I'm trying to show that there exists a functional in $l^\infty$ whose value lies between the infimum and supremum of a sequence. That is, there exists a functional $\phi:l^\infty\longrightarrow \mathbb R$ such that $\forall x=(x_n)\in l^\infty$, $\liminf x_n\leq\phi(x)\leq \limsup x_n$. My argument is as follows:
Since $l^1\subset l^\infty$, we first define $\phi(x)=\lim_{N\rightarrow \infty}\frac{x_1+x_2+...+x_N}{N}$ for $x=(x_1,x_2,...x_n)\in l^1$. Since $x\in l^1$, the partial sums converge and hence this is basically the zero functional. Now, the Hahn-Banach theorem (in some version) asserts that any continuous linear functional dominated by a sublinear function $p$ (i.e. $p(x+y)\leq p(x)+p(y)$ defined on a subspace can be extended to the whole space. Since $\limsup$ and $-\liminf$ are sublinear, the functional on $l^1$ can be extended to $l^\infty$ without violating the inequalities. Is this solution correct? If not, what modifications are necessary?
Yes, I think your argument works.
Here is another approach: Define, for any $x\in\ell^\infty$, functionals $\phi_n$ given by $$ \phi_n(x)=\frac{x_1+\cdots+x_n}n. $$ All these functionals satisfy your property, and moreover $\|\phi_n\|=1$ for all $n$. As the unit ball is weak$^*$-compact, the sequence $\{\phi_n\}$ has cluster points. Any such cluster point will be one of the functionals you are looking for.
The argument above is sometimes referred as a Banach Limit. The same can be achieved using free ultrafilters. Deep down, it is the same.
As Nate mentions, the functionals $\phi_n (x)=x_n $ can be used.