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
$$ <Tg,f>= <g,T^*f>$$
for all $f \in L ^1, \, g \in L^\infty$, where $<,>$ denotes the integral of an $L^\infty$ function times an $L^1$ function.
 A: 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 $\xi$.
A: Remark on weak* continuity
How do you prefer to define continuity in non-metric cases?  Sequences won't suffice, so I will do it in terms of nets.
Let $X$ be a Banach space.  The norm topology is given like this:  a net $x_k \in X$ converges to $x$ iff $\lim_k \|x_k-x\| = 0$.
The dual of $X$ is the set $X^*$ of all continuous linear functionals on $X$.  There is a pairing between $X$ and $X^*$,
$$
\langle f, x\rangle := f(x),\qquad \text{for } x \in X, f \in X^* .
$$
The weak topology on $X$ is given like this:  a net $x_k \in X$ converges weakly to $x$ iff
$$
\lim_k \langle f, x_k\rangle = \langle f, x\rangle\qquad\text{for all } f \in X^*
$$
The $ $weak*$ $ topology on $X^*$ is given like this: a net $f_k \in X^*$ converges weak* to $f$ iff
$$
\langle f_k, x\rangle = \langle f,x \rangle \qquad\text{for all } x \in X .
$$
Now let $X, Y$ be two Banach spaces. Let $T$ be an operator $T : X^* \to Y^*$ on their dual spaces.  We say $T$ is weak* continuous iff it is continuous where both spaces $X^*, Y^*$ have their weak* topologies.
It can be checked from the definitions that if there is an operator $S : Y \to X$ so that
$$
\langle T(f), y\rangle = \langle f, S(y)\rangle \qquad\text{for all }y \in Y, f \in X^* ,
$$
then $T$ must be weak* continuous.  So for my example, all I had to do is write down an operator that is not weak* continuous.  I used the Hahn-Banach theorem to get a linear functional $\xi$ that is not weak* continuous, and then souped it up into an operator.  
A: See also: http://en.wikipedia.org/wiki/Hermitian_adjoint
The operation $\langle\cdot\rangle$ denotes the scalar product in $L^2(\Omega,\Sigma,\mu)$.
