Space: The Case
[8/50%]Let , where . We know that by the Minkowski-like Inequality result given earlier.
We have for every .
From (a), we have .
If we write for the distance between and , then this distance function is truly a metric, if we identify functions equal a. e. on .
Writing for the distance between and does not define a metric on .
Consider the space and let . Write B for the Borel subsets of , and write for Lebesgue measure restricted to B. We will show that there exists no norm on such that forces .
Suppose that such a norm exists. Then there is such that for each .
Pick the minimal such from (a).
There is with .
Let and , so that . Then and . Use the minimality of to obtain , and deduce .
Conclude that for every , and hence the assumption in (a) entails a contradiction.
Let and let be nonnegative. Let , and .
.
.
.
Say , and let be nonnegative. Write .
if .
If , then under the assumption , so it must be the case that . Conclude that even if still exists and equals , giving the inequality in (a).
decreases to as , hence .
Verify the inequalities
Conclude that exists and equals . If , this is interpreted as .
Find conditions for equality in the Hölder-like Inequality.
Hölder-like Inequality: Let . Let be nonnegative, and let denote a nonnegative -measurable function such that exists and . Then .
Find conditions for equality in the Minkowski-like Inequality.
Minkowski-like Inequality: Let , and let be nonnegative. Then and .
For every and , we have . Use this to give a calculus-based proof of the Hölder-like Inequality that is similar to the proof of the original Hölder Inequality.
Let for , where is fixed. For each , we have that iff [here we use the fact that ]. Therefore, we have that for every , which gives for every .
Does the Riesz-Fischer Theorem necessarily hold in the case ?
Hint: note the relevance of the original Minkowski inequality in the proof of the Riesz-Fischer Theorem. When , the direction of the inequality is reversed.