$(L^\infty)^*$ is not isomorphic to $L^1$ I am working on this problem in Rudin:
Let $L^\infty=L^\infty(m)$, where $m$ is Lebesgue measure on $I=[0,1]$. Show
  that there is a bounded linear functional $\lambda\neq 0$ on $L^\infty$ that
  is $0$ on $C(I)$, and that therefore there is no $g\in L^1(m)$ that satisfies
  $\lambda f=\int_I fg dm$ for every $f\in L^\infty$. Thus
  $(L^\infty)^*\neq L^1$.
I've figured out all but the part that saids "
$(L^\infty)^*\neq L^1$". Most of the problem seems to be showing that the normal method of showing $(L^p)^* \cong L^q$ doesn't work. However, I don't see how that helps me do the last deduction. 
 A: Every $g \in L^1$ yields a linear functional $\lambda_g \in (L^\infty)^\ast$ defined by $\lambda_g(f) = \int fg\,dm$ on $L^\infty$. 
The estimate $\lvert \lambda_g(f)\rvert \leq
\lVert f \rVert_\infty \lVert g\rVert_1$ shows that $\lVert \lambda_g\rVert \leq \lVert g\rVert_1$, so $\lambda_g$ is continuous. Taking
$$
f(x) =
\begin{cases} 0 & \text{if } g(x) = 0 \cr
\frac{\overline{g(x)}}{\lvert g(x)\rvert} & \text{if } g(x) \neq 0
\end{cases}
$$
we get $\lambda_g(f) = \lVert g\rVert_1$, so $\lVert \lambda_g\lVert \geq \lVert g \rVert_1$. Thus, the linear map $L^1 \to (L^\infty)^\ast, g \mapsto \lambda_g$ is isometric and hence injective. We can therefore view $L^1 \subseteq (L^\infty)^\ast$  as a subspace.
The point of Rudin's exercise is to establish that $L^1 \subsetneqq (L^\infty)^\ast$, in words $L^1$ is a proper subspace of $(L^\infty)^\ast$.
Indeed, let $0 \neq \lambda \in (L^\infty)^\ast$ be a linear functional such that $\lambda(f) = 0$ for all $f \in C(I)$ which you say you have proved to exist.
Then $\lambda$ can't be of the form $\lambda = \lambda_g$ for any $g \in L^1$ since for $g \in L^1$ the vanishing $\int gf\,dm = 0$ for all $f \in C(I)$ implies that $g = 0$ and hence $\lambda_g = 0$. But we have $\lambda \neq 0$ and thus $\lambda \in (L^\infty)^\ast \setminus L^1$.
