solns5

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MATH 3224 Topology Problem Set 5, Solutions

1. (a) We must show that given any > 0 we can find N Z+ such that d(fn, fm) < for alln, m N. Now

d(fn, fm) = sup{|fn(x) fm(x)| : x [0, 1]} = sup{|xn

xm

| : x [0, 1]}

= (sup{|

x|

: x

[0, 1]}

) 1n

1

m = 1

n 1

m max{1/n, 1/m}

Hence, given > 0, we can choose any N Z+ such that N > 1/, since then for alln, m N,

d(fn, fm) 1N

< .

(b) Again, the main job is to estimate d(fn, fm):

d(fn, fm) = sup{|fn(x) fm(x)| : x [0, 1]} = sup{|(x + 1n

)3 (x + 1m

)3| : x [0, 1]}

= sup{|3x2

n+

3x

n2+

1

n3 3x

2

m+

3x

m2+

1

m3| : x [0, 1]}

sup

{3x2

|n1

m1

|+ 3x

|n2

m2

|+

|n3

m3

|: x

[0, 1]

} 3|n1 m1| + 3|n2 m2| + |n3 m3| 3max{n1, m1} + 3 max{n2, m2} + max{n3, m3} 3max{n1, m1} + 3 max{n1, m1} + max{n1, m1} = 7 max{n1, m1}.

Hence, given > 0, we can choose any N Z+ such that N > 7/, since then for alln, m N,

d(fn, fm) 7 1N

< .

2. (a) For all x, y [0, 1],

d((x), (y)) =

x3 + 1

4 y

3 + 1

4

=1

4|x3 y3| = 1

4|x y||x2 + xy + y2|

14|x y|{|x|2 + |x||y| + |y|2} 1

4|x y|{1 + 1 + 1} = 3

4|x y|.

Hence is a contraction mapping (with k = 34

).

(b) Since is differentiable, we can use the Mean Value Theorem. For all x, y R withx = y, there is some c between x and y such that

d((x), (y))

d(x, y)=

(x) (y)x y = (c) = 1 c

2

2(1 + c2)2 1 + c

2

2(1 + c2)2=

1

2(1 + c2) 1

2.

Hence, for all x, y, x = y, d((x), (y)) 12

d(x, y). The same inequality holds trivially ifx = y, so we conclude that is a contraction mapping (with contraction factor k = 1

2).

(c) For all f, g C([0, 1],R),d(f, g) = sup{|f(x) g(x)| : x [0, 1]}

and d((f), (g)) = sup{|(f)(x) (g)(x)| : x [0, 1]}.Now |(f)(x) (g)(x)| = 1

3|f(0) + xf(1 x) g(0) xg(1 x)|

13|f(0) g(0)| + |x|

3|f(1 x) g(1 x)|

13|f(0) g(0)| + 1

3|f(1 x) g(1 x)|,

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for all x [0, 1]. But |f(0) g(0)| d(f, g) and sup{|f(1 x) g(1 x)| : x [0, 1]} =d(f, g) since 1 x [0, 1] if and only ifx [0, 1]. Hence

d((f), (g)) 13

d(f, g) +1

3d(f, g) =

2

3d(f, g),

so is a contraction mapping (with contraction factor k = 23

).

(d) For all f, g

C([

1, 1],R), and all x

[

1, 1],

|(f)(x) (g)(x)| =x

1

t3 sin t(f(t) g(t)) dt

x

1

|t3|| sin t||f(t) g(t)| dt

x

1

|t3| 1 d(f, g) dt = d(f, g)x

1

|t3| dt d(f, g)11

|t3| dt

= 2d(f, g)

10

t3dt =1

2d(f, g).

Since this holds for all x [1, 1],

d(f, g) = sup{|(f)(x) (g)(x)| : x [1, 1]} 12

d(f, g),


).

(e) The important thing to note in this example is that f C([0, 1], [0, 1]) implies 0 f(t) 1 for all t [0, 1], so we have immediate pointwise bounds on f(t). So, for allf, g C([0, 1], [0, 1]), and all x [0, 1],

|(f)(x) (g)(x)| =x

0

t(f(t)sin f(t) g(t)sin g(t)) dt

x

0

t|f(t)sin f(t) g(t)sin g(t)| dt

=

x

0

t|f(t)sin f(t) f(t)sin g(t) + f(t)sin g(t) g(t)sin g(t)| dt

x

0

t{|f(t)|| sin f(t) sin g(t)| + |f(t) g(t)|| sin g(t)|} dt.

Since f, g : [0, 1] [0, 1] we know that |f(t)| 1 for all t. Furthermore sin is strictlyincreasing on [0, 1] and sin 0 = 0, so | sin g(t)| sin1 < 1. Using the Mean Value Theorem,we see that

| sin f(t) sin g(t)| |f(t) g(t)|since there is some c between f(t) and g(t) such that | sin f(t) sin g(t)| = |f(t) g(t)|| cos c|. Hence

|(f)(x) (g)(x)| x

0

t{|f(t) g(t)| + |f(t) g(t)|| sin1|} dt

(1 + sin 1)d(f, g) x

0t dt =

x2

2 (1 + sin 1)d(f, g).

Since this holds for all x [0, 1],

d(f, g) = sup{|(f)(x) (g)(x)| : x [1, 1]} 12

(1 + sin 1)d(f, g),


(1 + sin 1) < 1).

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3. We can show this directly by exhibiting a pair of points x, y such that |(x) (y)| |x y|.Heres a slicker argument. Since (R, d) is complete, if werea contraction mapping, it wouldhave a uniquefixed point, by the Contraction Mapping Theorem. But has threefixed points,0 and log 2 (you find these by solving (x) = x), so it cant be a contraction mapping.

