# Finding bounded linear functionals on $L^{\infty}(\mathbb{R})$

I'm trying to prove the next two propositions:

a) There is a nonzero bounded linear functional on $L^{\infty}(\mathbb{R})$ which vanishes on $C(\mathbb{R}).$

b) There is a bounded linear functional $\lambda$ on $L^{\infty}(\mathbb{R})$ such that $\lambda(f)=f(0)$ for each $f\in C(\mathbb{R}),$

where $C(\mathbb{R})$ is the space of all bounded real valued continuous functions on $\mathbb{R}$ equipped with the sup norm .

I'm stuck proving this. I guess Hahn-Banach could solve this:

Part b) with $g:C(\mathbb{R})\rightarrow\mathbb{C}$ defined by $g(f)=f(0).$ Such map is linear and I think is bounded. If this were the case, a Corollary of Hahn-Banach solves this.

With part a) I don't get any useful.

How to prove this?

• Are you defining $C(\mathbb{R})$ to consist of bounded continuous functions on $\mathbb{R}$? If not, then $C(\mathbb{R})$ is not contained in $L^{\infty}(\mathbb{R})$. – DisintegratingByParts Nov 4 '17 at 4:38
• Yes @DisintegratingByParts. $C(\mathbb{R})$ is the space of all real valued continuous functions on $\mathbb{R}$ equipped with the sup norm. . – Squird37 Nov 4 '17 at 5:58
• Not all functions in $C(\mathbb{R})$ are bounded, which is a problem for the sup norm. Perhaps you mean bounded continuous functions on $\mathbb{R}$? – DisintegratingByParts Nov 4 '17 at 14:22
• Part a) can be solved as follows: Find a space $S$ with $C(\mathbb R) \subsetneq S \subset L^\infty(\mathbb R)$. Then, define a nonzero functional on $S$ which vanishes on $C(\mathbb R)$. Extend by Hahn-Banach. – gerw Nov 4 '17 at 18:24
• Thanks @gerw. I'm working with your idea with $S$ the space of bounded functions and defining $T:S\rightarrow\mathbb{C}$ by $T(f)=f$ if $f\in B(\mathbb{R})/C(\mathbb{R})$ and zero in other case, with the property $T(f+g)=f$ if $f\in C(\mathbb{R}),$ but I'm not sure if this functional works because practically I'm asking the condition to be linear. – Squird37 Nov 4 '17 at 21:40

## 1 Answer

For (a) you could take the space of $L^\infty$ functions that have (a representive with) left and right limits at zero, define $L(f)$ to be the right limit minus the left limit on this space, and extend by Hahn-Banach. This will be nonzero on the Heaviside step function.