Is $J$ injective or surjective where $Jf(g):=\int d\mu f g$

Question

Let $X:=\{0, 1\}$ and $\mathcal{A}:=\mathcal{P}(X)$ and $\mu(\{0\})=1$ while $\mu(\{1\})=\infty$

I know that $f\in L^{1}(X,\mathcal{A}, \mu)$ iff $f\vert_{\{1\}}=0$ and $g \in L^{\infty}(X,\mathcal{A}, \mu)$ for any well-defined $g$

Now, I am asked to determine whether $J: L^{1} \to (L^{\infty})^{*}, Jf(g)=\int_{X}d\mu fg$ is injective and/or surjective.

I was told as a hint that $L^{\infty}$ is isometrically isomorphic to $\mathbb{C}$ but I have absolutely no idea, how this will help me

Answer 1

Do you consider $L^p(X,\mathcal{A},\mu)$ to be real or complex valued functions? Let us consider real functions. Then $g\in L^\infty$ just corresponds to $(g(0),g(1)) \in \mathbb{R}^2$ which you could identify with $\mathbb{C}$. Hence you can identify the dual space of $L^\infty$ with $\mathbb{R}^2$ (or $\mathbb{C}^2$) itself.

Let $f\in L^1$ and $g\in L^\infty$. Then we simply have $Jf(g)=\int_X fgd\mu = f(0)g(0)$.

$J$ is injective: $Jf=0$ iff $f(0)$ iff $f=0$ (since $f(1)=0$ because $f\in L^1$).

$J$ is not surjective: For instance $g\mapsto g(1)$ is an element of $(L^\infty)^*$ which does not belong to the image of $J$.