MA3421 (Functional Analysis 1) Tutorial sheet 7
[November 20, 2014]
Name: Solutions
1. Let f (x) = 1/xa (for 0 < x ≤ 1) where a > 0 is fixed (and f (0) = 0). For p with
1 ≤ p < ∞, find out which p have f ∈ Lp ([0, 1]) and find kf kp for those p.
R1
Solution: We need to look at 0 |f (x)|p dx. So
Z 1
1
dx
ap
0 x
That is an improper integral as a Riemann integral and so would be treated as
Z 1
1
dx
lim+
ap
δ→0
δ x
As a Lebesgue integral we can do more or less the same. To get a sequence, just look at
δ = 1/n. Then
1
χ[1/n,1] (x) a]
x
in an increasing sequence of (measurable) functions that converges to |f (x)|p pointwise.
So by the Monotone convergense theorem (the functions are also positive so that we can
use the theorem) the limit of their integrals is
Z 1
Z 1
1
1
dx =
dx
lim
ap
ap
n→∞ 1/n x
0 x
In any case
Z
δ
1
1
1
1
1
1
1
1
dx =
=
−
ap
ap−1
ap−1
x
−ap + 1 x
−ap + 1 −ap + 1 δ
δ
As δ → 0, if ap < 1, then the limit is 1/(1 − ap) < ∞, while if ap > 1 then the limit is
∞.
If ap = 1, then the calculation needs to be
Z 1
Z 1
1
1
dx =
dx = [log x]1δ = 0 − log δ → ∞
ap
x
x
δ
δ
(as δ → 0).
So we need ap < 1 or p < 1/a to get a finite integral and then kf kp is the pth root
kf kp =
1
(1 − ap)1/p
(1 ≤ p < 1/a)
2. Let f ∈ Lp ([0, 1]) where 1 ≤ p < ∞ and take r with 1 ≤ r < p. Show that f ∈ Lr ([0, 1])
and that kf kr ≤ kf kp . [Hint: Use Hölder’s inequality where one function is |f |r and the
other is the constant function 1, and one of the conjugate exponents is p/r.]
Solution: Write
Z
1
1
Z
r
Z
r
|f (x)| (1) dx =
|f (x)| dx =
g(x)h(x) dx
0
0
0
1
where g(x) = |f (x)|r and h(x) = 1.
Now Hölder’s inequality says
1
Z
g(x)h(x) dx ≤ kgkp khkq
0
where 1/p+1/q = 1 but that is not exactly what we want. We want Hölder with p replaced
by p/r and then q will need to satisfy
1
1
+ =1
p/r q
There will be such a q (with 1 < q < ∞) since p/r > 1 but we won’t actually need to
know what it is. Using Hölder with these new exponents, we get
Z 1
g(x)h(x) dx ≤ kgkp/r khkq
0
Now
Z
khkq =
1
1/q
=1
1 dx
q
0
while
Z
kgkp/r =
1
p/r
1/(p/r)
|h(x)|
Z
r/p
= (kf kpp )r/p = kf krp
0
So Hölder boils down to
Z 1
Z
g(x)h(x) dx =
0
p
|f (x)|
=
0
1
1
|f (x)|r dx ≤ kgkp/r khkq = kf krp
0
or kf krr ≤ kf krp . Taking rth roots, we get kf kr ≤ kf kp (and in particular f ∈ Lr ([0, 1])).
(See also Examples 2.6.6 (ii) in the notes.)
1 1
q
+ = 1 (where 1 ≤ p, q ≤ ∞) and b = (bn )∞
n=1 ∈ ` . Show that it makes sense to
p q
define T : `p → K by the formula
3. Let
T (x) = T
((xn )∞
n=1 )
=
∞
X
n=1
2
x n bn
Solution: By Hölder’s inequality (sequence version)
∞
X
an bn ≤ kakp kbkq
n=1
p
∞
q
for a = (an )∞
n=1 ∈ ` and b = (an )n=1 ∈ ` . Using x in place of a we see in particular that
∞
X
n=1
p
always converges for x = (xn )∞
n=1 ∈ ` .
Richard M. Timoney
3
x n bn