Show that $L^1$ is strictly contained in $(L^\infty)^*$

Question

How does one show that $L^1$ is strictly contained in $(L^\infty)^*$? Here, $(L^\infty)^*$ is the space of linear continuous functionals on $L^\infty$.

Answer 1

If the domain of our functions is $\mathbb{R}$, you can show in fact that there are elements of $(L^\infty)^\ast$ which are not in $L^1$, but given every $g\in L^1$ we can associate it with a linear functional $\psi_g:L^\infty\to \mathbb{R}$ by $\psi_g(f)=\int fgdx$, so in fact we have $L^1\subset (L^\infty)^\ast$, and the containment is strict.

To see that it is strict, first note that $C_b(\mathbb{R})$ of bounded continuous functions is a subspace of $L^\infty(\mathbb{R})$, and we can define a a continuous linear functional $\phi:C_b(\mathbb{R})\to \mathbb{R}$ on it by $\phi(f)=f(0)$. The Han-Banach theorem lets us then extend $\phi$ to a linear functional $\tilde{\phi}:L^\infty(\mathbb{R})\to\mathbb{R}$, which when restricted to $C_b(\mathbb{R})$ is still $\phi$. I don't know any good way to write this extension down as an actual formula, as its existence is guaranteed by Hahn-Banach but not constructively.

So, we have found a rather strange element of $(L^\infty)^\ast$, and it turns out that this element does not arise from an element of $L^1$. To see this, note that if $\tilde{\phi}$ did arise in this way there would be some $g\in L^1$ such that $\tilde{\phi}(f)=\int_\mathbb{R}fg dx$, and in particular if we restrict so that $f\in C_b(\mathbb{R})$ we know that $\tilde{\phi}\mid_{C_b}=\phi$, and so we have $$\tilde{\phi}(f)=\phi(f)=\int_\mathbb{R}f(x)g(x)dx=f(0)$$ for all $f$ in $C_b(\mathbb{R})$. But, there is no $g\in L^1(\mathbb{R})$ which satisfies this, so $\tilde{\phi}$ is in $(L^\infty)^\ast$ but not in $L^1$.