Let $f\in L^\infty(\mathbb{R})$. I know $\|f\|_\infty=\sup\{\int_Rfg \, dx:\int_R|g|\,dx=1\}$, but how to show there exists a $g$, with $\|g\|_1=1$ on which the supremum is attained?
I know $\|f\|_\infty\ge\sup\{\int_Rfg \, dx:\int_R|g|\,dx=1\}$ can be shown by Holder inequality, and $\|f\|_\infty\le\sup\{\int_Rfg \, dx:\int_R|g|\,dx=1\}$ can be done by taking $A:=\{x|f(x)\ge\|f\|_{\infty}-\epsilon\}$ for any $\epsilon$ and $g=\frac{\chi_A}{\mu(A)}$, by which we can get $\int fg\le (\|f\|_{\infty}-\epsilon)\int g=\|f\|_{\infty}-\epsilon$.
But I just can't find a $g$ with $\|g\|_1=1$ to get the supreme, can anybody give a hint?