Let $f_1$, $f_2$ be the functionals defined on the normed space $C[a,b]$ of all continuous functions defined on the closed interval $[a,b]$ with the maximum norm be defined as follows: $$f_1(x) \colon= \max_{t\in[a,b]} x(t), \; \; \; f_2(x) \colon= \min_{t\in[a,b]} x(t) \; \; \; \forall x \in C[a,b].$$
Then $f_1$ and $f_2$ are not linear because, for each $i = 1, 2$, $$f_i(-x) \not= -f_i(x) \; \; \; \forall x (\not= 0) \in C[a,b].$$ Am I right?
But what about boundedness of $f_2$?
I know that $f_1$ is bounded and that $\Vert f_1 \Vert = 1$. Let $t_0 \in [a,b]$. Then we have $$ x(t_0) \leq \vert x(t_0) \vert \leq \max_{t\in[a,b]} \vert x(t) \vert = \Vert x \Vert.$$ So $$\max_{t\in[a,b]} x(t) \leq \Vert x \Vert.$$ Also, $$-\max_t x(t) = \min_t \left(-x(t)\right) \leq -x(t_0) \leq \vert x(t_0) \vert \leq \max_t \vert x(t) \vert = \Vert x \Vert.$$ So $$\vert f_1(x) \vert = \vert \max_{t\in[a,b]} x(t) \vert \leq \Vert x \Vert \; \; \; \forall x \in C[a,b].$$ Thus it follows that $f_1$ is bounded with $\Vert f_1 \Vert \leq 1$.
For $x_0 \in C[a,b]$ defined as $x_0(t) \colon= 1$ for all $t \in [a,b]$, we see that $\vert f_1(x) \vert = \Vert x \Vert = 1$ so that $\Vert f_1 \Vert = 1$.
Now comes the turn of $f_2$: $$\min_{t\in[a,b]} x(t) \leq x(t_0) \leq \vert x(t_0) \vert \leq \max_{t\in[a,b]} \vert x(t) \vert = \Vert x \Vert $$ for all $x \in C[a,b]$. But $$ -\min_{t\in [a,b]} x(t) = \max_{t\in[a,b]} \left( - x(t) \right) ... $$
How to proceed from this point? That is, how to show that, for all $x \in C[a,b]$, whether or not the inequality $$ \vert f_2(x) \vert \leq \Vert x \Vert$$ holds? And if this inequality deos indeed hold, then how to determine if $\Vert f_2 \Vert$ is indeed equal to $1$?
$$ \left|\max_{t\in[a,b]}x(t)\right| \leq \max_{t\in[a,b]}\left|x(t)\right| = \|x\|_{C[a,b]}\\ \left|\min_{t\in[a,b]}x(t)\right| = \left|-\max_{t\in[a,b]}-x(t)\right| \leq \max_{t\in[a,b]}\left|x(t)\right| = \|x\|_{C[a,b]} $$ So $\|f_1\| \leq 1$ and $\|f_2\| \leq 1$. Assume $x_0 = 1, \|x_0\| = 1$. $f_1(x_0) = 1 = f_2(x_0)$. Hence $\|f_1\| = \|f_2\| = 1$.