Adjoint of Compact Operators in Normed spaces I have a little trouble understanding adjoint operators in spaces without inner products. So the definition of the adjoint operator is the following: 

Definition (Adjoint operator $T^\times$):
  Let $T: X \to Y$ be a bounded linear operator between normed spaces. Then the adjoint operator $T^\times: Y' \to X'$ is defined by
  $$(T^\times g)(x) = g(Tx) $$
  where $g \in Y'$.

So I wish to find some specific adjoint operators. In the examples we have that $1 \leq p < \infty$.
Example 1. Let $a_n$ be a bounded sequence with $\lim_{n\to\infty} a_n = 0$. Then find the adjoint of the operator $T_1: \ell^p \to \ell^p$ defined by $T_1(x_n)_{n=1}^\infty = (a_n x_n)_{n=1}^\infty.$
Example 2. Let $(a,b)$ be a non-empty interval. Find the adjoint operator of $T_2:L^p(a,b)\to L^p(a,b)$ defined by
$$ T_2x(t) = \int_a^b e^{t-s} x(s)\,\mathrm{d}s. $$
I know that if $p^{-1} + q^{-1} = 1$ then $(\ell^p)' = \ell^q$ and $(L^p(a,b))' = L^q(a,b)$ which I believe one should use at some point.
I don't necessarily need everything solved for me, but if someone could show the process once, maybe I could solve the rest myself.
Thanks for any help.
 A: Note that the isomorphism $S \colon \ell^q \to (\ell^p)'$ you know is given by 
$$ (Sx)(y) = \sum_n x_n y_n, \qquad x \in \ell^q, y \in \ell^p $$
Usually one omits $S$ in notation, that is identifies $\ell^q = (\ell^p)'$. For $T_1$ we have $y \in \ell^q$, $x \in \ell^p$
\begin{align*}
  (T_1^\times y)(x) &= y(Tx)\\
                    &= \sum_n y_n (Tx)_n\\
                    &= \sum_n y_n a_n x_n\\
                    &= \left(( a_n y_n)_n \right)(x)
\end{align*}
That is 
$$ T_1^\times y = (a_n y_n)_n $$
For $T_2$ note that the isomorphism is given by 
$$ (Sf)(g) = \int_a^b f(s)g(s)\, ds, \qquad f \in L^q, g \in L^p $$
we have for $g \in L^q$, $f \in L^p$:
\begin{align*}
  (T_2^\times g)(f) &= g(Tf)\\
                    &= \int_a^b g(t)(Tf)(t)\, dt\\
                    &= \int_a^b \int_a^b g(t)e^{t-s}f(s)\, ds\,dt\\
                    &= \int_a^b \int_a^b e^{t-s}g(t)\, dt\, f(s)\,ds
                    &= \int_a^b (T_2^\times g)(s)\, f(s)\, ds
\end{align*}
Hence
$$ (T_2^\times g)(s) = \int_a^b e^{t-s}\, g(t)\, dt $$
