It is obvious that $\Vert Tx\Vert_\infty\leq\Vert x\Vert_\infty$, hence
$\Vert T\Vert\leq 1$. Since $\Vert T(1,0,0,\ldots)\Vert_\infty=\Vert (1,0,\ldots)\Vert_\infty$, we conclude that $\Vert T\Vert=1$.
One can easily verify that
$$T(l^\infty)=\left\{(\xi_j):\exists C>0 \text{ such that } |\xi_j|\leq C/j\text{ for all j}\right\}$$
$T(l^\infty)$ is not closed in $l^\infty$: Let $x_n=(1,\dfrac{1}{\sqrt{2}},\ldots,\dfrac{1}{\sqrt{n}},0,0,\ldots)=T(1,\sqrt{2},\ldots,\sqrt{n},0,\ldots)$. The sequence $(x_n)$ obviously converges to $x=\left(1/\sqrt{j}\right)_{j=1}^\infty\in l^\infty$, which does not lie in $T(l^\infty)$ (if it did, what would be it's pre-image?).
The inverse operator $T^{-1}$ is not bounded: Consider the sequence $(x_n)\subseteq T(l^\infty)$ as above. This sequence is bounded but the image
$\left\{T^{-1}x_n\right\}$ is not, since $\Vert T^{-1}x_n\Vert_\infty=\sqrt{n}$.
You could also argue in this way: If $T^{-1}$ were bounded, then for every Cauchy sequence $(y_n)\in T(l^\infty)$, the sequence $(T^{-1} y_n)$ would also be Cauchy, and hence would converge to some $x\in l^\infty$ since $l^\infty$ is complete. But then, since $T$ is bounded, $y_n$ would converge to $Tx$. Then $T(l^\infty)$ would be complete, contradicting the fact that it is not closed in $l^\infty$ (this argument can actually be used to show that if $T:X\rightarrow Y$ is an isomorphism between a Banach space $X$ and some normed space $Y$ such that $T$ and $T^{-1}$ are bounded, then $Y$ is Banach).
Since for all $x\in C^1[a,b]$ we have $|f(x)|=|x'(t_0)|\leq \|x\|$ it follows that $\|f\|\leq1$.
In order to prove that in fact $\|f\|=1$ we may assume $a\leq0\leq b$ and $t_0=0$. Consider the functions
$$x_n(t):={t\over 1 + n^2 t^2}\ .$$
Then
$$x_n'(t)={1-n^2 t^2\over (1+n^2 t^2)^2}\ ,$$
and it is easy to see that $|x_n'(t)|\leq x_n'(0)=1$ for all $t\in\Bbb R$. Furthermore we can deduce that $|x(t)|$ takes its maximum value ${1\over 2n}$ at $t=\pm{1\over n}$. It follows that for sufficiently large $n$ we have
$$\|x_n\|=1+{1\over 2n}\ ,$$
so that $f(x_n)=x_n'(0)=1$ implies
$$\lim_{n\to\infty}{|f(x_n)|\over\|x_n\|}=1\ ,$$
as claimed.
Best Answer
Since every $x \in C[a, b]$ is continuous on a compact set, it's bounded.
For any $x \in C[a, b]$, we have:
$$ |f(x)| = \left|\int_a^b x(t) x_0(t) \, dt \right| \le \int_a^b |x(t)| \cdot |x_0(t)| \, dt \le \|x\| \int_a^b |x_0(t)| \, dt $$
Thus, $f$ is bounded and its norm satisfies:
$$ \|f\| \le \int_a^b |x_0(t)| \, dt \newcommand{sgn}{\operatorname{sgn}} $$
In fact, equality holds. To see this, consider the sign function $\hat x(t) = \sgn(x_0(t))$. By Lusin's theorem, there is exists a sequence of functions $x_n \in C[a, b]$ such that $\|x_n\| \le 1$ and $x_n(t) \to \hat x(t)$ as $n \to \infty$ for every $t \in [a, b]$. By the dominated convergence theorem, we have:
\begin{align} \lim_{n \to \infty} f(x_n) &= \lim_{n \to \infty} \int_a^b x_n(t) x_0(t) \, dt \\ &= \int_a^b \lim_{n \to \infty} x_n(t) x_0(t) \, dt \\ &= \int_a^b \sgn(x_0(t)) x_0(t) \, dt \\ &= \int_a^b |x_0(t)| \, dt \end{align}
For the second linear functional you added, approximate the following function via Lusin's theorem in a similar manner:
$$ \hat x(t) = \begin{cases} 1 & \text{if } t \le \frac{a+b}{2} \\ -1 & \text{if } t > \frac{a+b}{2} \end{cases} $$