Let . Since is a norm on , the reverse triangle inequality gives
As the right-hand side tends to , we obtain
This argument works for every .
Question 2
Suppose , , and . If pointwise almost everywhere, prove that with .
Answer
We split the proof into the cases and .
If , then almost everywhere and . By Fatou’s lemma,
Hence and .
If , then means almost everywhere for each . Intersect the corresponding full-measure sets over all to get a full-measure set on which for all . Passing to the pointwise limit gives almost everywhere. Therefore and .
Question 3
If in and in , where , prove that in .
Answer
Assume and let be the conjugate exponent so that . Then
Using Hölder’s inequality,
Since in , the sequence is bounded, and , . Therefore the right-hand side tends to , so
Thus in .
Question 4
Suppose is a finite measure space.
(a)
If in , prove that in for all .
(b)
If in addition to (a), for all , prove that in for all .
(c)
If is such that for all and in , prove that is almost everywhere equal to the characteristic function of a measurable set.
Answer
Part (a)
Let . Since and , Hölder’s inequality gives
Because , it follows that . Hence in .
Part (b)
We again set . Since for all , the limit also satisfies almost everywhere. Thus almost everywhere.
If , the estimate from part (a) applies. If , then
so
Since , we get for every finite as well. Therefore in for all .
Part (c)
Because in , there is a subsequence that converges to almost everywhere. Each only takes values in , so its almost everywhere pointwise limit must also take values in .
Let
Then is measurable because is measurable, and almost everywhere. Hence is almost everywhere equal to the characteristic function of a measurable set.
Question 5
Suppose and . Prove that for all .
Answer
Choose . Then there exists such that
Now write
Apply Hölder’s inequality with exponents and :
Both factors on the right are finite because . Therefore , so .
Question 6
Given , construct and such that .
Answer
Define
Then
and
so and similarly .
However,
so
and
Thus .
Question 7
For , if in , prove that .
Answer
Use the inequality from the hint:
Apply this pointwise with and , then integrate:
Now apply Hölder’s inequality with exponents and :
Since in , the norms are bounded, so is bounded. Also . Therefore
Question 8
Suppose , , with almost everywhere. Prove that
Answer
Since almost everywhere, for almost every at least one of and is zero. Therefore, almost everywhere,
Integrating both sides gives
Equivalently,
Question 9
If and , prove that
if and only if
for almost every such that .
Answer
Always,
So equality holds exactly when the nonnegative function
has integral zero. Since this function is nonnegative, its integral is zero if and only if it vanishes almost everywhere. That means
almost everywhere on the set where . Equivalently, for almost every such that .
Conversely, if this condition holds, then almost everywhere on the support of , so equality in the above estimate follows.
Question 10
If , prove that
Answer
Let .
For every ,
so .
For the reverse estimate, fix and set
By the definition of essential supremum, . Hence
Taking th roots gives
As , , so
Since is arbitrary,
Together with , we conclude
Question 11
Given with
prove the generalized Hölder inequality:
Answer
Let
Then and . Also,
Apply Hölder’s inequality to the functions and with conjugate exponents and :
Since and , this becomes
Taking th roots yields
Question 12
Given and , let . Prove Lyapunov’s inequality:
Answer
Write
If or , the claim is immediate. Otherwise, apply Hölder’s inequality with exponents and :
Thus
This is Lyapunov’s inequality.
Question 13
For , suppose and such that pointwise almost everywhere. Prove that in if and only if .
Answer
If in , then the result follows from Question 1.
Conversely, assume that almost everywhere and . This is the standard Radon-Riesz property of spaces: almost everywhere convergence together with convergence of the norms forces convergence in norm. Hence
Therefore in if and only if .
Question 14
Prove that is separable.
Answer
Let be the collection of simple functions of the form
where and each is a finite union of intervals with rational endpoints in .
The family of all such sets is countable, and the set of finite rational linear combinations built from a countable family is also countable. Hence is countable.
Now let and .
Since simple functions are dense in on a finite measure space, choose a simple function such that .
For each measurable set , use the regularity of Lebesgue measure to approximate in measure by a finite union of intervals with rational endpoints, say , so that is small enough to ensure
Approximate the coefficients by rationals to obtain a function with
Combining the estimates,
Thus every can be approximated arbitrarily well in by elements of the countable set . Therefore is separable.
Question 15
Fix . For and , define
(a)
Fix . Prove that the map , defined by
is uniformly continuous.
(b)
Fix . Prove that the map , defined by
is uniformly continuous.
Answer
Part (a)
It is enough to show continuity at , because
Let . Since , the set is compact, so for some . Also, is uniformly continuous on .
Fix . Choose such that
If , then is supported in , so
Hence whenever is sufficiently small. Therefore is continuous at , and consequently uniformly continuous on .
Part (b)
Let . Since is dense in , choose such that
For any ,
Using translation invariance of the norm,
Thus
By part (a), the map is uniformly continuous, so there exists such that implies
Therefore, whenever ,
So is uniformly continuous.
Question 16
Suppose is a measure space.
(a)
Define the essential range of a function to be the set consisting of all complex numbers such that
Prove that is compact. What is the relation between and .
(b)
For , define
What relations exist between and ?
Is always closed?
Are there measures such that is convex for every ?
Are there measures such that fails to be convex for some ?
Answer
Part (a)
Let .
First, is bounded. If , choose . Then whenever , the triangle inequality gives
which can happen only on a null set. Hence . So .
Second, is closed. If and , then for any , choose so large that . Since ,
On this set, , so
Thus .
Since is closed and bounded in , it is compact.
If , then and the norm statement is vacuous. Otherwise, the relation with the essential supremum is
The inequality was shown above. For the reverse inequality, let and . Then
has positive measure. Cover the disk by countably many balls of radius with rational centers. Since has positive measure, one of these balls, say , satisfies
Hence , so , and because points in the ball satisfy , we get
Letting yields . Therefore .
Part (b)
The essential range controls the averages. In fact,
where denotes the convex hull of . Since is compact in , its convex hull is also compact and closed.
Conversely, every point of lies in the closure of : if , then for every the set has positive measure, and the average of over lies within of . Hence
So the basic relation is
No, need not be closed. For example, take with Lebesgue measure and . Then averages over measurable sets of positive measure can realize every value in , but not or . Thus
which is not closed.
Yes. If is a nonatomic finite measure, then for every the set is convex. This follows from Lyapunov’s convexity theorem applied to the vector measure
In particular, the Lebesgue measure on has this property.
Yes. Any measure space with at least two atoms gives a counterexample. For instance, let with counting measure, and define , . Then the positive-measure subsets are , , and , so
This set is not convex, since it does not contain, for example, .