Solutions of Sixth Homework
Solution of 8.2 Ex. 2:
Assume that F is continuous. Let Z be a closed subset of Rq . The
complement Z c is open. Hence, since F is continuous, F −1 (Z c ) is a
relatively open set in D. On the other hand, F −1 (Z) = D − F −1 (Z c )
is relatively closed in D.
Conversely, assume that F −1 (Z) is relatively closed in D for any
closed set Z in Rq . Then, for any open set U in Rq , U c is closed in
Rq . Therefore, F −1 (U c ) is relatively closed set in D. On the other
hand, F −1 (U ) = D − F −1 (U c ) is relatively open set in D. Hence, F is
continuous on D.
Solution of 8.2 Ex. 5:
Let
1
1 + x2
for x ∈ R. Then F is a continuous function on R. The image of the
closed set R is (0, 1], i.e., it is not closed.
F (x) =
Solution of 8.2 Ex. 11:
Let > 0. Since F is uniformly continuous on D, there exists δ > 0,
such that x, y ∈ D such that kx−yk < δ implies that kF (x)−F (y)k < .
Let (xn ) be a Cauchy sequence in D. Then there exists N > 0 such
that n, m ≥ N implies kxn − xm k < δ. Hence kF (xn ) − F (xm )k < .
This in turn implies that (F (xn )) is a Cauchy sequence in Rq .
Solution of 8.3 Ex. 5:
Since F (x, y) = (x + 1, y + 1) for (x, y) ∈ R2 , we see that
p
kF (x, y)k = (x + 1)2 + (y + 1)2 .
Hence, kF (x, y)k is the distance from the point (−1, −1) to the point
(x, y). Since D is the closed unit disk in R2 centered at the origin,
we
√ √
2
see immediately that the most distant point in D is the point ( 2 , 22 )
on the unit circle centered at the origin on√the opposite side from
(−1, −1). The
√ distance of that point is 1 + 2. Hence, we see that
kF kD = 1 + 2.
Solution of 8.3 Ex. 12:
Assume that (Fn ) is a sequence of functions on D which converges
uniformly to 0. Then for any > 0 there exists N such that n ≥ N
implies that kFn (x)k < for all x ∈ D. In particular, for any sequence
1
2
(xn ) in D we have kFn (xn )k < for n ≥ N . Hence, the sequence
(Fn (xn )) converges to 0.
Therefore, if there exists a sequence (xn ) in D such that (Fn (xn ))
doesn’t converge to 0, the sequence of functions Fn cannot be uniformly
convergent to 0 on D.
Conversely, if the sequence Fn doesn’t converge uniformly to 0 on D,
there exists > 0 such that for any N > 0 there exists n ≥ N and x ∈ D
such that kFn (x)k ≥ . We construct a sequence (xn ) in the following
way. For N = 1 we pick n1 such that n1 ≥ 1 and xn1 ∈ D such that
kFn1 (xn1 )k ≥ . Assume that we consructed n1 < n2 < . . . < nm and
picked xni ∈ D such that kFni (xni )k ≥ for all 1 ≤ i ≤ m. Then, we
can pick nm+1 ≥ nm + 1 and xnm+1 ∈ D such that kFnm+1 (xnm+1 )k ≥ .
By induction, we form a sequence n1 < n2 < . . . < nm < . . . such that
kFnm (xnm )k ≥ for all m ∈ N. Picking arbitrary xn for other values of
n, gives us a sequence (Fn (xn )) which doesn’t converge to 0.