Struggling with non-Hilbert adjoint operators on $\ell^{p}$

Question

Say $p=1$ and we are considering the the Right-Shift operator $T: \ell^{1} \to \ell^{1}$ where $(s_{1},...) \mapsto (0,s_{1},...)$.

Compute the adjoint.

Now, I feel comfortable finding Hilbert adjoints, because we're looking at the same space. But when it comes to non-Hilbert spaces, then I struggle:

By definition, the adjoint $T^{\times}:{\ell^{1}}^{*} \to {\ell^{1}}^{*}$ and we identify ${\ell^{1}}^{*}$ with $\ell^{\infty}$, so we have $T: \ell^{\infty} \to \ell^{\infty}$

and thus $T^{\times}(a_{n})_{n}=(a_{n})_{n}T$. So clearly $(a_{n})_{n}T$ is a sequence in $\ell^{\infty}$ but how do I make sense of it? We know nothing about its components etc.

Answer 1

Take $a \in (\ell^{1})^\ast$ which corresponds to $(a_n)_n \in \ell^\infty$. By definition, we should have

$$ T^*(a)((s_n)_n) = aT(s_n)_n = \sum_{n \geq 1}a_nT(xs_n) = \sum_{n \geq 1}a_{n+1}s_n. $$

You also know that a functional $b \in (\ell^1)^{\ast}$ identifies with a bounded sequence via $b_i := b(e_i)$.

From these two facts, you should be able to identify $T^*(a)$ with its corresponding sequence in $\ell^\infty$.

By the remark above, simply note that $(T^*(a))_i = T^*(a)(e_i) = a_{i+1}$. Thus, the corresponding operator $T^* : \ell^\infty \to \ell^\infty$ sends $(a_n)_n$ to $(a_{n+1})_{n}$.