Answers to Homework 11, Math 4121
(1) For 0 < r < p < s < ∞, prove that Lr ∩ Ls ⊂ Lp . Further, if
µ(X) < ∞, prove that Ls ⊂ Lr if 0 < r < s < ∞.
Writing p = (1 − t)r + ts with t ∈ (0, 1) and using the fact
that φ(x) = ax is a convex function for any a > 0 on R1 , we see
that if f ∈ Lr ∩ Ls , |f |p = |f |(1−t)r+ts ≤ (1 − t)|f |r + t|f |s , and
integrating, we see that f ∈ Lp .
The other case is equally easy. Notice that |f |0 = 1 ∈ L1 (X)
since µ(X) < ∞. Writing r = ts with 0 < t < 1, we get for any
f ∈ Ls , |f |r = |f |(1−t)0+ts ≤ (1 − t)|f |0 + t|f |s . Notice that this
implies, f ∈ Lr and ||f ||r ≤ (1 − t)µ(X) + t||f ||s .
(2) If f, g are positive measurable functions on X Rwith µ(X)
R =1
such that f (x)g(x) ≥ 1 for all x ∈ X, prove that X f dµ X gdµ ≥
1.
√
√
Consider F = f , G = g. Then these are positive measurable functions and by Hölder’s inequality for p = q = 2, we
have,
Z
2 Z
Z
gdµ .
f dµ
F Gdµ ≤
X
X
X
R
R
Since F (x)G(x) ≥ 1 for all x, we get that X F Gdµ ≥ X 1dµ =
1.
(3) Let X = (0,
Define
R ∞) and let f ∈ Cc (X) which is positive.
1 x
p
F (x) = x 0 f (t)dt for x ∈ X. Prove that F ∈ L (X) for any
p
p, 1 < p < ∞ and ||F ||p ≤ p−1
||f ||p . (Same is true for any
p
f ∈ L (X).)
First, we check that F is measurable. F is continuous (in fact
differentiable) in (0, ∞) by Fundamental Theorem of Calculus
and hence measurable. Let [a, b] ⊂ (0, ∞) contain the support
of f where 0 < a < b < ∞. Then support of F is contained in
[a, ∞) and thus to check that F ∈ Lp , it suffices to check that
F ∈ Lp ([a, ∞)). Since F is continuous, suffices to check that
F ∈ RLp ([b, ∞). But for all x ≥ b, we have F (x) = A/x where
∞
A = 0 f dx and since x1 ∈ Lp ([b, ∞)), we are done.
Integrating by parts, we get,
Z ∞
Z ∞
p
p
||F ||p =
F dx = −p
F p−1 F 0 xdx.
0
0
The derivative of xF (x) is just f (x). So, we get, xF 0 (x) +
F (x) = f (x) and thus substituting for in the above integral, we
1
2
get,
||F ||pp
∞
Z
F p−1 (f − F )dx,
= −p
0
which yields,
(p −
1)||F ||pp
∞
Z
F p−1 f dx
=p
0
By Hölder’s inequality, we have,
||F p−1 f ||1 ≤ ||f ||p ||F p−1 ||q
where q is conjugate to p. We have,
Z ∞
1q
p−1
(p−1)q
||F ||q =
F
dx .
0
p
Since (p − 1)q = p, we get ||F p−1 ||q = ||F ||pq . Putting these
together, we get,
p
(p − 1)||F ||pp ≤ p||f ||p ||F ||pq
p
Canceling |F ||pq from both sides (it is a finite positive quantity)
and noting that p − pq = 1, we get the desired inequality.
(4) Suppose Rµ(X) = 1 and f : X → [0,
√∞] a measurable
R p function.
2
Let A = X f dµ. Then prove that, 1 + A ≤ X 1 + f 2 dµ ≤
1 + A. If X = (0, 1) with the Lebesgue measure and f = g 0
for a differentiable function, this must be familiar to you from
calculus
√ The first inequality is obvious by Jensen’s inequality, since
1 + x2 is easily seen to be convex onR (0,p∞). The second
is equally easy, since it is the same as X ( 1 + f 2 − 1)dµ ≤
p
R
2
f
dµ.
But
1 + f2 − 1 = √ f 2 ≤ f.
X
1+f +1
p
(5) Let p, q be conjugate with 1 < p, q < ∞. For
R any f ∈ L , we
q
have a map Tf : L → R defined as Tf (g) = X f gdµ.
(a) If L : Lq → R is a linear functional, prove that L is continuous if and only if there exists a non-negative number l
such that |L(g)| ≤ l||g||q for all g.
Assume that L is continuous. Then we can find a δ > 0
such that |L(h)| = |L(h) − L(0)| < 1 whenever ||h||q =
||h−0||q ≤ δ. Let l = 1δ . The inequality is obvious for g = 0
with any choice of l, so let us assume that g 6= 0. Then
δ
letting a = ||g||
, we have ||ag||q = δ and thus |L(ag)| < 1
q
and so |L(g)| <
1
a
=
||g||q
δ
= l||g||q .
3
Conversely, if we have such an l, for any g, h ∈ Lq , |L(g) −
L(h)| = |L(g − h)| ≤ l||g − h||q . So, given > 0, we can
choose δ = /l (and if l = 0, any choice of δ will do), to
get continuity.
(b) Prove that |Tf (g)| ≤ ||f ||p ||g||q . Deduce that Tf is a continuous linear functional.
This is just Hölder’s inequality.
(c) Denoting by H the set of all continuous linear functionals
on Lq , for any L ∈ H, define ||L||H = inf{l ≥ 0||L(g)| ≤
l||g||q } and prove that this makes H a normed linear space.
(The last phrase just means H is a vector space and the
norm has all the basic properties of a norm in Lp -spaces.)
All are easy. It is clearly a linear space, since if L, M ∈ H,
then aL+bM is both linear and continuous for any a, b ∈ R.
||L||H ≥ 0 and zero if and only if L = 0. If a ∈ R, then
||aL||H = |a|||L||H and ||L + M ||H ≤ ||L||H + ||M ||H .
(d) Prove that the map A : Lp → H defined by f 7→ Tf is
continuous linear with ||f ||p = ||Tf ||H for all f .
From part (b), we see that ||Tf ||H ≤ ||f ||p and thus ||A(f )|| ≤
||f ||p . Now continuity follows exactly as in part (a). Linearity of A is clear, since Taf +bg = aTf + bTg .
Only thing left to prove is the reverse inequality. Suffices
to check these for positive functions and so, let f ∈ Lp and
f ≥ 0. Define g = f p−1 , then g is measurable, positive pand
g q = f q(p−1) = f p . So, we see that g ∈ Lq and ||f ||pq =
||g||q . Also,
Z
Z
p− p
f p dµ = ||f ||pp = ||g||q ||f ||p q
f gdµ =
A(f )(g) = Tf (g) =
X
Since p −
p
q
X
= 1, we are done.