General Measures and the Lebesgue -Spaces: the Importance of Lebesgue's Ideas in Functional Analysis
[9/0%]Prove that the triples given in Examples 4 and 6 at the beginning of Section 6 are, in fact, measure spaces. For the measure in Example 3, show that
whenever implies .
For , we can define the -space and in the same way that we did for . Prove, by giving a suitable example, that does not satisfy the triangle inequality, and hence is not a norm.
Prove, as outlined in the text, that
for and Hölder conjugates and with .
Prove that
for and . (This equality is used in the proof of Theorem 3.19.)
Prove that , with norm , is a normed linear space.
In this exercise you will investigate relations between the various -spaces.
Let . Consider Lebesgue measure on . Construct examples to show that neither nor holds.
Next, suppose that . Show that . (This is for any measure space .)
Now assume that is a finite measure space, i.e., that is measurable and that . Prove that for , and give an example to show that this is a proper inclusion. Now prove that
whenever is a probability measure.
Prove that for , and give an example to show that this is a proper inclusion. Now prove that
Show that for each almost everywhere.
Let be an interval in .
Assume that . Prove that is dense in whenever is closed and bounded.
Now drop the assumption that is closed and bounded. We certainly cannot expect to be dense in , since it is not even contained in it. We define the "continuous functions with compact support" on to be
The set is often called the "support" of . Again, assume that . Prove that is dense in . (Note that this result subsumes the result of (a).)
Prove that is not dense in .
(The results of Exercise 8 give an alternative proof of Theorem 4.7, and give a way to define that is independent of measure theory. See the paragraph following the proof of Theorem 4.7.)
The point of this exercise is to show that the norm on , for , does not come from an inner product. We start by considering , with Lebesgue measure. Set and .
Show that
Using part (a), show that the parallelogram equality asserts that
Verify that this equality holds for .
Prove that the parallelogram equality does not hold for values . Hint: Show that the function is a strictly decreasing function of and thus takes on the value zero for at most one value of .
Modify the functions given in part (a) to prove the result for where is any interval, bounded or unbounded, in .