I'm having trouble trying to understand what does means the first expression in particular the last term in it
should we add $\|f\|_{\infty} \leq \infty$ or what i can't see what is his role ($\|f\|_{\infty}$) here
\begin{align*} {C}(\mathbb{R}^{n})&=\{f:\mathbb{R}^{n}\longmapsto \mathbb{R}: \text{continue},\lim_{\|{x}\|\to \infty }f(x)=0,\|{f}\|_{\infty}\}\\ {C}_{0}(\mathbb{R}^{n})&=\{ f \in {C}: \text{of compact support } \mathbb{R}^{n} \} \\ Lip_{0}(\mathbb{R}^{n})&=\{f \in C_{0}: \exists M: |f(x)-f(y)|\leq M|x-y| \} \\ \overline{Lip_{0}}(\mathbb{R}^{n})&={C}(\mathbb{R}^{n}) \\ \mathcal{D}(\Omega)=\mathcal{C}_{c}^{\infty}(\mathbb{R})&=\{f \in \mathcal{C}^{\infty} : \text{ of compact support } \} \end{align*}
would you please also explain how $\lim_{\|{x}\|\to \infty }f(x)=0,$ implies $\|{f}\|_{\infty} \leq \infty$
Well, to answer your question:
let $$f:\Bbb R^n\to \Bbb R, \quad \text{continuous},\quad \lim_{|x|\to\infty}f(x)=0.$$
Fix $\varepsilon >0$; then by definition of the limit there exists $R>0$ such that $|f(x)|<\varepsilon$ whenever $|x|>R$. On the compact set $\{|x|\le R\}$ the function $|f|$ is continuous, hence attains its supremum and infimum (Veierstrass theorem). Therefore, $$\|f\|_\infty\le \max(\varepsilon, \sup_{|x|\le R}|f(x)|)<\infty.$$
As for notations, it's a bit weird to see $C(\Bbb R^n)$ defined as a space of continuous functions vanishing at infinity. In most cases, this notation is reserved for the space of continuous functions. Then you can define $C_b$ as the space of continuous bounded functions, $C_0$ as a space of continuous vanishing functions, $C_c$ as a space of continuous function with compact support, etc.