I am preparing for my exam in functional analysis and came across the following question:
Let $n\in \mathbb{N}$. Find the adjoint to A: $\ell^1\to\ell^1 $ (where $\ell^1=\{x=(x_1,x_2,x_3,...),\,\sum^{\infty}_{i=1}|x_i|<\infty \}$ if:
a) $A(x)=(x_1,x_2,x_3,..,x_n,0,0,...)$
b) $A(x)=(0,x_1, x_2,...)$
c) $A(x)=(x_2,x_3,...)$
Is there some sort of general way to do this? Thanks in advance!
Recall that for linear spaces $X, Y$ and a linear operator $A: X \mapsto Y$ the adjoint $A'$ is a linear map from the dual space $Y'$ to $X'$ and it maps a functional $f: Y \rightarrow \mathbb{C}$ to the functional $f \circ A: X \rightarrow \mathbb{C}$.
To get a grip on the adjoint operator, it helps to find an explicit description of the involved dual spaces. In your example we have $(\ell^1)' = \ell^\infty$ and $(y_n)_n \in \ell^\infty$ applied to $(x_k)_k \in \ell^1$ is $\sum_{k=1}^\infty x_k y_k$.
Now, start with $(y_k)_k \in \ell^\infty$ and ask: What is $(y_k)_k$ applied to $Ax$ for some $x = (x_k)_k \in \ell^1$? Then finally you have to find the $(\tilde{y}_k)_k \in \ell^\infty$ that corresponds to this map and the adjoint of $A$ will map $(y_k)_k$ to $(\tilde{y}_k)_k$.
For example, in (a) we have $Ax = (x_1, \dots, x_n, 0, \dots)$. Hence $(y_n)_n$ applied to $Ax$ is just $\sum_{k=1}^n x_k y_k$. This is exactly the result we would have obtained by applying $(\tilde{y}_k)_k$ to $x$, where $\tilde{y}_k = y_k$ if $k \leq n$ and $\tilde{y}_k = 0$ otherwise. Hence the adjoint of $A$ coincides with $A$ itself, apart from the fact that the adjoint is a function from $\ell^\infty$ to $\ell^\infty$.
The same reasoning shows for (b) that $(y_k)_k$ applied to $Ax$ is $\sum_{k=1}^\infty x_k y_{k+1}$. Hence the adjoint of $A$ is the left-shift on $\ell^\infty$ which maps $(y_1, y_2, \dots )$ to $(y_2, y_3, \dots)$.