Trying to understand adjoint operators and the dual space of $l_2(\Bbb C)$ Consider the operator: $A: l_2(\Bbb C) \to l_2(\Bbb C)$
$$A(x_1,x_2,\cdots, x_m,x_{m+1},x_{m+1}, \cdots) = (x_1,x_2,\cdots,x_m,0,0,\cdots)$$
I.e. it's the identity map on the first $m$ values, and $0$ on all others.

I want to find the adjoint operator $A^*$, but I don't really understand it.

So $A^*:l_2(\Bbb C)\to l_2(\Bbb C)$ being the adjoint operator means that for every $f\in l_2(\Bbb C)^*$, the dual space, and $x\in l_2(\Bbb C)$
$$f(Ax) = (A^* f)(x)$$
Firstly I am not sure what an arbitrary function of the dual space of $l_2(\Bbb C)$ looks like. 

An example of an adjoint operator on an infinite dimensional space would be nice!
 A: If you have the inner product $<x, Ay>$, the adjoint is the operator $B$ such that $<Bx,y> = <x,Ay>$ (think of transposes/hermitian conjugates). 
So, in this case, $<x,Ay> = \sum_{i=1}^m x_i y_i^* = <Bx,y> = \sum_{i=1}^\infty (Bx)_i y_i^*$. 
From this, I can read off $Bx = (x_1,\ldots,x_m,0,\ldots,0)$ to make the statement true. 
So, the adjoint of $A$ is $A$ itself. 
A: Note that on a Hilbert space $H$ (as $\ell^2(\def\C{\mathbf C}\C)$ is), the (Hilbert) adjoint of an operator $A \colon H \to H$ is (as you wrote in the first line) an operator $A^* \colon H \to H$ (note the lack of dual stars on the space). The (Hilbert) adjoint is given by the property that $\def\(#1){\left(#1\right)}$
$$\(Ax,y) = \(x,A^*y), \quad x,y\in H$$
where $\(-,-)$ denotes the inner product of $H$. To see the connection with the usual (Banach) adjoint, let $\Theta\colon H\to H^*$ denote the operator $\Theta x = \(-,x)$. $\Theta$ is a conjugate linear isomorphism, the so called Riesz isomorphism. Then, we have, if $A^{*_B} \colon H^* \to H^*$ denotes the Banach adjoint, that 
$$ A^{*_B}\Theta = \Theta A^*. $$
To give an example for a Hilbert adjoint, consider $A \colon \ell^2(\def\Z{\mathbf Z}\Z) \to \ell^2(\Z)$ given by 
$$ (Ax)_n = x_{n+1}, \qquad x \in \ell^2(\Z), n \in \mathbf Z $$
Then 
\begin{align*}
  \(Ax,y) &= \sum_n x_{n+1} \bar y_n\\
          &= \sum_n x_n \bar y_{n-1}\\
          &= \(x,A^* y)
\end{align*}
with $(A^*y)_n = y_{n-1}$.
