# How to show that $L_p(\mu)_+$ is not solid (i.e., $L_p(\mu)_+$ has empty interior)?

Let $(\Omega, \Sigma, \mu)$ be a probability space.

Let $L_p (\mu)$ with fixed $p \geq 1$ be defined to be the collection of all ($\mu$- equivalence classes of) Borel measurable functions $f \colon \Omega \to \mathbb{R}$ for which $\int_{\Omega} |f|^{p} \mathrm{d} \mu < \infty$. Denote by $L_p (\mu)_+$ the positive cone of $L_p (\mu)$.

We say that a subset $E$ of an ordered Banach space (or a Banach lattice) is solid if $E$ contains at least one interior point.

When I read some textbook, it states that "the positive cone $L_p(\mu)_+$ is not solid". Therefore, I'm struggling to figure out how to verify this statement and is this statement always true for any $p \geq 1$? Could anyone help me out please?

If it is convenient to set $\Omega = \mathbb{R}$, please feel free to do it then.

Sincerely appreciate your comment or idea in advance!

• Probably a good idea to define positive cone. – zhw. May 12 '17 at 3:38
• Do you mean the functions that are ae. non negative? – copper.hat May 12 '17 at 4:38
• @copper.hat For the case of the positive cone of $L_p$ space, the functions are non-negative a.e.. While in the case for $bc(\Omega)_+$, the functions are non-negative. – Paradiesvogel May 12 '17 at 4:56
• You need more conditions. If you take $\Omega = \{1,...,n\}$ and $\mu$ the uniform measure then the positive cone has a non empty interior. – copper.hat May 12 '17 at 5:05
• Thanks @copper.hat. What if we take $\Omega = \mathbb{R}$ and $\mu$ be some Gaussian distribution, then does $L_p(\mu)_+$ have a non-empty interior? Could you give me some idea or hint to show whether it contains a non-empty interior? Many thanks again – Paradiesvogel May 12 '17 at 5:15

## 1 Answer

Let assume $\Omega = N$,sequence of natural number, then we are dealing with ${l^p}$. Let $x=(x_n)_{n=1}^{\infty}$ with $x_n > 0$ for all $n$ is an interior point. And take $\lambda >0$ such that $x+y \in l^P_+$ for all $|| y || = \lambda$. There is such a positive $\lambda$ since $x$ is an interior point of $l^p_+$. Let $n$ is large enough such that $x_n < \lambda$. Now define $a= -\lambda e_n$ then $||a|| = \lambda$ but $x+a \notin l^P_+$ since its n-th component is negative! which is a contradiction .