Say $X = \C$ and $U = \C^*$, $j: U \to \C$. Do you know what is $j^*$ and $j^!$ of $O_X$?

https://arxiv.org/pdf/1607.02064.pdf

This paper discusses general stuff. Very readable, except the notation needs some getting used to. The setup is as following : $X$ is some stable $\infty$-category. No need to assume compactly generated. $U$ is some subcategory which is both reflexive and coreflexive, which means $j_*$ admits both left-adjoint $j^*$ and right-adjoint $j^!$ (I don't know which term corresponds to which condition)

We define $Z^L$ and $Z^R$ as subcategory of $X$. $Z^L = Z^\vee$, is something $\Hom(Z^L, U) = 0$. Similarly, $Z^R = Z^\wedge$ satisfies $\Hom(U, Z^R)=0$.

Consider the original example. Consider $\C[x]/(x) = O_0$. Does $U$ map to it? This module $\C[x][1/x]$ is a colimit of $\C[x]$. To define a map out of a colimit, is to define a map from every member to it, in a consistent way. Say, if $1 \in \C[x,1/x]$ maps to $1 \in \C[x]/x$, then we need to ask where does $1/x$ maps to? No place, so we are dead, so $U$ does not map to it.

Does this map to $U$? I don't think so, this would give $Hom(\C[x]/(x), U) = U \xto{x} U = 0$. So, either way, this guys is in $Z^L$ and $Z^R$. How about $\C[[x] ] =\lim \C[x]/(x^n)$? It is complete. It must be in $Z^R$. We should have $$ j_!j^! \to id \to i_\wedge i^\wedge $$ where $i^\wedge -| i_\wedge$, so $i_\wedge$ only admit left-adjoint, so we say $Z^\wedge= Z^R$ is reflexive, meaning 'pullback to it' is a left-adjoint.

Similarly, we get $$ i_\vee i^\vee \to id \to j_* j^* $$ consist of things set theoretically supported on $Z$.

I don't know the difference.