Adjoint operator of $L^\infty$ Lets denote with $(\Omega,\Sigma,\mu)$ a $\sigma$-finite measurble space with a linear, continuous operator $$T : L^\infty \to L^\infty.$$ Does this always imply the existence of a linear, continuous operator $$T^*: L^1 \to L^1 $$
that solves the integral equation
$$ \int_\Omega Tg (x) f(x) \mu(dx) = \int_\Omega g(x) T^*f (x) \mu(dx)  $$
or in short notation
$$ &lt;Tg,f&gt;= &lt;g,T^*f&gt;$$
for all $f \in L ^1, \, g \in L^\infty$, where $&lt;,&gt;$ denotes the integral of an $L^\infty$ function times an $L^1$ function.
Here is a brief example.  This is the type of thing you will learn in a course on Banach spaces.  Or a course on functional analysis with a good chapter on Banach spaces.   The general fact is that space $L^1$ is not reflexive.  I have adapted this to the particular case of $L^1, L^\infty$.

Let's use measure space $[0,1]$ with Lebesgue measure $\lambda$.  For $g \in L^\infty$ and $f \in L^1$, define the  pairing $\langle g, f\rangle = \int fg\,d\lambda$.  This pairing identifies $L^\infty$ with the isometric dual space of $L^1$.  Topological words used will refer to the norm toplogies of the spaces.

Let the sequence $r_n \in L^\infty$ be the functons $1, \sin(n \pi x), \cos(n \pi x)$ in some order.  This sequence separates points of $L^1$.  That is, if $f \in L_1$ and $\langle r_n,f\rangle = 0$ for all $n$, then $f=0$.  

Now the space $L^\infty$ is non-separable, so the closed span $N$ of $\{r_1, r_2, r_3, \cdots\}$ is not the whole space.  By the Hahn-Banach theorem, there is a continuous linear functional $\xi$ on $L^\infty$ such that $\xi$ vanishes on $N$ but $\xi$ is not identically zero.  Define an operator $T \colon L^\infty \to L^\infty$ by $T(g) = \xi(g)\mathbb 1$.  Here I wrote $\mathbb 1$ for the constant function $1$.  Thus $T(g)$ is a certain scalar $\xi(g)$ times the constant $1$.

Now suppose there is $T^* \colon L^1 \to L^1$ with
$$
\langle g, T^*(f)\rangle = \langle T(g), f\rangle \qquad\text{for all $f \in L^1, g \in L^\infty$}
$$
Write $\mathbf 1$ for the constant $1$ function in $L^1$.  What is $T^*(\mathbf 1)$ ?  (We claim it has contradictory properties.)  Write $F = T^*(\mathbf 1)$.

For all $g$ in $L^\infty$,
$$
\langle g, F\rangle =\langle g, T^*(\mathbf 1)\rangle = \langle T(g), \mathbf 1\rangle = \langle \xi(g)\mathbb 1, \mathbf 1 \rangle = \xi(g) .
$$
Now for all $n$, we have $\langle r_n, F\rangle = \xi(r_n) = 0$.  Since the $r_n$ separate points of $L^1$, we have $F = 0$.  But then for all $g \in L^\infty$,
$$
\xi (g) = \langle g, F\rangle = 0 ,
$$
so $\xi$ is the zero linear functional.  This contradicts the choice of $\xi$.
See also: http://en.wikipedia.org/wiki/Hermitian_adjoint

The operation $\langle\cdot\rangle$ denotes the scalar product in $L^2(\Omega,\Sigma,\mu)$.