MA3421 (Functional Analysis 1) Tutorial sheet 5
[October 30, 2014]
Name: Solutions
Let E = C([0, 1]) = {f : [0, 1] → K} and define norms k · k1 and k · k∞ on E by
Z 1
kf k1 =
f (t) dt,
kf k∞ = sup |f (t)|.
t∈[0,1]
0
Let fn (t) = max(0, 1 − nt) for t ∈ [0, 1].
1. Sketch the graphs of fn for n = 1, 2, 3.
Solution:
2. Show that limn→∞ kfn k1 = 0
Solution:
Z
kfn k1 =
1
1/n
Z
2
|fn (t)| dt =
0
1 − nt dt = [t − nt
1/n
/2]0
0
1
= 0 − −n 2
2n
=
1
2n
Hence limn→∞ kfn k1 = 0.
3. Show that the pointwise limit function f (t) = limn→∞ fn (t) exists for all t ∈ [0, 1] but is
not continuous.
Solution: For 0 < t ≤ 1 we have fn (t) = 0 for n > 1/t and so limn→∞ fn (t) = 0. But for
t = 0, fn (0) = 1 for all n and so limn→∞ fn (0) = 1.
Thus
and it is not continuous at 0.
(
1
f (t) =
0
if t = 0
for 0 < t ≤ 1
4. Show that the sequence (fn )∞
n=1 is not a Cauchy sequence in the metric arising from the
norm k · k∞ .
Solution: (#1) If it was a Cauchy sequence then it would converge in the complete space
(Banach space) (C[0, 1], k · k∞ ), which means it would converge uniformly to some limit
g ∈ C[0, 1]. But uniform convergence implies pointwise convergence and so g(t) =
limn→∞ fn (t) = f (t). But that is not possible since f is not in the space C[0, 1].
Solution: (#2) fn (1/(2n)) = max(1 − n(1/(2n)), 0) = 1/2 but f2n (1/(2n)) = 0.
If it was a Cauchy sequence then there would be N so that kfn − fm k∞ < 1/4 holds for
all n, m > N . If we take n = N + 1 and m = 2n we get a contradiction from
1
1
= |fn (1/(2n)) − f2n (1/(2n))| ≤ kfn − f2n k∞ <
2
4
Richard M. Timoney
2