$A_1,A_2$ fulfill property, but their sum $A_1+A_2$ does not

Question

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 $(A_1+A_2)u \neq 2b?$

Answer 1

Here is a counterexample. Let $V = \mathbb{R}$ and $$ A_1(x)= \begin{cases} \frac1x & \text{if } x \ne 0 \\ 42 & \text{if } x = 0 \end{cases} $$ and $$ A_2(x)= \begin{cases} -\frac1x & \text{if } x \ne 0 \\ 23 & \text{if } x = 0 \end{cases} $$ Then, it is easy to check that $A_1$ and $A_2$ satisfy your property, but $A_1 + A_2$ does not satisfy it.

Answer 2

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.