MATH 321 - HOMEWORK #6
Due Friday, March 4.
We defined equicontinuity for sequences of functions. The same definition can be used
to define equicontinuity for any set F of functions on a domain E. We simply replace the
condition for all fn with he same condition for all f ∈ F. Similarly, one can define pointwise
boundedness and uniform boundedness for a set F.
P ROBLEM 1. (a) Let f(x, y) be a continuous function on [0, 1] × [0, 1]. For each fixed y, let
fy (x) = f(x, y). Prove that the set of functions
{fy }y∈[0,1]
is uniformly bounded and equicontinuous. (Hint: f is a continuous function on a
compact set, hence uniformly continuous.)
(b) Let k(x, y) be a continuous function on [0, 1] × [0, 1]. For any continuous function f(x)
on [0, 1], define
Z1
Kf(x) = k(x, y)f(y)dy.
0
If F is any uniformly bounded set of continuous functions on [0, 1], prove that the set
KF = {Kf|f ∈ F}
is uniformly bounded and equicontinuous.
P ROBLEM 2. Let {fn } be a sequence of differentiable functions on [a, b], satisfying:
(1) |fn0 (x)| ≤ 1 for any n and any x.
Rb
(2) a |fn (x)|dx ≤ 100 for any n.
Prove that the sequence contains a uniformly convergent subsequence.
P ROBLEM 3. Let F be the set of all continuous functions f on [0, 1] satisfying:
(1) f(0) = 0.
p
(2) |f(x) − f(y)| ≤ |x − y| for all x, y ∈ [0, 1].
Let g be a continuous function on [0, 1]. Prove that there exists a function f∗ ∈ F that
R1
maximizes the integral 0 |f(x) − g(x)|dx:
Z1
Z1
∗
|f (x) − g(x)|dx = max |f(x) − g(x)|dx.
0
f∈F
0
(Hint: Try to adapt the proof that every continuous function L on a compact set achieves
its maximum at some point. One proof goes as follows. We can find a sequence of points
1
pn such that L(pn ) approaches supp L(p). Now the sequence {pn } contains a convergent
subsequence and the limit point is the maximum point.)
P ROBLEM 4. Recall that a subset C of a metric space is compact if and only if every sequence
in C contains a convergent subsequence. Consider the vector space C[a, b] of continuous
functions on [a, b] with the supremum norm giving the metric. Prove that a subset C ⊂ V
is compact if and only if C is closed, bounded, and equicontinuous. (Here bounded in
the supremum norm means uniformly bounded. Similarly, closed with respect to the
supremum norm means that every sequence in C that converges uniformly has its limit
also in C.)
(Hint: To prove that compactness implies equicontinuity, see the hint to problem 7.19 in
the textbook.)