I am trying to show that $L^1(\mathbb{R})$ is not isometrically isomorphic to the dual of a Banach space. I know that one can show this by showing that the unit ball in $L^1(\mathbb{R})$ has no extremal points. Then the result follows by the Krein-Milnam and Banach-Alaoglu theorems.
As suggested in this question, if the measure space was finite you could find a constant such that $f\chi_{[0,c]}$ has exactly half the $L^1$ norm of $f$. Does this proof generalise to the infinite measure space with $\mathbb{R}$ as domain?
My idea was that you consider $F(t) := \int_{-t}^t|f(x)|\ dx$. This function is continuous and even though the intermediate value theorem does not necessarily apply here, the function converges to $||f||_{L^1}$ and is monotone, so there must be a $c$, such that $F(c) = \frac12 ||f||_{L^1}$. Then again I would just take $g_1 = 2*f\chi_{[-c,c]}$ and $g_2 = 2* f\chi_{[-c,c]^C}$ and I would have found a strict convex combination of $f$. Is this argument correct, or can you not simply copy the proof from finite intervals?
Any help would be appreciated
