As we know from basic functional analysis, the sequence spaces $l^p$ and $(l^q)'$ are isometrically isomorphic.
Are they considered to be canonically isomorphic or just isomorphic?
Based on the replies to this question, I'd guess that they are indeed canonically isomorphic because the map between them doesn't seem to require a choice of a basis. But I'm not entirely sure.
The usual (isometric) isomorphism between $\ell^p$ and $(\ell^q)^*$, where $1/p+1/q=1$, is given by $$\Phi:\ell^p\to(\ell^q)^*$$ $$\ell^p\ni(x_n)_{n=1}^\infty:=x\mapsto\Phi(x):\ell^q\to\mathbb{C}$$ where $$\Phi(x)(y_n)_{n=1}^\infty=\sum_{n=1}^\infty x_ny_n $$ for all $(y_n)\in\ell^q$.
Based on the discussion in the linked post, since there were no "choices" involved in the definition of this isomorphism, we can say that this is a canonical isomorphism. Also, as mentioned already, this term does not have a precise meaning, so we usually say that something is "canonical" when it is "expected" to be defined in this way. This map is definitely of that form, the reason behind that being measure theory:
We can actually see $\ell^p$ as the $L^p$ space of $\mathbb{N}$ with the counting measure and $(x_n)\in\ell^p$ is nothing but a function $f$ such that $\int|f|^p<\infty$. We send such an element of $L^p$ to the linear functional $g\mapsto\int f\cdot g$, which is the most "canonical" thing to do. By the famous H"older inequality, this is bounded on $L^q$, which again in our case is identified with $\ell^q$. This is precisely what our map $\Phi$ does.
