We know that for an immersion $j:U \to X$ the restriction functor $j^*:{\cal O}_X-mod \to {\cal O}_u-mod$ has a left adjoint $j!$. I am looking for some condition to deduce that $j!$ takes its values in Qco(X) that is to be a left adjoint for the functor $j^*:Qco(X) \to Qco(u)$.

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By the way, it is more interesting to write down the right adjoint of $j^*$ (which exists by general nonsense). For example, you can do this when $j$ is quasi-compact, i.e. $U$ is retrocompact in $X$. Then the right adjoint is just $j_*$. – Martin Brandenburg May 13 '12 at 18:47

The restriction functor $\mathrm{Qcoh}(X) \to \mathrm{Qcoh}(U)$ doesn't preserve infinite products in general (which always exist, by the way). Therefore it cannot have a left adjoint.
Well it is true when $U$ is a clopen subset of $X$, because then $\mathrm{Qcoh}(X) = \mathrm{Qcoh}(U) \times \mathrm{Qcoh}(X \setminus U)$. But otherwise probably it's only true in pathological cases. This is already visible in the affine case, say $X=\mathrm{Spec}(A)$ and $U=D(f)$ for some $f \in A$. When does $M \mapsto M_f$ preserve infinite products? Almost never. – Martin Brandenburg May 13 '12 at 9:05
Yes, but this is a special case. When $X \setminus U$ is a disjoint union of opens, then it is open ... – Martin Brandenburg May 14 '12 at 9:30