Dual space of $L^p([0,1])$ for $0 < p < 1$.

Question

I convinced myself that in the space $L^p([0,1])$ there is no other non-empty, open, convex subset different from $L^p([0,1])$ itself when $0 < p < 1$.

I was told that I can conclude from this statement that the topological dual of $L^p([0,1])$ if $\{0\}$, but I can not see it. I suppose that Hahn-Banach theorem would be useful here but I am not sure how.

Thanks in advance.

Answer 1

Any nontrivial element $\lambda\in X'\setminus\{0\}$ divides the space into the two halves $\{x\in X \mid \lambda(x)>0\}$ and $\{x\in X \mid \lambda(x)<0\}$ both of which are non-empty (because $\lambda$ is non-zero), open (because $\lambda$ is continuous), convex, and proper subsets of $X$. By your prior reasoning, no such sets exist.

Answer 2

Denote by $\mathbb D$ the open unit disk in the complex plane. Let $\varphi:L^p[0,1]\to\mathbb C$ be linear and continuous. Then $\varphi^{-1}(\mathbb D)$ is

• open, because $\varphi$ is continuous;

• nonempty, because it contains $0$;

• convex, because $\varphi$ is linear.

Then $\varphi^{-1}(\mathbb D)=L^p[0,1]$. This means that $|\varphi(f)|<1$ for all $f$. As $L^p[0,1]$ is a vector space, this implies that $\varphi=0$.