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
Suppose $Tx=y$, we have: $$y(t)=(Tx)(t)= \int_0^t x(r) dr$$ Since $x$ is continuous, $y$ is differentiable and we have: $$x=y'$$ So $$T^{-1}(y)=y'$$
About range of $T$, it's clear that each $Tx$ is differentiable and $$(Tx)(0)=0$$ Let $y\in C[0,1]$ be a differentiable function such that $$y(0)=0$$ we have: $$(Ty')(t)=\int_0^t y'(r) dr=y(t)$$ provided this integral exists. So which integral do you use? e.g if this definition is used then the integral always exists and the range of $T$ is all differentiable functions that are 0 at 0. About Riemann integration I think the range is more limited.