$A_1,A_2$ fulfill property, but their sum $A_1+A_2$ does not Let $V$ be a real, reflexive, separable Banach space. 
Are there operators $A_1,A_2: V \to V^*$ that fulfill the property 
\begin{cases}
    u_n \rightharpoonup u \\
    A_iu_n \rightharpoonup b \\
    \limsup_{n \to \infty} \langle A_iu_n,u_n\rangle\leq\langle b,u\rangle
\end{cases}implying $A_iu=b$
for a sequence $(u_n)_{n \in \mathbb N}$ in $V$, $b \in V^*$, but their sum $A_1+A_2$ doesn't?

My work:

$(A_1+A_2)u_n=A_1u_n+A_2u_n \rightharpoonup2b \in V^*$ and
$\limsup\, \langle (A_1+A_2)u_n,u_n \rangle=\limsup \langle A_1u_n,u_n \rangle+\limsup \langle A_2u_n,u_n \rangle \leq 2 \langle b,u\rangle$
So I think I need to find $A_1,A_2$ fulfilling the property but $(A_1+A_2)u \neq 2b?$
Here is a counterexample. Let $V = \mathbb{R}$ and
$$
A_1(x)=
\begin{cases}
\frac1x &amp; \text{if } x \ne 0 \\
42 &amp; \text{if } x = 0
\end{cases}
$$
and
$$
A_2(x)=
\begin{cases}
-\frac1x &amp; \text{if } x \ne 0 \\
23 &amp; \text{if } x = 0
\end{cases}
$$
Then, it is easy to check that $A_1$ and $A_2$ satisfy your property, but $A_1 + A_2$ does not satisfy it.
No, this is not possible.

Using the linearity of $A$ in these properties and an inequality involving $\limsup$,
it can be shown that $A_1+A_2$ fulfills the same properties.