I am working with conjugate operators acting between Banach spaces. I am doing the following exercise.
Let $(\beta_{n})_{n \in \mathbb{N}}$ be a bounded sequence of complex numbers. Define the operator $T: \ell_{p} \to \ell_{p}$, for $1 \leq p < \infty$, by $$T(x_{1},x_{2},\dots) = (\beta_{1}x_{1}, \beta_{2} x_{2}, \dots)$$ Find the conjugate $T^{*}:Y^{*} \to X^{*}$.
Definition (Conjugate operator)
Let $A: X \to Y$ be a bounded linear operator where $X$ and $Y$ are Banach spaces. The conjugate operator of $A$ is the operator $A^{*}: Y^{*} \to X^{*}$, where $X^{*}$ and $Y^{*}$ are the dual spaces of $X$ and $Y$ respectively. The operator $A^{*}$ is defined in the following way. For any $f \in Y^{*}$ $$f \mapsto A^{*}f$$ where $A^{*}f$ is the linear functional acting on $X$ by $$(A^{*}f)(x) = f(Ax)$$ for each $x$ in $X$
I am struggling to find an explicit description of $A^{*}$. Can someone please help?
The key to these kind of questions is how the duality is implemented. When we say that $\ell_q$ is the dual of $\ell_p$, what we mean is that any continuous linear functional $f:\ell_p\to\mathbb C$ is of the form $$\tag1 f(x)=\sum_nx_ny_n $$ for some $y\in\ell_q$. Note that we have $y_n=f(e_n)$, where $\{e_n\}$ denotes the canonical basis.
Now $T^*$ needs to be seen as a map $\ell_q\to\ell_q$. Given $z\in\ell_q$, we need to see $T^*z$ as an element of $\ell_q$ via $(T^*z)x=z(Tx)=\sum_nz_n(Tx)_n$. We then have $$\tag2 (T^*z)x=\sum_nz_n(Tx)_n=\sum_n\beta_n\,z_nx_n=\sum_n(\beta_nz_n)x_n. $$ As we also have $$\tag3 (T^*z)x=\sum_n(T^*z)_nx_n, $$ direct comparison between $(2)$ and $(3)$ (concretely considering the case where $x=e_n$ for each $n$) gives us that $$ T^*z=(\beta_1z_1,\beta_2z_2,\ldots). $$ We cannot really write $T^*=T$ (though the formula does say so) because their domains and codomains are different. But they both do the same: multiply each entry by the corresponding entries in $\beta$.