$ \newcommand{C}{\mathcal{C} } \newcommand{D}{\mathcal{D} } \newcommand{I}{\mathcal{I} } $ (Full) subcategory $\D$ of the category $\C$ is called reflexive if there exists a localization functor $L : \C \to \D$ and it is left adjoint to the inclusion functor $I : \D \hookrightarrow \C$.
There is theorem:
If $\D$ is a reflective subcategory of $\C$, then $I$ creates all limits that $\C$ admits.
But if this theorem is true, then every diagram $x: \I \to\D$, which admits a limit $y = \lim_{i \in \I} Ix_i $, will also have a limit in $\D$. But as inclusion $I$ is right adjoint, it preserves limits, and so $y = I\Big(\lim_{i \in \I} x_i \Big)$. Then, as $y$ is included into $\C$, it must hold that $\lim_{i \in \I} x_i = Ly$ and $ ILy = y$. But, I don't know how to prove this fact, as $L$ only preserves colimits.
I can prove that $Ly$ indeed forms a cone over $x$ and for any other cone $c$ there exists a morphism $\psi: c \to Ly$, but I don't know how to prove uniqueness.
I construct $\psi$ as $L\phi$, there $\phi: Ic \to y$, which exists by universal property of limit $y$.
What am I missing here? Сan you help me prove uniqueness?