Figured I might be able to come up with an interesting way to show that $L^1$ is non reflexive by finding an operator in the dual space that isn't bounded in the unit ball in $L^1$ and using some corollaries of Hahn-Banach. My question is as follows:
Is there some $f \in L^\infty$ such that the operator: $$ g\in L^1 \longmapsto \int f \cdot g $$ does not have a maximum value on the closed ball $\{g \in L^1 \: | \: ||g||_{L^1} \leq 1 \}$?
On
Here is an example on a compact interval, i.e., on $L^1(0,1)$. Define the functional $$ f(g):=\int_0^1 (1-x)g(x)dx. $$ Then it is clear that $\|f\|\le 1$. In addition, for $g_n=n \chi_{0,1/n}$ we have $f(g_n)\to 1$, $\|g_n\|_{L^1}=1$, which proves $\|f\|=1$.
I will prove that for every $g\in L^1(0,1)$ with $\|g\|_{L^1}$ we can construct $\tilde g$ with $\|\tilde g\|_{L^1}=1$ and $f(\tilde g)<f(g)$.
First, if $g$ is negative on a set of positive measure, then $\tilde g=|g|$ does the job.
Second, let $g\ge0$. Then there is $t>0$ such that $\int_t^1 g \ne0$. Define $$ \tilde g(x) = \chi_{(0,t)}(x) (g(x) + t^{-1} \int_t^1 g(s)ds). $$ Then we get $$ f(\tilde g)=\int_0^t (1-x)\left(g(x) + t^{-1} \int_t^1 g(s)ds \right) \ dx\\ = \int_0^t (1-x)g(x) dx + (1-t/2)\int_t^1 g(s)ds \\ > \int_0^t (1-x)g(x) dx + \int_t^1 (1-s) g(s)ds =f(g). $$ Since for $g\in L^1(0,1)$ with $\|g\|_{L^1}=1$ one of the both cases is true it follows that the supremum is not attained.
Actually, the supremum is 'attained' at the Dirac functional $\delta_0$, which is not an integrable function.
On $L^1(\mathbb R)$, you can take $f(x)=1-e^{-|x|}$. Then $$ |Tg|=|\int_{\mathbb R}fg|\leq\|g\|_1. $$ For $g(x)=n\,1_{[n,n+1/n]}$, $$ |Tg|=n\,\int_n^{n+1/n}(1-e^{-x})\,dx\geq1-e^{-n-1/n}. $$ So $\|T\|=1$. But the norm cannot be achieved; this can be easily seen by noticing that, for any $f\geq0$ supported on $[0,\infty)$, $$ Tf<Tf_k\leq \|f_k\|_1=\|f\|_1, $$ where $f_k$ is obtained from $f$ by "moving mass to the right": $f_k=f(x-k)\,1_{[k,\infty)}$.