Previously asked on this site: for $p\in(0,1)$, there are no bounded linear functionals on $L^p(\mathbb{R})$. I want to follow-up about what happens if we consider a general measure $\mu$; I do not see how to generalize the given solution to that problem.
The question, which is exercise 1.2 in Stein and Shakarchi's Functional Analysis is as follows:
Consider $L^p(\mathbb{R})$ where $0<p<1$. Prove that there are no bounded linear functional on $L^p(\mathbb{R})$. That is, prove that if a $l$ linear functional $l:L^p(\mathbb{R})\to \mathbb{C}$ is such that $|l(f)|\leq M|f|_{p}$ for all $f$ and some $M>0$, then $l=0$. Hint: Let $F(x)=l(\mathcal{X}_{[0,x]})$, and consider $F(x)-F(y)$.
Here, of course, I do not want to assume that the measure for $L^p(\mathbb{R})$ is Lebesgue.
I tried to adapt the proof given in the other post, but could not get it to work out. Any help would be greatly appreciated!
Best Answer
Here I present two results that address the question the OP is considering. Key to this is the notion of atomic (nonatomic) measures.
Recall that if $(X,\mathscr{F},\mu)$ is a measure space, then a set $A\in\mathscr{F}$ is an atom if $\mu(B)\in\{0,\mu(A)\}$ for all $B\in \mathscr{F}$ with $B\subset A$. A measure $\mu$ is nonatomic is there are not atoms.
We have the following result
Our first result concerns nonatomic measures.
The particular case of $(X,\mathscr{F},\mu)=([0,1],\mathscr{B}([0,1]),\lambda)$, where $\lambda$ is Lebesgue's measure is discussed in many textbooks, for example Rudin W. Functional Analysis, 2ns Edition, McGraw Hill 1968. The ideas from this example can be carried out to the setting of nonatomic measures almost verbatim with the aid of the following result:
Proof for Theorem 1: Suppose $\phi\in (L_p(\mu))^*$, and choose $f\in L_p(\mu)$. We will show that $\phi(f)=0$. Suppose $\phi(f)\neq0$. Then either $\phi(f_+)$ or $\phi(f_-)$ is positive. This means that we may assume without loss of generality that $f\geq0$ and $\phi(f)\geq1$. By Lemma 2, the measure $A\mapsto \int_Af^p\,d\mu$ is nonatomic. Hence, there exists $A_1\in\mathscr{F}$ such that $\int_{A_1}f^p\,d\mu=\frac12\int f^p>0$. Let $g_1=f\mathbb{1}_{A_1}$ and $g_2=f-g_1=f\mathbb{1}_{A_2}$ where $A_2=X\setminus A_1$. Since $\phi(f)\geq1$, then there is $i_1\in\{1,2\}$ for which $\phi(g_{i_1})\geq\frac12$. Define $f_1=2g_{i_1}$. Then $\phi(f_1)\geq1$ and $$ d(f_1,0)=2^p\int g_1^p\,d\mu=2^{p-1}\int f^p\,d\mu$$ Iterating this argument, we obtain a sequence $(f_n:n\in\mathbb{Z}_+)$ such that
As $0<p<1$, $\lim_nd(f_n,0)=0$, and so, by the continuity of $\phi$, $\lim_n\phi(f_n)=0$, which is not possible by (2). Consequently, $\phi(f)=0$.
For a measure $\mu$ that admits an atom, we have the following result:
Proof of Theorem 3: Let $B $ be an atom with finite measure. Every measurable function $f : X \rightarrow\mathbb{R}$ is constant almost everywhere on $B$. Denote the $\mu$ almost-everywhere common value of $f$ on $B$ by $\phi(f)$. It is left to the OP to check that $\phi$ is a non-zero continuous bounded linear functional on $L_p(\mu)$.
Proof of Lemma 2: Suppose $\mu$ is a nonatomic measure. Let $f\in L_1^+(\mu)\setminus\{0\}$ and define $\mu_f(dx)=f(x)\cdot\mu(dx)$. We argue by contradiction by asuming there is atom $A\in\mathscr{F}$ for $\mu_f$. Then, there is $A\in\mathscr{F}$ with $\mu_f(A)>0$ such that $\mu_f(B)\in \{0,\mu_f(A)\}$ for all $B\in\mathscr{F}$ with $B\subset A$. Without loss of generality, suppose that $f\mathbb{1}_{\Omega\setminus A}=0$. Let $$s=\sum^n_{j=1}a_j\mathbb{1}_{A_j}$$ be any nonnegative simple dominated by $f$ and such that $0<\mu(A_j)<\infty$ and $a_j>0$. Fix $0<\varepsilon<\max_{1\leq j\leq n}\mu(A_j)$ and choose $\delta>0$ such that $\mu(B)<\delta$ implies $\mu_f(B)<\varepsilon$. Since $\mu$ is nonatomic, for each $1\leq j\leq n$, there is $B_j\in\mathscr{F}$ with $B_j\subset A_j$ such that $0<\mu(B_j)<\delta$. Hence $\mu_f(B_j)<\varepsilon$. As $A$ is an atom, it follows that $\mu_f(B_j)=0$. Since $\mu_f(\mathbb{1}_{B_j}s)=a_j\mu(B_j)=0$, we conclude that $a_j=0$. As $f$ is the monotone limit of simple functions, it follows that $f=0$ $\mu$-a.s. which contradicts $\|f\|_1>0$. This means that $\mu_f$ is nonatomic.
A nice treatment of the properties of $L_p$ with $0<p<1$, one may take a look at