Show that the functional is sublinear $$ \begin{array}{rlll} p:\ell^{\infty}(\mathbb{C})&\longrightarrow&\mathbb{R}\\ \{t_{n}\}&\longmapsto&p\big(\{t_{n}\}\big):=\lim\limits_{n\rightarrow+\infty}\sup\textrm{Re}(t_{n}) \end{array} $$ is sublinear, where $$ \ell^{\infty}(\mathbb{C})=\left\{\{t_{n}\}_{n\in\mathbb{N}}\subsetneq\mathbb{C}:\sup\limits_{n\in\mathbb{N}}|t_{n}|<+\infty\right\} $$
Show that there is a linear function $f:\ell^{\infty}(\mathbb{C})\rightarrow \mathbb{R}$ such that for all $\{t_{n}\}_{n\in\mathbb{N}}\in\ell^{\infty}(\mathbb{C})$ $$ \lim\limits_{n\rightarrow+\infty}\inf\textrm{Re}(t_{n})\leq f\big(\{t_{n}\}\big)\leq\lim\limits_{n\rightarrow+\infty}\sup\textrm{Re}(t_{n}) $$ Also, show that $f$ is necessarily bounded.
What I do not understand is, what Hahn Banach theorem's analytic-algebraic form should be applied, the real case or complex case?
What is it that I could consider, $W:=\{0\}$ together with the functional $$ \begin{array}{rlll} g:W&\longrightarrow&\mathbb{R}\\ \{t_{n}\}&\longmapsto&g\big(\{t{n}\}\big):=\lim\limits_{n\rightarrow+\infty}t_{n} \end{array} $$ but I do not know how to continue
Let $$ C = \{ (c_n)_{n\in \mathbb{N}} \in \ell^\infty: \lim_{n \to \infty} Re (c_n) \text{ exists }\}. $$ One can show that $C$ is a subspace of $\ell^\infty$. We define $\Lambda : C \to \mathbb{R}$ by $\Lambda(c) = \lim_{n\to \infty} Re (c_n)$. As you can see $\Lambda$ satisfies all your requirements, but the domain of definition is not large enough. I think this is also what you were trying to do at the end of your post. To proceed you now need to extend $\Lambda$ to all of $\ell^\infty$. Naturally, one wants to use the theorem of Hahn-Banach here. A suitable sub-linear functional on $\ell^\infty$ governing $\Lambda$ is certainly the functional $p$ that you defined at the beginning. Clearly for all $c, c' \in \ell^\infty$ $$ p(c + c') = \lim \sup Re(c_n + c_n') \leq \lim \sup (Re(c_n) + Re(c_n')) $$ and $$ \lim \sup (Re(c_n) + Re(c_n')) \leq \lim \sup (Re(c_n) + \lim \sup (Re(c_n') = p(c) + p(c'). $$ Moreover, by construction $p(\lambda c) = \lambda p(c)$ for all positive $\lambda$. Hence $p$ is sublinear. Applying Hahn-Banach to $\Lambda$ and $p$ yields a functional $\widetilde{\Lambda}$ on $\ell^\infty$ such that $$ \widetilde{\Lambda}(c) \leq \lim_{n\to \infty} \sup Re(c_n) $$ for all $c \in \ell^\infty$. Especially, this holds for $-c$ and by linearity this implies that $$ -\widetilde{\Lambda}(c) = \widetilde{\Lambda}(-c)\leq \lim_{n\to \infty} \sup Re(-c_n) = -\lim_{n\to \infty} \inf Re(c_n) $$ we conclude that $\widetilde{\Lambda}(c) \geq \lim_{n\to \infty} \inf Re(c_n)$ for all $c \in \ell^\infty$. Finally, it is clear by definition that $$ |\widetilde{\Lambda}(c)| \leq |\lim_{n\to \infty} \sup Re(c_n)| \leq \sup_{n\in \mathbb{N}} |c_n| = \|c\|_{\ell^\infty}, $$ for all $c \in \ell^\infty$. This yields the boundedness.
I hope this helps. Please tell me if some steps need clarification.