# How do I prove this inequality involving $\liminf$ and $\limsup$?

I'm trying to solve the following question.

Prove that there exists a bounded linear functional $F:l^{\infty}\rightarrow\mathbb R$ satisfying the following conditions:

(i) $|F(x)|\leq \sup_{n} |x_n|$
(ii) $F(x)=\lim_{n}x_n$, if limit exists
(iii) $\liminf x_n\leq F(x)\leq \limsup x_n$.

I proved the first two parts by defining the functional $F$ for the space of convergent sequences and then extending it to the whole of the Banach space $l^{\infty}$ using the Hahn-Banach theorem. However, I can't prove part (iii). Does it follow simply by using the definitions of $\liminf$ and $\limsup$? This seems to be not too difficult, but I can't seem to get this part. I'm not sure this particular part needs any functional analysis and so the tag need not be justified. Thank you for any help.

Suppose first $|\limsup x_n| \geq |\liminf x_n|$. Note that then $\limsup x_n \geq 0$. For all $\epsilon>0$, there exists $N>0$ such that if $n \geq N$, $x_n \leq \limsup x_n + \epsilon$. We can write $x_n = y_n + z_n$, where $y_n = x_n$ for $n < N$, and $0$ for $n \geq N$, and $z_n = 0$ for $n<N$ and $z_n = x_n$ for $n \geq N$. Now, $\lim y_n = 0$, so $F(y) = 0$, and $\sup |z_n| \leq \limsup x_n + \epsilon$, so $|F(x)| \leq |F(z)| \leq \limsup x_n + \epsilon$. Sending $\epsilon$ to zero gives us one side of the result. Clearly it is also the case that, if $|\limsup x_n| \leq |\liminf x_n|$, $F(x) \geq \liminf x_n$.
Now let's remove the hypotheses about the relative sizes of the limsups and the liminfs. Consider the sequence $w_n = x_n - \limsup x_n$. Then, $F(w) = F(x) - \limsup x_n$. Also, $|\liminf w_n| \geq |\limsup w_n| = 0$, so the above gives us $F(w) \geq \liminf x_n - \limsup x_n$, whence $F(x) \geq \liminf x_n$.
According to the theorem, if we can define a linear functional $\phi$ on a subspace of a Banach space, which is bounded by a sublinear functional $\rho$, then we extend $\phi$ to a linear functional $\tilde{\phi}$ defined all of the Banach space. Furthermore, $\tilde{\phi}$ is still bounded by $\rho$.
As you correctly observed, $|| \cdot ||_{\infty}$ is sublinear functional, so extending from the space of convergent sequences gives (i). For (iii), you need to notice that the limsup operator is also sublinear (see this SE question). This will give you the limsup part of the inequality chain. Applying the same procedure to the negative of the sequence then gives the liminf inequality.