Understanding dual normed spaces: Is $L^\infty$ exactly $(L^1)^*$ or just a "proxy" to $(L^1)^*$? My issue can be best described using a concrete example as below.
As we know, $L^{\infty}$ is dual to $L^1$ in the following sense:
For any continuous linear functional $x^*$ on $L^1[0,1]$, i.e. $x^*\in (L^1)^*$, there exists a unique $u \in L^{\infty}[0,1]$ such that $\langle x,x^*\rangle=\int_0^1 x(t)u(t)dt$ for all $x \in L^1[0,1]$. Moreoever, $||x^*||=||u||_{L^\infty}$. Conversely, any $u \in L^\infty[0,1]$ determines a unique $x^* \in (L^1)^*$ in the above way.
I have no problem in understanding the above result. I see that it tells such an important fact: There is a one-to-one, onto, and norm preserving map from $(L^1)^*$ to $L^\infty$. However, I am not comfortable with the saying (which appears to be popular in many authors) that "The dual of $L^1$ is $L^\infty$". I think $(L^1)^*$ and $L^\infty$ are not the same space, rigorously speaking.
[I admit that because of the existence of the one-to-one, onto, and norm preserving map from $(L^1)^*$ to $L^\infty$, when we attack practical problems (e.g. optimization under normed spaces) using duality theories, we can apply $L^\infty$ as if it were really $(L^1)^*$. Having said this, I think $L^\infty$ can be viewed as a "proxy" to $(L^1)^*$, so "proxy" that to work with $L^\infty$ is the same as to work with $(L^1)^*$, in many pratical problems.]
Could anybody kindly help me by commenting on my understanding above? Thank you!
 A: You are right that they are not really the same objects. But the mapping
$$I : L^\infty \to (L^1)^*,(I(f))(g)=\int f(x) g(x) dx$$
is an isomorphism of Banach spaces. Its inverse $I^{-1} : (L^1)^* \to L^\infty$ is given by the equality
$$(I^{-1}(\phi))(x)=\lim_{r \to 0^+} \frac{1}{m(B(x,r))} \phi \left ( 1_{B(x,r)} \right )$$
which holds almost everywhere for any particular representative. Here $B(x,r)$ is the ball of radius $r$ centered at $x$ and $m$ is the Lebesgue measure. (A general expression for the inverse in a general measure space is harder to write down, but it always has this kind of flavor of "convolving" $\phi$ with a "Dirac delta".)
$I$ and $I^{-1}$ preserve everything that you could care about from the Banach space point of view in either space. This means that there is really no loss in thinking of the two as actually being identical. This is a "categorical" view, which is of more limited application in analysis than in algebra or topology, but it is still useful in this setting. 
Just as an example, another place where not being pedantic about such matters is useful is in bouncing back and forth between thinking of $L^p$ as containing equivalence classes vs. containing functions. There are reasons we want to think of elements of $L^p$ as equivalence classes (e.g. uniqueness of limits does not hold in a seminormed space) vs. functions (e.g. regularity results for Sobolev functions only strictly hold for the "most continuous" representative of a $W^{k,p}$ function). In practice the distinction is not particularly important, provided that we know how our theorems get modified when we change perspective (e.g. $L^p$ limits aren't unique among functions, but any two limits are equal a.e.)
