval, then
f · g satisfies a Lipschitz condition. Give a counterexample to show that it is necessary to assume
boundedness.
Proof. Let f and g be bounded Lipschitz function on an interval I. Since f and g are bounded, then
there exists some M1 > 0 and M2 > 0 such that
|f (x)| ≤ M1 ,
and |g(x)| ≤ M2 ,
∀x ∈ I.
Since f and g satisfy the Lipschitz condition on I, then there exists some C1 > 0 and C2 > 0 such
that
|f (x) − f (y)| ≤ C1 |x − y|,
and |g(x) − g(y)| ≤ C2 |x − y|,
∀x, y ∈ I.
Now for any x, y ∈ I, we have
|f (x)g(x) − f (y)g(y)| = |f (x)g(x) − f (y)g(x) + f (y)g(x) − f (y)g(y)|
≤
|f (x)g(x) − f (y)g(x)| + |f (y)g(x) − f (y)g(y)|
= |g(x)||f (x) − f (y)| + |f (y)||g(x) − g(y)|
≤ M2 C1 |x − y| + M1 C2 |x − y|
=
[M2 C1 + M1 C2 ]|x − y|.
So f · g is also Lipschitz.
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2
MINGFENG ZHAO
Cunterexample I: Let f (x) = g(x) = x on R, then |f (x) − f (y)| = |g(x) − g(y)| = |x − y| for all
x, y ∈ R, that is, f and g are Lipschitz on R. But h(x) = f (x) · g(x) = x2 in R, then for all x 6= 0, we
know that
|h(x) − f (0)|
|x − 0|
When |x| → ∞, then
=
|x2 − 0|
|x − 0|
=
|x|.
|h(x) − h(0)|
→ ∞. So h is not Lipschitz.
|x − 0|
2. [10 Points] Let f be a monotone function on an interval. Show that if the image of f is an interval,
then f is continuous. Give an example of a non-monotone function on an interval whose image is an
interval but that is not continuous.
Proof. Without loss of generality, we assume f is a monotone increasing function on an open interval
I = (a, b). By the assumption, we know that f (I) = J for some interval J. For any x0 ∈ I, if f is
not continuous at x0 , then there exists some sequence xn ∈ I such that xn → x0 , as n → ∞, but
f (xn ) 6→ f (x0 ), as n → ∞. Up to take a subsequence of xn , without loss of generality, we assume
xn > x0 for all n ≥ 1. Then there exists some 0 > 0 such that for any k ≥ 1, there exists some
nk > k such that
f (xnk ) − f (x0 ) ≥ 0 .
Since f (xnk ) is decreasing and bounded below, then
inf f (xnk ) = lim inf f (xnk ) ≥ f (x0 ) + 0 .
k≥1
k→∞
For any x > x0 , since xnk → ∞ as k → ∞, then there exists some k0 such that x > xnk0 , which
implies that f (x) ≥ f (xnk0 ) ≥ f (x0 ) + 0 . Hence we get
f (x) ≥ f (x0 ) + 0 ,
∀x ≥ x0 .
SOLUTION OF HW7
3
But for all x ≤ x0 , we have f (x) ≤ f (x0 ). So there is no point x ∈ I such that f (x) ∈ (f (x0 ), f (x0 )+
), which contradicts with the assumption that f (I) is an interval.
Therefore, we know that f is continuous on I.
Counterexample I: Define




x + 1, ∀x < 0





f (x) =
0, x = 0







 x − 1, ∀x > 0.
Then f is not continuous at 0, but f (R) = R, that is, the image of f is the whole line.
3. [10 Points] Let f = p + g where p is a polynomial of odd degree and g is a bounded continuous
function on the line. Show that there is at least one solution of f (x) = 0.
Proof. Since g is a bounded continuous function on R, then there exists some M > 0 such that
|g(x)| ≤ M for all x ∈ R. Since p is a polynomial of odd degree, then
lim p(x) = ∞,
x→∞
and
lim
x→−∞
p(x) = −∞.
Since f (x) = p(x) + g(x) for all x ∈ R, then we know that
lim f (x) = ∞,
x→∞
and
lim
x→−∞
f (x) = −∞.
Since both p and g are continuous, then f is also continuous. So we know that there exists some
x0 ∈ R such that f (x0 ) = 0 .
4. [10 Points] If f is a continuous function on a compact set, show that either f has a zero or f is
bounded away from zero.
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MINGFENG ZHAO
Proof. Let f be a continuous function on a compact set K. If there exists some x0 ∈ K such that
f (x0 ) = 0, then x0 is a zero of f , we are done. If there is no point x ∈ K such that f (x) = 0. Since f
is continuous on a compact set K, then there exists some x1 ∈ K such that
f (x1 ) = inf f (x).
x∈K
By the assumption, we know that f (x1 ) > 0. So we get f (x) ≥ f (x1 ) > 0 for all x ∈ K, that is, f
is bounded away from zero.
.
5. [10 Points] Given an example of a function on R that assumes its sup and inf on every compact
interval and yet is not continuous.
Proof. Define
f (x) =




 1,
∀x ∈ Q



 0,
∀x ∈ R\Q.
For any compact interval [a, b], if a = b, then sup
f (x) = f (a) = f (b) = inf
then there exists some x0 ∈ Q
T
f (x). If a < b,
x∈[a,b]
x∈[a,b]
T
[a, b] and y0 ∈ R\Q [a, b], which implies that
1 = f (x0 ) = sup f (x),
and
0 = f (y0 ) = inf
f (x).
x∈[a,b]
x∈[a,b]
On the other hand, it is easy to see that f is not continuous at any point on R. In fact, for any
x ∈ R, we can find rational numbers x1 , x2 , · · · ∈ Q such that lim xi = x, and f (xi ) = 1 for all i ≥ 1.
k→∞
Also we can find irrational numbers y1 , y2 , · · · , ∈ R\Q such that lim yi = x, and f (yi ) = 0 for all
k→∞
i ≥ 1. So f is not continuous.
SOLUTION OF HW7
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Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu