"Duality" for weak $L^p$ spaces Let $1<p<\infty$. Denote by $L^{p,\infty}$ the weak $L^p$ space in $\mathbb{R}^n$ and let $f\in L^{p,\infty}$ where we define the weak $L^p$ quasinorm as
$$\|f\|_{p,\infty} = \sup_{\lambda >0} \lambda\cdot m(\{ |f|>\lambda \})^{1/p}$$
where $m$ denotes the Lebesgue measure on $\mathbb{R}^n$.

Question: Is it true that
  $$\|f\|_{p,\infty}= \sup_{E} |E|^{-1/p^\prime} |\langle f,1_E\rangle|$$
  where the supremum goes over all measurable sets $E$ of finite measure, $1_E$ denotes the characteristic function of $E$ and $\langle f,g\rangle=\int fg$ and $\frac{1}{p}+\frac{1}{p^\prime}=1$.

I would already be very happy with "$\le$" and I also don't care if "$\le$,$\ge$" are maybe only true with additional multiplicative constants.
To show "$\le$" I tried plugging in $E=\{|f|>\lambda\}$ for $\lambda>0$, but then I get
$$|\langle f,1_E\rangle|=\left| \int_{\{|f|>\lambda\}} f \right|$$
The idea was to estimate this against $\lambda\cdot m(\{ |f|>\lambda\})$, but that doesn't work because the absolute value signs are outside instead of inside the integral. So I tried assuming that $f$ is positive, then its fine, but the application I have in mind needs $f$ complex-valued.
Remark:
The question is motivated by the fact that for (normal) $L^p$ spaces we have the duality statement
$$\|f\|_p = \sup_{g\in L^{p^\prime}, g\not\equiv 0} \frac{|\langle f,g\rangle|}{\|g\|_{p^\prime}}$$
 A: First of all, there is no reason to expect $$\|f\|_{p,\infty}= \sup_{E} |E|^{-1/p^\prime} |\langle f,1_E\rangle|\tag{0}$$ to hold as equality. For example, $f(t)=t^{-1/2}\chi_{\{t>0\}}$  is in weak $L^2$, with  $\|f\|_{2,\infty}=1$. On the other hand, $\sup_{E} |E|^{-1/2} |\langle f,1_E\rangle| \ge \int_0^1 f(t)\,dt = 2$.
Proof of $\le $ in (0), with a multiplicative constant. Write $f = (f_1-f_2)+i(f_3-f_4)$ where all four functions are nonnegative. The quasinorm of $f$ is  bounded by a multiple of the maximum of quasinorms  of $f_k$. 
Also, for every $k=1,\dots,4$ we have 
$$\sup_{E} |E|^{-1/p^\prime} |\langle f_k,1_E\rangle|\le \sup_{E} |E|^{-1/p^\prime} |\langle f,1_E\rangle|\tag{1}$$
by specializing the supremum on the right to the subsets of $\{f_k\ne 0\}$. (For example, take $k=1$. Then $f_2$ vanishes on the set $\{f_1\ne 0\}$, and the imaginary components only increase the right hand side of (1).)
Basically, the above says that $\sup_{E} |E|^{-\alpha} |\langle f,1_E\rangle|$ is comparable to $\sup_{E} |E|^{-\alpha}  \langle |f|,1_E\rangle$. This has nothing to do with weak $L^p$ spaces.
Proof of $\ge $ in (0), with a multiplicative constant. Since both sides of (0) are degree-1 homogeneous in $f$, we 
can make $\|f\|_{p,\infty} = 1$ by dividing $f$ by $\|f\|_{p,\infty}$. Then
$$\begin{split}
\int_E |f|&\le \sum_{k=-\infty}^\infty 2^{k+1} |\{x\in E: 2^k \le |f(x)|\le 2^{k+1}\}| \\
&\le  \sum_{k=-\infty}^\infty 2^{k+1} \min( 2^{-pk}, |E|)  \end{split}
\tag{2}$$
The right  hand side of (2) is a decaying geometric series starting with the first term where $2^{-pk}<|E|$; up to then it's a growing geometric series. Therefore, the sum is comparable to this first term, which is about 
$|E|^{-1/p}|E| = |E|^{1/p'}$, as desired. 
