Do we have $f(x)\le\limsup x_n$ for every functional extending limit? Let us consider functionals $f\colon\ell_\infty\to\mathbb R$ such that:


*

*$f$ is linear and continuous,

*$f$ extends the usual limit, i.e., if $x=(x_n)$ is convergent and $\lim\limits_{n\to\infty} x_n=a$, then also $f(x)=a$.


Question. Do these conditions already imply that $f(x)\le\limsup\limits_{n\to\infty} x_n$? What are some counterexamples showing that this is not the case?
This came up in the discussion related to a recent question which is based on T. Tao's blog post Generalisations of the limit functional (Wayback Machine). You can see some discussion related to that question in the comments on the linked post and also in chat.
In that case we have worked with some additional conditions which assure that $f(x)$ is always between $\liminf x_n$ and $\limsup x_n$. (For example, it is sufficient to add $\|f\|=1$ or that $f$ is positive, i.e., $f(x)\ge0$ whenever $x_n\ge0$ for each $n$.) One of the problems in that discussion was that it was unclear why the additional conditions are needed. Perhaps it is reasonable to make a separate question out of it, where this can be presented more clearly.
It might be worth mentioning that one such example can be obtained from this answer: Continuous extension of the limit functional. (Although purpose of that question was different.) 
I have also posted my attempt to give such example as an answer.
 A: Let us consider any functional $f$ which fulfills the conditions in the question. Then we define a new functional $g$ by setting
$$g(x)=2f(x_{2n+1})-f(x_{2n})$$
for any sequence $x=(x_n)_{n=1}^\infty$. I.e., we apply $f$ to the subsequences given by odd/even terms of the original sequence and then we combine these values. For the sake of consistency, let us assume that all sequences are numbered from $n=1$, i.e., $x=(x_1,x_2,x_3,\dots)$ 
It is easy to see that $g$ is a linear functional. We also have
$$\|g\|\le 3\|f\|,$$
i.e., $g$ is continuous.
The functional $g$ extends limit. Indeed, if $x=(x_n)$ is convergent and $\lim\limits_{n\to\infty} x_n=a$, then we also have $\lim\limits_{n\to\infty} x_{2n}=\lim\limits_{n\to\infty} x_{2n+1}=a$. Consequently, $f(x_{2n})=f(x_{2n+1})=a$ and $g(x)=2a-a=a$.
For this functional we have:


*

*The sequence $x=(1,-1,1,-1,1,-1,\dots)$ is mapped to $g(x)=3$. So for this sequence the value is bigger than $\limsup\limits_{n\to\infty} x_n$.

*This also implies $\|g\|\ge3$. (And if we take $f$ such that $\|f\|=1$, then we get $\|g\|=3$.)

*The sequence $y=(0,1,0,1,0,1,\dots)$ is mapped to $g(y)=-1$, so $g$ is not positive. 


I will add that "limits" of odd/even terms were also used in this answer: Hahn–Banach Theorem for Normed Spaces: not unique extension. (Although for different purpose.)
