For $f\in L^\infty(\mathbb{R})$, can I find the $f^*\in (L^\infty(\mathbb{R}))^*$ such that $$\|f^*\|_* = 1 \text{ and } \langle f^*, f \rangle = \|f\|_\infty.$$
I know that $(L^\infty(\mathbb{R}))^*$ is the space of bounded finitely additive measures that are absolute continuous with respect to the Lebesgue measure on $\mathbb{R}$. So I need to find a $\lambda\in ba(\mathbb{R})$ such that $$\int_\mathbb{R} f d\lambda = \|f\|_\infty \quad \text{ and } \quad \|\lambda\|_{var}(\mathbb{R}) = 1.$$
Is this a well known result, specific for $L^\infty$?
I stared looking at simple functions, for $$f = \sum_{i=1}^n c_i \chi_{A_i}$$ As Martini said, we could define our $$\lambda(A) = \frac{1}{\mu(A_i)} \lambda (A\cap A_I)$$ for $i$ such that $|c_i|$ is maximal.
But then I am stuck at the next step again when generalizing this to continuous functions because when approximate a $L^\infty$ function with simple functions, the measure of the maximal set $A_i$ might go to zero.
The answer is yes. This follows from the Hahn-Banach theorem.
Indeed, for each non-zero element $x$ in a Banach space $X$, there is a norm-one functional $f\in X^*$ such that $\langle f,x\rangle = \|x\|$. To see this, consider the one-dimensional subspace spanned by $x$. Let $\langle f,cx\rangle = c\|x\|$ ($c$ is a scalar). Then $f$ is a norm-one functional that can be extended to the whole of $X$ in a norm-preserving way.