About the adjoint operator and weak operator topology. Let $X,Y$ be Banach spaces. Let $\lbrace{S_n\rbrace}\subset\mathcal{L}(X,Y)$, and $T\in\mathcal{L}(X,Y)$, such that $S_n\xrightarrow[n\to\infty]{WOT}T$, that is:
$$\langle y^*,S_nx\rangle\xrightarrow[]{}\langle y^*,Tx\rangle\;\;\forall x\in X,\;\forall y^*\in Y^*.\quad(1)$$
Now we consider the sequence of adjoint operators $\lbrace{S_n^*\rbrace}\subset\mathcal{L}(Y^*,X^*)$. 
Is it true that $S_n^*\xrightarrow[n\to\infty]{WOT}T^*$?

What I observed: 
$S_n^*\xrightarrow[n\to\infty]{WOT}T^*$ is the same as:
$$\langle x^{**},S_n^*y^*\rangle\xrightarrow[]{}\langle x^{**},T^*y^*\rangle\;\;\forall x^{**}\in X^{**},\;\forall y^*\in Y^*.\quad (2)$$
But, from (1), using the definition of adjoint operator, we only get the following condition (weaker, because $X\subset X^{**}$)
$$\langle S_n^*y^*,x\rangle\xrightarrow[]{}\langle T^*y^*,x\rangle\;\;\forall x\in X,\;\forall y^*\in Y^*.$$

The question seems to be false, but I can't find a counterexample. 
Thanks a lot in advance for any help.
 A: Note you've proved that your result is true for reflexive Banach spaces X. 
I think you can actually prove it's true in general, though it's been a few years since I've done any functional analysis so you might want to check this carefully. The basic idea is to follow your argument, noting we're fine if $x^{**}$ is in $J_X(X)$ (where $J_X$ is canonical embedding of $X$ into  $X^{**}$). We can then use the fact that the weak* closure of $J_X(X)$ is all of $X^{**}$ (this is Goldstine's theorem). 
More precisely, firstly note that the $S_n$ are uniformly bounded (the hypothesis implies that for any $y^*$, $S_n^* y^*$ converges pointwise to $T^*y^*$ so uniform boundedness thm. implies $$
\sup_n || S_n^* y^* || < \infty $$
and uniform boundedness theorem again implies $$ \sup_n || S_n^* || = \sup_n ||S_n || < \infty $$. Let's call $\sup_n || S_n || = M$.
Now, fix $x^{**}$ and $y^*$, $\epsilon > 0$; we need to see $|S_n^{**}x**(y*) - T^{**}x^{**}(y^*)| < \epsilon$ for $n$ sufficiently large. Note by the Goldstine result, and the fact that $T^**$ is weak* to weak* continuous, we can find an $x in X$ with 
$$ | x^{**}(y^*) - J_X(x)(y^*) | < \frac{\epsilon}{3M} $$ and $$ |T^{**} x^{**}(y^*) - T^{**}J_X(x)(y^*) | < \frac{\epsilon}{3}$$.
Now, 
\begin{align} 
|S_n^{**}x^{**}(y*) - T^{**}x^{**}(y^*)| < 
|S_n^{**}x^{**}(y*) - S_n^{**}J_X(x)(y^*)| & + |S_n^{**}J_X(x)(y^*) - T^{**}J_X(x)(y^*)| \\
& + |T^{**}J_X(x)(y*) - T^{**}x^{**}(y^*)|
\end{align}
The first and third terms are less than $\frac{\epsilon}{3}$ by choice of $x$ and uniform boundedness, the middle term is less than $\frac{\epsilon}{3}$ for $n$ large enough since we know the result holds for $x^{**} \in J_X(x)$.
A: I'm very sorry but I realised there is a mistake in my "proof" above (also for some reason I was logged in only as a guest account when I posted my answer which is why I'm writing under a different profile now). The mistake is as follows: in estimating the first term in the right hand side of the final inequality, I'd estimated as follows
$$|S_n^{**} x^{**}(y^*) - S_n^{**}J_X(x) (y^*) | \leq ||S_n^{**} || \, ||x^{**}(y^*) - J_X(x)(y^*) || < M||x^{**}(y^*) - J_X(x)(y^*) ||  $$
This is total nonsense - for one thing $x^{**}$ is in $X^{**}$ so acts on elements in the dual of $X$, not elements in the dual of $Y$; I can't pull out the $||S_n^* ||$ term. If it's still not clear what is wrong with this step, I'd encourage you to write things out in your notation, at which point it should become clearer to see what's going.
(Note in response to your questions (as it could be of use to you for further questions in FA): note that straight by defn of the adjoint, $\langle x^{**} , T^* y^* \rangle = \langle T^{**} x^{**}, y^* \rangle$ - I had just passed to the double adjoint, as it seemed convenient to me at the time. It is a general result that if $T : X \to Y$ is continuous linear operator between Banach spaces, then $T^* : Y^* \to X^*$ is weak* to weak* continuous. The proof of this is very easy if you know about nets.)
Going back to your original question, perhaps the following is a counterexample that works - I'm afraid I haven't had time to check all the details completely, so again, check this through...
I hope you are familiar with the Banach space $c_0 = c_0(\mathbb{N})$ - it is the subspace of $\ell_{\infty}$ consisting of those sequences converging to 0 (and we equip $c_0$ with the $|| \cdot ||_{\infty}$ norm). There are canonical identifications of the dual of $c_0^*$ with $\ell_1$. The second dual of $c_0^{**}$ is therefore identifiable with $\ell_1^*$ which can be identified with $\ell_{\infty}$.
Take $S_n : c_0 \to c_0 $ to be equal to $L^n$ where $L : c_0 \to c_0$ is the left shift operator: $(a_1, a_2, a_3, \dots) \mapsto (a_2, a_3, \dots)$. Note $S_n$ converges in the WOT to the $0$ operator on $c_0$ (i.e. the operator which maps everything to 0). 
Now, the adjoint of $S_n$ should be $R^n$ where $R$ is the right shift operator on $c_0^* = \ell_1$. The adjoint of the $0$ operator on $c_0$ is the $0$ operator on $\ell_1$. Note that the right shift operator on $\ell_1$ does not converge to the $0$ operator in the WOT. To see this, take the vector $\mathbb{1} := (1,1,1,1, \dots) \in \ell_{\infty} = \ell_1^* = c_{0}^{**}$ (where "=" really means isomorphic here), and $\mathbf{b} = (1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \frac{1}{25}\dots) \in \ell_1$. Then, for every value of $n$,
$$
\langle \mathbb{1}, R^n \mathbf{b} \rangle = \sum_k \frac{1}{k^2} 
$$
so that $\langle \mathbb{1}, R^n \mathbf{b} \rangle$ does not converge to $0$. 
Let me know if I've made a mistake, or you need any help filling in more of the details.
