The multiplier algebra of a Reproducing Kernel Hilbert Space and its commutant In my research in the theory of Reproducing Kernel Hilbert Spaces I was concerned with this topic which came up but I could not find a reference on:

If $ \mathbb{H} $ is an RKHS and we denote the multiplier algebra by $ Mult(H) $ the algebra of multiplication operators on $ \mathbb{H} $ by multipliers of H, that is, those special functions f such that $ \forall h \in H : fh \in H $ and now we have $ Mult(H)' $ the commutant of $ Mult(H) $, that is all bounded operators on $ \mathbb{H} $ that commute with every multiplier operator on H, $  Mult(H)' = \{ T \in B(H) | \forall M \in Mult(H): MT=TM \}$

My question is can we deduce something (given that H is a RKHS) about the inclusion or intersection relation between the multiplier algebra and its commutant? It seems there should be something nice holding here, but would I be so lucky as to have inclusion or equality? It seems to good to be true but intuitively I can think of a few cases for this argument. Could someone please help me resolve this?
 A: I only have an answer for a special case (I should've put this as a comment but being new to MO, I cannot comment yet): when $1\in H$ and $Mult(H)$ is a dense subset of $H$. In this case $Mult(H)' = Mult(H)$. An example of such an $H$ is $H=H^2$ as in Chris Ramsey's example.  
The following argument is standard in showing $Mult(H)' = Mult(H)$ under the above assumptions. For each $f\in Mult(H)$, we denote by $M_f$ the multiplication operator by $f$. Suppose $T$ is a bounded operator on $H$ that commutes with $M_f$ for all $f\in Mult(H)$. We'll show that $T$ is also a multiplication operator. Well, if it is, then the corresponding multiplier must be $\varphi=T(1)$ so we try to show that it is the case. By assumption, $TM_{f} = M_{f}T$ for all $f\in Mult(H)$. This shows that $TM_{f}(1) = M_{f}T(1)$, that is, $$T(f) = f\varphi.$$
Since $Mult(H)$ is dense in $H$, the above identity will in fact imply that $T(h)=h\varphi$ for all $h\in H$. To see this, take a sequence $\{f_m\}\in Mult(H)$ converging to $h$ in norm. Then $T(f_m)\rightarrow T(f)$ in norm. Since $H$ is an RKHS, ${f_m}\rightarrow h$ pointwise and $T(f_m)\rightarrow T(h)$ pointwise. Since $T(f_m)=f_m\varphi$ for all $m$, pointwise convergence implies that $h\varphi = T(h)$ as desired.
If we remove the assumption that $1$ belongs to $H$ but require that $Mult(H)\cap H$ be dense in $H$, then a similar argument shows the following: for any $g\in Mult(H)\cap H$ and any $h\in H$, $$T(hg) = h\,T(g).$$
Unfortunately, I was not able to say anything more.
A: Problem 147 in Halmos's A Hilbert space problem book is that the commutant of the unilateral shift in $H^2$ is $H^\infty$ considered as multipliers on $H^2$. That is, $(H^\infty)' = H^\infty \subset B(H^2)$. 
Moreover, for $H^2_d$, the Drury-Arveson space, an RKHS on $\mathbb B_d$, a survey paper by Orr Shalit "Operator theory and function theory in Drury-Arveson space and its quotients", section 3.9, gives that $Mult(H^2_d) = M_d = M_d'$. 
In general, $Mult(H) \subseteq Mult(H)'$ for any RKHS $H$ since the multiplier algebra is always a commutative algebra of functions. I am not sure whether it must be an equality.
A: The answer is no. 
Let $H$ the Segal-Bargmann space in $n=1$ variables. Every multiplier is an entire function, and since a multiplier on any reproducing kernel Hilbert space is a bounded function, we get that every multiplier is a constant: $Mult(H) = \mathbb{C}1$. Thus $Mult(H)' = B(H) \neq Mult(H)$. 
