Bounded linear functionals on $L^p(\mathbb{R})$, $0

Question

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!

Answer 1

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

Theorem (Lyapunov-Sacks): If $(X,\mathscr{F},\mu)$ is nonatomic, then for any $E\in\mathscr{F}$ with $0<\mu(E)<\infty$, and any $0<\alpha<\mu(E)$, there exists $A\in\mathscr{F}$ such that $A\subset E$, and $\mu(A)=\alpha$.

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Our first result concerns nonatomic measures.

Theorem 1: If $(X, \mathscr{F}, \mu)$ is a nonatomic measure space and $0 < p < 1$, then $(L_p(\mu))^* = \{0\}$.

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:

Lemma 2: If $(X,\mathscr{F},\mu)$ is a notatomic measure and $f\in L_1(\mu)$, then $\mu_f(A):=\int_A f\,d\mu$ defines a non atomic measure on $\mathscr{F}$.

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

1. $d(f_n,0)=\int f^p_n=2^{p-1}\int f^p_{n-1}\,d\mu=2^{n(p-1)}\int f^p\,d\mu$, $n\geq1$, and
2. $\phi(f_n)\geq1$

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, $\phi(f)=0$.

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For a measure $\mu$ that admits an atom, we have the following result:

Theorem 3: If the measure $\mu$ contains an atom with finite measure, then $(L_p(\mu))^*\neq \{0\}$.

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 $L_p(\mu)$.

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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.

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A nice treatment of the properties of $L_p$ with $0<p<1$, one may take a look at

1. Conrad, K. $L_p$-Spaces for $0<p<1$. Link here
2. M. M. Day, The spaces $L_p$ with $0 < p < 1$, Bull. Amer. Math. Soc. 46 (1940), 816–823.

Answer 2

To complement the accepted answer, I would like to mention that two more general pathologies about $L^p(\mu)=L^p(X,\mathcal{F},\mu)$ spaces with $0<p<1$ occur when the measure $\mu$ is non atomic. The first concerns to $L^p(\mu)$ itself, and the second concerns bounded linear transformations from $L_p(\mu)$ into a Frechet space.

A Frechet space $Y$ is a topological linear space whose topology is locally convex and generated by a complete translation invariant metric $d$: $d(x+z,y+z)=d(x,y)$ for all $x,y,z\in Y$. For example, Banach spaces are Frechet spaces.

Suppose then that $\mu$ is a non atomic measure on some measure space $(X,\mathcal{F})$.

The first pathology is:

Proposition 1: $L^p(X,\mathcal{F},\mu)$ with $0< p <1$ is not locally convex.

The second pathology is

Proposition 2: If $Y$ is a Frechet space and $T:L^p(X,\mathcal{F},\mu)\rightarrow Y$ is a bounded continuous transformation, then $T=0$.

Observe that the problem in the OP corresponds to the case where $Y=\mathbb{R}$ with the usual topology.

It turns out that Proposition 2 follows easily from Proposition 1. Supposed $T$ is as in the hypothesis of proposition 2. Let $B(0;\varepsilon)$ the ball of radius $\varepsilon$ in $Y$ and let $W\subset B(0;\varepsilon)$ be an open convex neighborhood of $0\in Y$. Then $V=T^{-1}(W)$ is a nonempty open convex neighborhood of $0\in L^p$. Proposition 1 implies that $V=L^p(X,\mathcal{F},\mu)$ which leads to $T(L^p)\subset B(0;\varepsilon)$. Taking $\varepsilon$ as small as we want gives that $T\equiv 0$.

The rest of this posting is dedicated to presenting a proof of Proposition 1. Taking a queue from the answer by Oliver Diaz, I adapt the case $L^p([0,1], \mathcal{B}([0,1]),m)$ discussed in Rudin's book on functional analysis to the present setting.

Suppose $V$ is an open convex neighborhood of $0$ in $L^p$ and let $f\in L^p$ and $f\neq0$. The conclusion of Proposition 1 will follow if we show that $f\in V$. Let $B(0;r)$ be a ball centered at $0\in L^p$ of radius $r>0$ such that $B(0;r)\subset V$. Choose $n$ large enough so that $n^{p-1}\int_X|f|^p\,d\mu <r$, which is possible since $0<p<1$. By Lemma 2 in Oliver's posting the measure $$\mu_f(A)=\int_A|f|^p\,d\mu\qquad A\in\mathcal{F}$$ is nonatomic. This implies that there are pairwise disjoint sets $A_1,\ldots,A_n$ such that $\int_{A_k}|f|^p \,d\mu =\frac1n\int_X|f|^p \,d\mu$. Define $g_k=n f \mathbf{1}_{A_k}$ for each $k=1,\ldots, n$.

Notice that $$d(0,g_k)=n^p\int_{A_k}|f|^p\,d\mu < n^{p-1}\int_X|f|^p\,d\mu < r$$ that is, each $g_k\in B(0;r)\subset V$. Then $$f=\sum^n_{k=1}\frac1n g_k\in V$$ by virtue of $V$ being convex. This concludes the proof of Proposition 1.