4. Note that x solves the equation if and only if it is a fixed point of the mapping (x) = 15

(2x3).This maps [0, 1] into [0, 1], and [0, 1] equipped with the metric d(x, y) = |x y| is a completemetric space, so the result follows from the Contraction Mapping Theorem if we can show that : [0, 1] [0, 1] is a contraction mapping. Now

|(x) (y)| = 15|x3 y3| = 1

5|x y||x2 + xy + y2| 3

5|x y|,


). Hence has a unique fixedpoint in [0, 1], so the equation has a unique solution in [0 , 1].

5. (a) Let f : X Y be locally Lipschitz and xn x in X. There exists a neighbourhoodU of x and a constant k > 0 such that (f(x), f(x)) kd(xx) for all x, x U.Since xn x, there exists N Z+ such that n N implies xn U. But then, for alln N, (f(xn), f(x)) kd(xn, x), and d(xn, x) 0, so f(xn) f(x) by the SqueezeRule. Hence f is sequentially continuous. Since X is a metric space, it follows that f iscontinuous (Theorem 136).

(b) Let X = Y = R with the usual metric, d(x, y) = (x, y) = |x y|. Then f(x) = x 13is continuous but is not locally Lipschitz. To see this, imagine f were locally Lipschitz.Then there would exist some neighbourhood U of 0 and some constant k > 0 such thatfor all x U, |x 13 0 13 | k|x 0|. But 1/n 0, so there is some N Z+ such that1/n U for all n N, and hence |(1/n)23 | = n 13 k for all n N. But his is false (thesequence n

2

3 is unbounded above).

(c) Let X = Y = R with the usual metric as above, and let f(x) = x2. Then |f(x)f(x)| =|x + x||x x| so for each x R we may take the neighbourhood U = (x1, x +1) andthe constant k = 2|x| + 2 > 0. Hence f is locally Lipschitz. It is not globally Lipschitzhowever since, for example |f(x) f(0)|/|x 0| = |x| is an unbounded function on R\{0}(so there is no k > 0 such that |f(x) f(0)| k|x 0| for all x R).

(d) By assumption f

: [a, b] R is continuous, hence bounded ([a, b] is compact, so f([a, b])is a compact subset of R, which is a metric space, so f([a, b]) is bounded). Let k =1+sup{|f(x)| : x [a, b]} > 0. Then for all x, x [a, b], ifx = x then, by the MeanValue Theorem, there is x between x, x such that

f(x) f(x)x x

= |f(x)| k.Hence f is globally Lipschitz.

6. We seek a mapping : C([0, 1],R) C([0, 1],R) such that y = f(x) solves the differentialequation if and only if f is a fixed point of . In this case, the correct is

(f) = a + x

0

(t2f(t) + (t)) dt.

Check: if f is continuous, then (f)(x) exists for all x [0, 1] and is differentiable and hencecontinuous. So is a well-defined map C([0, 1],R) C([0, 1],R). Also (f)(0) = a for all f.Furthermore

d

dx(f)(x) = x2f(x) + (x)

by the Fundamental Theorem of Calculus, so y = f(x) solves the initial value problem if andonly iff is a fixed point of . Since C([0, 1],R) (with the sup metric) is complete, the desired

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result follows immediately from the Contraction Mapping Theorem once we show that is acontraction mapping.

Now for all f, g C([0, 1],R) and all x [0, 1],

|(f)(x) (g)(x)| =x

0

(t2f(t) + (t)) dt x

0

(t2g(t) + (t)) dt

=x

0

t2(f(t) g(t)) dt

x

0 t

2

|f(t) g(t)| dt d(f, g) x

0 t

2

dt =

x2

2 d(f, g).

Since this holds for all x [0, 1], we have that

d((f), (g)) = sup{|(f)(x) (g)(x)| : x [0, 1]} 12

d(f, g)


). The result now follows.

7. We seek a mapping : C([0, 1],R) C([0, 1],R) such that y = f(x) solves the differentialequation if and only if f is a fixed point of . In this case, the correct is

(f) = +1

2

x

0

sin(t2 + f(t)) dt.

Check: if f is continuous, then (f)(x) exists for all x [0, 1] and is differentiable and hencecontinuous. So is a well-defined map C([0, 1],R) C([0, 1],R). Also (f)(0) = for all f.Furthermore

d

dx(f)(x) =

1

2sin(x2 + f(x))

by the Fundamental Theorem of Calculus, so y = f(x) solves the initial value problem if andonly iff is a fixed point of . Since C([0, 1],R) (with the sup metric) is complete, the desiredresult follows immediately from the Contraction Mapping Theorem once we show that is acontraction mapping. Establishing this will be slightly harder work than in previous examples.

First, I claim that for all a, b R| sin a sin b| |a b|. ()

By the Mean Value Theorem applied to the continuously differentiable function (x) = sin x,there is, given any a, b R, with a = b, some c between a and b such that(a) (b)a b = |(c)| = | cos c| 1.

() follows immediately.Now for all f, g C([0, 1],R) and all x [0, 1],

|(f)(x) (g)(x)| = 12

x

0

[sin(t2 + f(t)) sin(t2 + g(t))] dt

12

x

0

|(t2 + f(t)) (t2 + g(t))| dt by ()

= 12x0

|f(t) g(t)| dt 12 d(f, g)x0

dt = x2 d(f, g).

Since this holds for all x [0, 1], we have that

d((f), (g)) = sup{|(f)(x) (g)(x)| : x [0, 1]} 12

d(f, g)


). The result now follows.

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