functional-analysislp-spacesmeasure-theory
Consider $L^p(\mathbb{R})$ where $0<p<1$.
Why there are no bounded linear functional on $L^p(\mathbb{R})$?
i.e.
If $l$ is linear functional $l:L^p(\mathbb{R})\to \mathbb{C}$ such that $|l(f)|\leq M|f|_{p}$ for all $f$ and some $M>0$, then $l=0$.
Why?
Hint: Let $F(x)=l(\mathcal{X}_{[0,x]})$, and consider $F(x)-F(y)$
Suppose $A$ is positive, then $${\big\|}\,\|A\|e_\mathcal{A}-A\,\big\|≤\|A\|,$$ which you can see by considering the spectrum of $A$. Specifically $\sigma(A)$ consists only of numbers $≥0$ and we have $\|A\|=\sup_{\lambda\in\sigma(A)}|\lambda|$ for normal elements.
If you assume $\varphi(A)<0$ then $\varphi(\|A\|e_\mathcal{A}-A)=\|A\|-\varphi(A)>\|A\|$, contradicting $\|\varphi\|=1$.
So you see that for any positive $A$, $\varphi(A)$ must also be positive or admit a non-vanishing imaginary component. But positive elements are self-adjoint and here your hint tells you that $\varphi(A)$ can never have non-vanishing imaginary component if $A$ is positive.
To see how the hint can be proven, suppose $A$ is self-adjoint with $\varphi(A)= x+ iy$ with $y\neq0$. Note that for real $\lambda$ $$\|A+i\lambda\, e_{\mathcal A}\|= \sqrt{\|A\|^2+\lambda^2}$$ and that $$|\varphi(A+i\lambda\, e_\mathcal{A})|=\sqrt{x^2+(\lambda+y)^2}.$$ It follows that $$\frac{\varphi(A+i\lambda\,e_\mathcal{A})}{\|A+i\lambda\,e_\mathcal{A}\|}=\sqrt{\frac{x^2+(\lambda+y)^2}{\|A\|^2+\lambda^2}}.$$ If you do some analysis you will find that for very large $\lambda$ this is roughly $\left|\mathrm{sign}(\lambda)+\frac y\lambda\right|$, once again contradicting $\|\varphi\|=1$.
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 $\mu(A)=\alpha$.
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
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:
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 $L_p(\mu)$.
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
Best Answer
This question surely was already answered here. In the case $0 <p <1$ the usual definition of the $L^p$-norm is not a norm, because the $\Delta$-inequality isn't satisfied. If $0< p <1$ the space $L^p(\mathbb{R})$ is only a metric space with invariant metric given by $$\|f\|_p := \int |f|^p d \lambda.$$ If $l$ is a continuous functional, we know that $|l(f)| < 1$ for $f \in B_\delta(0)$ and some $\delta >0$. By rescaling and using linearity, we see that $$|l(f)| \le M \|f\|_p^{1/p}$$ with $M= \delta^{-1}$. Now decompose $[a,b]$ in $n$ disjoint intervals $I_1,\ldots,I_n$ with lenght $(b-a)/n$. Then we have $$|l(1_{[a,b]})| \le \sum_{k=1}^n |l(1_{I_i})| \le nM \|l(1_{I_i})\|_p = (b-a)^{1/p} n^{1-1/p}.$$ Since $0<p<1$, we see that $1-1/p$ is negative. Thus, letting $n \rightarrow \infty$, we get $l(1_{[a,b]}) =0$. Since all linear combinations of such intervals are dense in $L^p$ (also for $0<p<1$), we see that $l$ is trivial.