Proving continuity of a monotone map from a Banach space to its dual with the weak* topology I'm following Brezis' book on functional analysis, and there is this problem. Let $E$ be a Banach space and $A: E \to E^{*}$ a monotone map defined in $D(A)=E$. Assume that $ \forall x,y \in E$ the map:
$$t \in \mathbb{R} \mapsto \langle A(x+ty),y\rangle $$
is continuous at $t=0$. We need to show that $A$ is continuous from $E$ to $E^{*}$ with the $σ(E^{*},E)$ topology (this is the weak* topology).
I was given the following proof, but I can't understand some of the steps.
Proof:
It suffices to show this for sequences. Lets take $(x_n)$ in $E$  and assume assume that $x_n \to x$ (converges strongly in $E$), but that  $Ax_n \not\rightharpoonup Ax$ for  $σ(E^{*},E)$.
Meaning we have a $y \in E$ such that $\langle Ax_n,y\rangle \not\to \langle Ax,y\rangle$.
We know $Ax_n$ is bounded.
So we have a sub-sequence such that $\langle Ax_{nk},y\rangle \to l \not=\langle Ax,y\rangle$, given that $A$ is monotone we have that:
$$\langle Ax_n-A(x+ty),x_{nk}-x-ty\rangle \geq 0.$$
Taking limits we get:
$$-tl+t\langle A(x+ty),y\rangle \geq 0.$$
Which would mean that $l =\langle Ax,y\rangle $; a contradiction. Hence we are done.
I don't see very clearly why $Ax_n$ would be bounded, but more importantly how do we know that $l \not = \langle Ax,y\rangle $, is this because $Ax_n$ does converges but just not to $Ax$? 
Taking the limits at the end would mean evaluating the limit of $\langle Ax_{nk},x_n\rangle $ and $\langle Ax_{nk},x\rangle$, how do we know these two are the same limits?
And finally how are we using the continuity at $t=0$ of the second map given?
 A: I don't see very clearly why $Ax_n$ would be bounded.
$A$ is locally bounded (and thus bounded in a neighborhood of $x$). See Problem 2.6, page 50.

How do we know that $l \not = \langle Ax,y\rangle$?
Using that $Ax_n \not\rightharpoonup Ax$, we can take $(x_{nk})$ such that
$$|\langle Ax_{nk},y_0\rangle-\langle Ax,y_0\rangle|\geq \varepsilon_0$$
for some $y_0\in E$ and some $\varepsilon_0>0$. Therefore, 
$$\left\{\begin{align*}
&\langle Ax_{nk},y\rangle \not\to \langle Ax,y\rangle\\
&\langle Ax_{nk},y\rangle \to l
\end{align*}\right.$$
which implies that $l \not = \langle Ax,y\rangle$.

Taking the limits at the end would mean evaluating the limit of $\langle Ax_{nk},x_n\rangle $ and $\langle Ax_{nk},x\rangle$, how do we know these two are the same limits?
Since $Ax_{nk}$ is bounded and $x_{nk}\to x$ strongly, the difference of these terms goes to zero:
$$|\langle Ax_{nk},x_n\rangle-\langle Ax_{nk},x\rangle|\leq \|Ax_{nk}\|\|x_{nk}-x\|\to 0.$$

And finally how are we using the continuity at $t=0$ of the second map given?
In order to obtain $l =\langle Ax,y\rangle$ from the last inequality, we have to take the limit as $t\to 0$.
