Pullback functor preserves Day convolution? let $A$ und $B$ be small monoidal categories enriched over a bicomplete closed symmetric monoidal category $V$. 
On the functor categories $V^A$ und $V^B$ we have the Day convolution as a tensor product, see for instance 
ncatlab.org/nlab/show/Day+convolution
Now let $f: A \to B$ be a functor and $f^* : V^B \to V^A$ the pullback functor.
What are conditions that ensure that $f^*$ is monoidal? One should probably require that $f$ is monoidal and respects the enrichment. However, even in that case I find it not too easy to see monoidality. But it is easy to see that $f^*$ preserves the external product (see also the nlab article). Maybe that helps.
Is it maybe easier to prove something about the left-adjoint $f_! : V^A \to V^B$ of $f^*$.
I would also be happy to see that $f^*$ is only lax monoidal (and hence $f_!$ oplax monoidal). 
Thanks in advance for any hints. 
 A: It is always wise to consider examples first. Consider two commutative rings $A,B$ as one-object monoidal categories enriched in abelian groups (Remark: Commutativity is important for the monoidal structure!). Then an enriched strict monoidal functor $A \to B$ is the same as a ring homomorphism, and the associated pullback functor identifies with the "restriction of scalars" or "forgetful" functor $U : {}_B \mathsf{Mod} \to {}_A \mathsf{Mod}$. This is well-known to be lax monoidal, but not strong monoidal. This can be already seen for the unit morphism $A \to U(B)$.
In the general setting, let $f : A \to B$ be an enriched lax monoidal functor between $V$-enriched monoidal categories, and $f^* : V^B \to V^A$, $X \mapsto X \circ f$ be the pullback functor. The unit object of $V^B$ is $\hom(1_B,-)$. We have a morphism
$$\hom(1_A,-) \to f^* \hom(1_B,-) = \hom(1_B,f(-))$$
which corresponds, by Yoneda, to the given morphism $1_B \to f(1_A)$.
Recall the following description of Day convolution: For $X,Y,Z \in V^A$ we have
$$\hom(X \otimes Y,Z) = \int_{a,a' \in A} \hom(X(a) \otimes Y(a'),Z(a \otimes a')). ~~~ (\dagger)$$
This bijection is induced by natural morphisms $X(a) \otimes Y(a') \to (X \otimes Y)(a \otimes a')$. $(\star)$
Now let $X,Y \in V^B$. Let $a,a' \in A$. Then we have natural morphisms
$$X(f(a)) \otimes Y(f(a')) \xrightarrow{~(\star)~} (X \otimes Y)(f(a) \otimes f(a')) \xrightarrow{~f~ \text{lax}~} (X \otimes Y)(f(a \otimes a')),$$
i.e.
$$f^*(X)(a) \otimes f^*(Y)(a') \to f^*(X \otimes Y)(a \otimes a')$$
These induce by $(\dagger)$ a morphism $f^*(X) \otimes f^*(Y) \to f^*(X \otimes Y)$.
It is easy to check the coherence conditions; of course they follow from the ones given for $f$.
