The dual space of $L^1[0,1]$ Is a proof known that the dual space of $L^1[0,1]$ coincides with $L^\infty[0,1]$ that does not use the Radon-Nikodym theorem? May be  a proof that uses only Fourier analysis?
 A: The book Sobolev Spaces by Robert Adams and John Fournier proves this using a different approach in chapter 2 section "The norm dual of $L^p(\Omega)$.
A: For this specific case of $\Phi \in (L^1)^*$, consider
$$
    \omega(x) = \Phi(\chi_{[0,x]}).
$$
The function $\omega$ is continuous because $x\mapsto\chi_{[0,x]}$ is a continuous function from $[0,1]$ into $L^1$ by the dominated convergence theorem. $\omega$ is absolutely continuous because, for disjoint intervals $(x_{n-1},x_n]$, one has
\begin{align}
    \sum_{n=1}^{N}|\omega(x_n)-\omega(x_{n-1})|&=\sum_{n=1}^{N}e^{i\theta_n}(\omega(x_n)-\omega(x_{n-1})) \\
   &= \left|\Phi\left(\sum_{n=1}^{N}e^{i\theta_n}\chi_{(x_{n-1},x_n]}\right)\right| \\
  &\le \|\Phi\|\|\sum_{n=1}^{N}e^{i\theta_n}\chi_{(x_{n-1},x_n]}\|_{{L^1}} \\
  & \le \|\Phi\|\sum_{n=1}^{N}|x_n-x_{n-1}|
\end{align}
So $w(x)=\int_{0}^{x}\rho(t)dt$ for a function $\rho$ is that is uniformly bounded by $1$. Therefore, if $f\in C[0,1]$,
\begin{align}
       \Phi(f)&= \lim_{n\rightarrow\infty}\Phi\left(\sum_{k=1}^{n}f(k/n)\chi_{[(k-1)/n,k/n]}\right) \\
   &= \lim_{n\rightarrow\infty}\sum_{k=1}^{N}f(k/n)\int_{k/n-1/n}^{k/n}\rho(t)dt \\
   &= \lim_{n\rightarrow\infty}\int_{a}^{b}\left(\sum_{k=1}^{n}f(k/n)\chi_{[k/n-1/n,k/n]}(t)\right) \rho(t)dt \\
   &= \int_a^b f(t)\rho(t)dt
\end{align}
Now a limiting argument allows you to extend the above to $f\in L^1$ by the density of $C[0,1]$ in $L^1[0,1]$, and by the boundedness of $\rho$.
This is a weaker form of the conventional Radon-Nikodym argument that may be applied in this special case, but it is very obviously related.
