Math 3520
Homework 10 Solutions
1. (p138,#3) If the domain of a continuous function is an interval, show that the image is an
interval. Give examples where the image is an open interval.
Solution: Let I be the (interval) domain of f . Pick y1 , y2 ∈ f (I), without loss of generality assume y1 < y2 . We would like to show that for all y ∈ (y1 , y2 ), there exists an x ∈ I
such that y = f (x). This is a direct consequence of the Intermediate Value Theorem: f
is a continuous function on the closed interval [x1 , x2 ] (with y1 = f (x1 ), y2 = f (x2 )), so
there exists an x ∈ (x1 , x2 ) such that f (x) = y.
2. (p138,#4) If a continuous function on an interval takes only a finite set of values, show that the
function is constant.
Solution: Let I be the (interval) domain of f , and f (I) = {y1 , y2 , . . . , yk } with y1 < y2 <
· · · < yk . Take xi ∈ I such that yi = f (xi ), then f is a continuous function on the closed
interval [xi , xi+1 ] (or [xi+1 , xi ]). By the Intermediate Value Theorem, for all y ∈ (yi , yi+1 )
there exists an x ∈ (xi , xi+1 ) such that f (x) = y. This is a contradiction since y 6∈ f (I).
3. (p138,#5) Suppose f and g both satisfy a Lipshitz condition on an interval (| f (x) − f (y) | ≤
M | x − y | for all x and y in the interval). Show that f + g also satisfies a Lipshitz condition.
Solution: Suppose the functions f and g satisfy a Lipshitz condition on an interval I:
there exist positive real constants M1 and M2 such that
| f (x) − f (y) | ≤ M1 | x − y |
| g(x) − g(y) | ≤ M2 | x − y |
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for all x, y ∈ I. In particular, with a simple application of the Triangle Inequality:
| (f + g)(x) − (f + g)(y) | = | (f (x) + g(x)) − (f (y) + g(y)) |
= | f (x) − f (y) + g(x) − g(y) |
≤ | f (x) − f (y) | + | g(x) − g(y) |
≤ M1 | x − y | + M2 | x − y |
= (M1 + M2 ) | x − y |
for all x, y ∈ I.
4. (p138,#6) Show that if f and g are bounded and satisfy a Lipshitz condition on an interval,
then f · g satisfies a Lipshitz condition. Give a counterexample to show that it is necessary to
assume boundedness.
Solution: Suppose the functions f and g satisfy a Lipshitz condition on an interval I:
there exist positive real constants M1 and M2 such that
| f (x) − f (y) | ≤ M1 | x − y |
| g(x) − g(y) | ≤ M2 | x − y |
for all x, y ∈ I. Since f and g are also bounded, there exists an N1 , N2 such that
| f (x) | ≤ N1
| g(x) | ≤ N2
for all x ∈ I. In particular,
| (f g)(x) − (f g)(y) | = | 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) |
= N2 M1 | x − y | + N1 M2 | x − y |
= (N2 M1 + N1 M2 ) | x − y |
for all x, y ∈ I.
The function h(x) = x2 is not Lipshitz on the interval I = (−∞, ∞). Indeed
| h(x) − h(y) | = x2 − y 2 = |x + y||x − y|.
There is no constant M such that | x + y | ≤ M for all x, y ∈ (−∞, ∞). The constant
function f (x) = x is Lipshitz on I, and h = f · f .
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5. (p138,#7) 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.
Solution: Suppose that f : I → J is a monotone function on intervals I, J ⊂ R. Constant
functions are automatically continuous, hence without loss of generality we may assume
f is strictly increasing. Suppose that U ⊂ J is an open set, we would like to show
f −1 (U ) ⊂ I is open. For each f (x) ∈ U , there exists an ε such that (f (x) − ε, f (x) + ε) ⊂
U . Since f strictly increasing, there exist x1 , x2 ∈ f −1 (U ) such that f (x1 ) = f (x) − ε and
f (x2 ) = f (x) + ε with x ∈ (x1 , x2 ) ⊂ f −1 (U ). Hence f −1 (U ) is open. For an example of
a non-monotone function on an interval whose image is an interval yet is not continuous:
(
1 − 2x 0 < x < 1/2
f (x) =
2 − 2x 1/2 < x < 1
6. Prove: Let f be defined in a neighborhood of x0 , and differentiable at x0 . If f monotone
increasing at x0 then f 0 (x0 ) ≥ 0. Similarly, if f monotone decreasing at x0 , then f 0 (x0 ) ≤ 0.
Solution: Suppose that f is monotone increasing at x0 . Then for x < x0 , f (x) ≤ f (x0 ),
so that x − x0 < 0 and f (x) − f (x0 ) ≤ 0. Hence
f (x) − f (x0 )
≥0
x − x0
since the numerator and denominator are both negative. Similarly, if x0 < x the f (x0 ) ≤
f (x) so that 0 < x − x0 and 0 ≤ f (x) − f (x0 ). Hence
f (x) − f (x0 )
≥0
x − x0
since the numerator and denominator are both positive. In particular,
f 0 (x0 ) = lim
x→x0
f (x) − f (x0 )
≥ 0.
x − x0
Now suppose that f is monotone decreasing at x0 . Then for x < x0 , f (x) ≥ f (x0 ), so that
x − x0 < 0 and f (x) − f (x0 ) ≥ 0. Hence
f (x) − f (x0 )
≤0
x − x0
since the numerator is positive and the denominator is negative. Similarly, if x0 < x then
f (x0 ) ≥ f (x) so that 0 < x − x0 and 0 ≥ f (x) − f (x0 ). Hence
f (x) − f (x0 )
≤0
x − x0
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since the numerator is negative and the denominator is positive. In particular,
f 0 (x0 ) = lim
x→x0
f (x) − f (x0 )
≤ 0.
x − x0
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