Let $X$ be a normed vector space. Let $f: X \rightarrow \mathbb{R}$ be a bounded linear functional. We define: $$\|f\|_{X^*} = \sup_{x \in X \\ x \neq 0} \dfrac{|f(x)|}{\|x\|_X}$$ Prove that:
$$\|f\|_{X^*} = \sup_{\|x\| \leq 1} |f(x)| = \sup_{\|x\| = 1} |f(x)| = \sup_{\|x\| \leq 1} f(x) = \sup_{\|x\| = 1} f(x).$$
Assume $x\ne 0$ and $X$ be of finite-dimensional, if not then please comment.
Since $f$ is bounded linear functional so, $$|f(x)|\le ||f||\cdot||x||\implies \sup_{||x|\le 1}|f(x)|\le ||f||.....(1)$$
Next, note that one of the definition of $\displaystyle||\cdot||_{X^*}$ is $||f||=\inf_{M\in \mathcal M}\{M\}$. Where $\mathcal M =\{M>0:|f(x)||\le M ||x||,\forall x\in X\}$.
Now, $|f(x)|=|f(\frac{x}{||x||}||x||)|=||x||\cdot|f(\frac{x}{||x||})|\implies f(\frac{x}{||x||})\in \mathcal M$
So, $||f||\le f(\frac{x}{||x||})$ for all nonzero $x\in X$. Hence, $||f||\le \sup_{||x||\le 1}|f(x)|.....(2)$, the norm of $x$ is omitted as the maximum is attained at $||x||=1$
So $(1),(2)$ gives the equality. Moreover, in every finite dimensional space, any two norms are equivalent [Proof here], so this completes the proof. Hope it works.