Uniforme Continuità

8/3/2019 Uniforme Continuit

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HOMEWORK 3 SOLUTIONS

MATH 171, 2009

Book 31.5. Prove that f has a limit at a if and only if for every > 0, there exists > 0 such that if 0 < |xa| < and 0 < |y a| < , then |f(x) f(y)| < .

Solution. First suppose f has limit L at a, and let > 0. Then there exists some such thatif 0 < |x a| < and x X, then |f(x) L| < 2 . Therefore if 0 < |x a| < and 0 < |y a| < ,then we have

|f(x) f(y)| | f(x) L| + |L f(y)| < 2

+

2 .

Conversely, suppose that for all there exists > 0 such that 0 < |x a| < and 0 < |y a| < imply |f(x) f(y)| < . Let > 0 and let be as prescribed above. Let {an} be a sequence

with limn an = a. Then there exists some N1 with |an a| < for all n N1, and hence|f(an) f(am)| < for all n, m N1. Hence {f(an)} is a Cauchy sequence, and so the limitlimn f(an) = L exists. So, there exists some N2 such that n N2 implies |f(an) L| < .Now, given any sequence {bn} with limn bn = a, we can pick N3 such that n N3 implies0 < |bn a| < . Hence, for all n max{N1, N2, N3}, we have

|f(bn) L| |f(bn) f(an)| + |f(an) L| < + = 2.Since was arbitrary, we have limn f(bn) = L. Hence by Theorem 31.2, we have limxa f(x) = L.

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Book 32.4. Prove that if limx f(x) = L, then limn f(n) = L.

Solution. Let > 0. By Definition 32.3, since limx f(x) = L, there exists a number M such thatif x > M, then |f(x) L| < . Let N = M + 1. Then for all n N, we have n > M, and so|f(n) L| < . Hence by Definition 10.2, we have limn f(n) = L

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List 1 #2. Let f : R R be a function such thatx, y R, f(x + y) = f(x) + f(y), f(xy) = f(x)f(y).

Show that f is either the zero function or the identity x x.

Solution. First note that f(0) = 0, since for every x R we have

f(x) = f(x + 0) = f(x) + f(0).Next, note that

f(1) = f(1 1) = f(1)f(1) = f(1)2.

But the only numbers satisfying x = x2 are 0 and 1, so we have either f(1) = 0 or f(1) = 1.

If f(1) = 0, then for all x R, we havef(x) = f(1 x) = f(1)f(x) = 0 f(x) = 0,

and f is the zero function.

If f(1) = 1, we prove that f(x) = x for all x N using induction. The base cases x = 0, 1 areclear. Assuming f(k) = k, we get that f(k + 1) = f(k) + f(1) = k + 1. Hence, f(x) = x for allnN. Next note that for x

N, we have

f(x) + f(x) = f(x + x) = f(0) = 0 = f(x) = f(x) = x.So f(x) = x for all x Z. Next note that for x Z with x = 0, we have

f(x)f 1

x

= f(1) = 1 = f

1x

=

1

f(x).

Hence for x, y Z with y = 0, we have

fx

y

= f(x)f

1y

=

f(x)

f(y)=

x

y.

So f(x) = x for all x Q.

Now, suppose x R with x 0. Then x = x x and so

f(x) = f(x)f(x) = f(x)2

.Hence f maps any nonnegative real number to a nonnegative real number. We use this to show thatf is order-preserving: if y x, then since y x 0, we have

f(y) f(x) = f(y x) 0 = f(y) f(x).

Finally, consider any x R. Pick a sequence {an} such that an Q and x 1n < an < x for alln, which is possible since Q is dense in R. Similarly, pick a sequence {bn} such that bn Q andx < bn < x +

1n

for all n. So for all n, we have an < x < bn. Since f is order-preserving, we havef(an) f(x) f(bn). Since an and bn are rational, this gives an f(x) bn. We then take thelimit as n goes to infinity and use Theorem 14.3 to get

x = limn

an f(x) limn

bn = x.

Hence f(x) = x for all x R.

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List 1 #21. Let f : R R be a continuous function such thatx, y R : f(x + y) = f(x) + f(y).

Show that there exists R such that f(x) = x for any x.

Solution. First note that f(0) = 0, since for every x R we have

f(x) = f(x + 0) = f(x) + f(0).Let f(1) = .

We prove that f(nx) = nf(x) for all n N and x R using induction on n. The base casesx = 0, 1 are clear. Assuming f(kx) = kf(x), we get that f((k + 1)x) = f(kx + x) = kf(x) + f(x).Hence, f(x) = nf(x) for all n N.

Next note that for n N, we havef(nx) + f(nx) = f(nx + nx) = f(0) = 0 = f(nx) = f(nx) = nf(x).

So f(zx) = zf(x) for all z Z.

Plugging in x = 1, we get f(n) = n for all z

Z. And for x, y

Z with y

= 0, we have

x = f(x) = f

y xy

= yf

xy

= f

xy

= x

y.

So f(x) = x for all x Q.

Finally, consider any x R. Pick a sequence {an} such that an Q for all n and limn an = x.Since Q is dense in R, such a construction is possible (for each n N, there exists some an Q withx < an < x +

1n

). We have

f(x) = limyx

f(y) since f is continuous, by Definition 33.1

= limn

f(an) by Theorem 31.2

= limn

an since an Q

= limnan since an Q= x.

Hence f(x) = x for all x R.

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List 2 #2. Let fn : R R be defined by fn(x) = xn2 where n 1.(1) Show that the sequence of functions (fn) converges pointwise to the zero function. Is the con-vergence uniform on R?(2) Show that the convergence is uniform on any [a, b].(3) Show that the series of functions

n=1 fn(x) converges normally on any [a, b].

Solution. (1) Let > 0 and x R. If x = 0, then we have |fn(x) 0| = 0 < for all n, and so wecan assume x = 0. Pick N such that 1

N2< |x| , which is possible since limN

1N2

= 0. Then for all

n N, we have |fn(x)0| = | xn2 | |x|N2 < . Hence the sequence (fn) converges pointwise to the zerofunction. However, the convergence is not uniform, because supxR |fn(x) 0| = supxR | xn2 | = does not tend to 0 as n .

(2) The convergence is uniform on any [a, b] because we have

supx[a,b]

|fn(x) 0| =max{|a|, |b|}

n2

converges to 0 as n .

(3) The series of functions

n=1 fn(x) conveges normally on any [a, b] because we have

supx[a,b]

|fn(x)| =max{|a|, |b|}

n2

withn=1

max{|a|, |b|}

n2= max{|a|, |b|}

n=1

1

n2< .

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A. Prove the the square root mapping is uniformly continuous on [0,).

Solution. Let > 0 be arbitrary. First we show the square root is continuous on [0, 2]. Ifx = 0, let = 2. Then if |x y| < , then either y = 0 so |xy| = 0 < or y = 0 so

|x y| < = y < 2 = |xy| = y b implies that f(x) > f(0). Since f is continuous, by Theorem 34.4 there exists a pointc [a, b] such that f(c) f(x) for all x [a, b]. Additionally, we have f(c) f(0) < f(x) for allx / [a, b]. Hence we have f(c) f(x) for all x R, so f has a global minumum on R and attains itat c.

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