Section 3.6

General Measures and the Lebesgue LpL^p-Spaces: the Importance of Lebesgue's Ideas in Functional Analysis

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Problem 3.6.1

Prove that the triples (X,R,μ)(X, \mathcal{R}, \mu ) 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

μ(⋃n=1∞[an,bn])=∑n=1∞μ([an,bn]) \mu \left(\bigcup _{n=1}^{\infty }\left[a_{n}, b_{n}\right]\right)=\sum _{n=1}^{\infty } \mu \left(\left[a_{n}, b_{n}\right]\right)

whenever n≠mn \neq m implies [an,bn]∩[am,bm]=∅\left[a_{n}, b_{n}\right] \cap \left[a_{m}, b_{m}\right]=\emptyset.

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Problem 3.6.2

For 0<p<10<p<1, we can define the LpL^{p}-space and ∥⋅∥p\left\| \cdot \right\|_{p} in the same way that we did for 1≤p<∞1 \leq p<\infty. Prove, by giving a suitable example, that ∥⋅∥p\left\| \cdot \right\|_{p} does not satisfy the triangle inequality, and hence is not a norm.

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Problem 3.6.3

Prove, as outlined in the text, that

xy≤xpp+yqq x y \leq \frac{x^{p}}{p}+\frac{y^{q}}{q}

for x,y≥0x, y \geq 0 and Hölder conjugates pp and qq with 1<p<∞1<p<\infty.

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Problem 3.6.4

Prove that

∥∣f+g∣p−1∥q=(∥f+g∥(p−1)q)p−1=(∥f+g∥p)p−1 \left\| |f+g|^{p-1}\right\| _{q}=\left(\left\| f+g\right\| _{(p-1) q}\right)^{p-1}=\left(\left\| f+g\right\| _{p}\right)^{p-1}

for f,g∈Lp(μ)f, g \in L^{p}(\mu ) and 1<p<∞1<p<\infty. (This equality is used in the proof of Theorem 3.19.)

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Problem 3.6.5

Prove that L∞L^{\infty }, with norm ∥⋅∥∞\left\| \cdot \right\|_{\infty }, is a normed linear space.

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Problem 3.6.6

In this exercise you will investigate relations between the various LpL^{p}-spaces.

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(a)

Let 1≤p<q≤∞1 \leq p<q \leq \infty. Consider Lebesgue measure on Rn\mathbb {R}^{n}. Construct examples to show that neither Lp⊆LqL^{p} \subseteq L^{q} nor Lq⊆LpL^{q} \subseteq L^{p} holds.

(b)

Next, suppose that 1≤p<r<q<∞1 \leq p<r<q<\infty. Show that Lp∩Lq⊆LrL^{p} \cap L^{q} \subseteq L^{r}. (This is for any measure space (X,R,μ)(X, \mathcal{R}, \mu ).)

(c)

Now assume that XX is a finite measure space, i.e., that XX is measurable and that μ(X)<∞\mu (X)<\infty. Prove that Lq⊆LpL^{q} \subseteq L^{p} for 1≤p<q<∞1 \leq p<q<\infty, and give an example to show that this is a proper inclusion. Now prove that

∥f∥1≤∥f∥p≤∥f∥q≤∥f∥∞,1≤p<q<∞ \left\| f\right\| _{1} \leq \left\| f\right\| _{p} \leq \left\| f\right\| _{q} \leq \left\| f\right\| _{\infty }, \quad 1 \leq p<q<\infty

whenever μ\mu is a probability measure.

(d)

Prove that ℓp⊆ℓq\ell^{p} \subseteq \ell^{q} for 1≤p<q≤∞1 \leq p<q \leq \infty, and give an example to show that this is a proper inclusion. Now prove that

∥f∥∞≤∥f∥q≤∥f∥p≤∥f∥1,1≤p<q<∞ \left\| f\right\| _{\infty } \leq \left\| f\right\| _{q} \leq \left\| f\right\| _{p} \leq \left\| f\right\| _{1}, \quad 1 \leq p<q<\infty
Problem 3.6.7

Show that for each f∈L∞,∣f(x)∣≤∥f∥∞f \in L^{\infty },\left|f(x)\right| \leq \left\| f\right\|_{\infty } almost everywhere.

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Problem 3.6.8

Let II be an interval in R\mathbb {R}.

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(a)

Assume that 1≤p<∞1 \leq p<\infty. Prove that C(I)C(I) is dense in Lp(m)L^{p}(m) whenever II is closed and bounded.

(b)

Now drop the assumption that II is closed and bounded. We certainly cannot expect C(I)C(I) to be dense in LP(m)L^{P}(m), since it is not even contained in it. We define the "continuous functions with compact support" on X⊆RX \subseteq \mathbb {R} to be

Cc(X)={f∣f∈C(X) and {x∈X∣f(x)≠0}‾ is compact } C_{c}(X)=\left\{ f \mid f \in C(X) \text{ and } \overline{\left\{ x \in X \mid f(x) \neq 0\right\} } \text{ is compact }\right\}

The set {x∈X∣f(x)≠0}‾\overline{\left\{ x \in X \mid f(x) \neq 0\right\} } is often called the "support" of ff. Again, assume that 1≤p<∞1 \leq p<\infty. Prove that Cc(I)C_{c}(I) is dense in Lp(m)L^{p}(m). (Note that this result subsumes the result of (a).)

(c)

Prove that Cc(I)C_{c}(I) is not dense in L∞(m)L^{\infty }(m).

(The results of Exercise 8 give an alternative proof of Theorem 4.7, and give a way to define LpL^{p} that is independent of measure theory. See the paragraph following the proof of Theorem 4.7.)

Problem 3.6.9

The point of this exercise is to show that the norm on LpL^{p}, for p≠2p \neq 2, does not come from an inner product. We start by considering Lp([−1,1])L^{p}([-1,1]), with Lebesgue measure. Set f(x)=1+xf(x)=1+x and g(x)=1−xg(x)=1-x.

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(a)

Show that

∥f∥pp=2p+1p+1=∥g∥pp∥f+g∥pp=2p+1∥f−g∥pp=2p+1p+1 \begin{gathered} \left\| f\right\| _{p}^{p}=\frac{2^{p+1}}{p+1}=\left\| g\right\| _{p}^{p} \\ \left\| f+g\right\| _{p}^{p}=2^{p+1} \\ \left\| f-g\right\| _{p}^{p}=\frac{2^{p+1}}{p+1} \end{gathered}
(b)

Using part (a), show that the parallelogram equality asserts that

(p+1)2p=3 (p+1)^{\frac{2}{p}}=3

Verify that this equality holds for p=2p=2.

(c)

Prove that the parallelogram equality does not hold for values p≠2p \neq 2. Hint: Show that the function (p+1)2p−3(p+1)^{\frac{2}{p}}-3 is a strictly decreasing function of p≥1p \geq 1 and thus takes on the value zero for at most one value of pp.

(d)

Modify the functions given in part (a) to prove the result for Lp(I)L^{p}(I) where II is any interval, bounded or unbounded, in R\mathbb {R}.