The existence of a bounded linear functional

Question

Prove the existence of a bounded linear functional $\Delta:L^{\infty} (\mathbb R) \rightarrow \mathbb{R}$ such that $||\Delta||=1$ and $\Delta f=f(0)$ for every $f\in L^{\infty}(\mathbb{R}) \cap C^0(\mathbb{R})$ Show however that there exists no function $g\in L^1(\mathbb{R})$ such that $\Delta f = \int f(x)g(x)dx$. Deduce that $(L^{\infty}(\mathbb{R}))^*$ cannot be identified with $L^1(\mathbb{R})$

Answer 1

Consider the linear functional on $L^\infty (\mathbb R) \cap C(\mathbb R)$ that sends $f \mapsto f(0)$. This linear functional is bounded: it has norm $1$, since $| f(0) | \leq \sup_{x \in \mathbb R} |f(x) |$ and since the inequality is saturated by $f = 1$.

By Hahn-Banach, you can extend this linear functional to a functional $\Delta$ on the whole of $L^\infty (\mathbb R)$, while preserving the property that $|| \Delta || = 1$.

The reason why you can't find a $g \in L^1 (\mathbb R)$ such that $\int fg = f(0)$ for all $f \in L^\infty (\mathbb R) \cap C(\mathbb R)$ is as follows. Consider this sequence of "triangle" functions, $$ f_n (x) = \begin{cases} 0 & x \leq -\frac 1 n \\ nx+1 & -\frac 1 n < x \leq 0 \\ -nx + 1 & 0 < x \leq \frac 1 n \\ 0 & x > \frac 1 n \end{cases}$$ If $g$ has the claimed property, then each $f_n$, we have $$ \int f_n g = 1.$$ This sequence of functions converges pointwise to the function, $$ f(x) = \begin{cases} 1 & x = 0 \\ 0 & x \neq 0 \end{cases}$$ By the dominated convergence theorem, $$ \int f g = \lim_{n \to \infty} \int f_n g = 1.$$ But $fg$ is zero almost everywhere, so $\int fg = 0$, and we have a contradiction.

This shows that you can't identify $L^1(\mathbb R)$ with $L^\infty(\mathbb R)^\star$ by identifying $g \in L^1(\mathbb R)$ with the linear functional $ f \mapsto \int fg $ on $L^\infty(\mathbb R)$. However, I don't think this kind of argument proves that $L^1(\mathbb R)$ can't be identified with $L^\infty(\mathbb R)^\star$ in any other way...