What is the dual of a sum of Banach function spaces Given two Banach function spaces, e.g. $L^1(\mathbb{R})$ and $L^2(\mathbb{R})$, define the sum $X=L^1+L^2$ as the class of functions $f=f_1+f_2$ having $f_i\in L^i$. Define a norm by
$$\|f\|_X=\inf\{\|f_1\|_{L^1}+\|f_2\|_{L^2}; f=f_1+f_2\} $$
where the infimum is taken over all decompositions $f=f_1+f_2$. Is there any (simple) way to see what the dual of $X$ is? As far as I can see it must be a subset of $L^\infty\cap L^2$. Could it be that this is actually the dual space? 
 A: Sorry that I post another answer, but I cannot comment. As pointed out by GEdgar in the comments there is a problem with the above answer as the dual of $L^\infty$ is not $L^1$. In fact the authors of [1] use the notation $X'$ for the space associated to $X$, where the dual space is written as $X^*$. However the answer seems to be given in [1], Chapter 3, Exercise 3:

Let $(X_0,X_1) $ be a conjugate couple of complex Banach spaces. Then 
  $(X_0 \cap X_1)^* $ is isometrically isomorphic to $X_0^* +X_1^*$. 
  (HINT: Embed $X_0\cap X_1$ as a closed subspace of the direct sum 
  $X_0\oplus X_1$ and use the Hahn-Banach theorem).

I still have a problem with this, as this seems to require that the spaces are complex, but I don't know why. The proof seems to work without this assumption. Or is it just meant as an addition (i.e. that this works even for complex spaces)? In particular $L^p(\mathbb{R})$ is not complex so for a complete answer this needs to be clarified.
EDIT: Just noticed, that what I posted is exactly the wrong direction. So the actual question is answered by [1], Chapter 3, Exercise 2:

Let $(X_0,X_1) $ be a conjugate couple of Banach spaces. Then 
  $(X_0 + X_1)^* $ is isometrically isomorphic to $X_0^* \cap X_1^*$. 
  (HINT: Bounded linear functionals on $X_0, X_1$ and $X_0 + X_1$ may be identified with their restrictions to $X_0 \cap X_1$).

So there complexity is not required. But I am still wondering whether complexity is actually necessary for the other statement.
[1] Bennett, Colin, and Robert C. Sharpley. Interpolation of operators. Vol. 129. Academic press, 1988.
A: There is a problem with this answer. See comment below and the other answer.
From [1], Chapter 3,  Exercise 6:

If $1\leq p_0,p_1\leq \infty$, the complex spaces $L^{p_0}\cap L^{p_1}$ and $L^{p_0}+ L^{p_1}$ are rearrangement-invariant Banach functions spaces and
  $$
 (L^{p_s0}\cap L^{p_1})' = L^{p'_0}+ L^{p'_1};\quad
 (L^{p_0}+ L^{p_1})' = L^{p'_0}\cap L^{p'_1}, 
$$
  isometrically.

In your case, this indeed implies 
$
(L^{1}+ L^{2})' =L^{\infty}\cap L^{2}.
$

[1] Bennett, Colin, and Robert C. Sharpley. Interpolation of operators. Vol. 129. Academic press, 1988.
A: See also: Theorem 2.7.1 in [2] for a relatively straightforward proof of the relation for the case of spaces $X$ and $Y$ such that ($X$,$Y$) is a conjugate couple in the sense of [1] (i.e. that $X\cap Y$ is dense in both $X$ and $Y$). This is the context of [3], Chapter 3, Exercise 2 as previously discussed above.
[1] Bennett, Colin. Banach Function Spaces and Interpolation Methods. I: The abstract Theory. J Funct. Anal. 17:4, 1974. pp 409-440. doi:10.1016/0022-1236(74)90050-0
[2] Bergh, Joran and Lofstrom, Jorgen.  Interpolation Spaces, An Introduction. Springer-Verlag. 1976
[3] Bennett, Colin, and Robert C. Sharpley. Interpolation of operators. Vol. 129. Academic press, 1988.
