A question about the dual operator of an operator in a Banach space

Question

Let $c=(c_1,c_2....) \in l^{\infty}$ and $p,q \in [1,+ \infty)$ such that $\frac{1}{p}+ \frac{1}{q}=1$ and $T: l^p \longrightarrow l^p$ such that $Tx=(c_1x_1,c_2x_2,....)$ . Find the dual operator $T'$ of $T$

The dual operator of an operator $T: X \longrightarrow Y$ is defined as $T':Y^* \longrightarrow X^*$ such that $T'(y^*)(x)=y^*(Tx)$

I proved that $||T||=||c||_{\infty}$ and i pressume that we must use the Schauder basis of $l^p$ which is the set $\{e_n|n \in \mathbb{N} \}$ with $e_n=(0.0...1.0.0...)$ in the $n-$th position, but a i did not come up with a way to compute the dual operator of $T$

Can someone help me with some examples or a general method,if it exists,to find the dual operator?

Thank you in advance!

Answer 1

You know that $(\ell^p)^* = \ell^q$ where $y \in \ell^q$ acts on $\ell^p$ via the map $x \mapsto \sum x_i y_i$. So $T'(y)(x) = y(Tx) = \sum c_i x_i y_i = \sum x_i (c_i y_i)$ and since $c \in \ell^\infty$ we have $(c_i y_i) \in \ell^q$ so $T'y = (c_i y_i)$

Answer 2

You are close i think.

Using $T'(y^*)(x)=y^*(Tx)$ with $y^*=e_n$ and $x=e_k$ yields $T'(e_n)(e_k) = \langle e_n, T e_k \rangle = \delta_{kn} c_n$.

By linearity and continuity this results in $T' y^* = (c_1 y_1, c_2 y_2, \dots )$.