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).
I don't think you can find $x$. the idea is that the perfect $x$ would be the function thqt has value -1 on $[-1,0[$ and 1 on $[0,1]$. The problem is that this function is not continuous...
Never mind! try to approximate it with well chosen partially linear functions. You don't need to find an $x\in C[-1,1]$ such that $\frac {\vert f(x)\vert}{\Vert x \Vert}\geqslant 2$.
It suffices to prove that, for any $\epsilon>0$, you can design a continuous function $x$ such that $\frac {\vert f(x)\vert}{\Vert x \Vert}\geqslant 2-\epsilon$.
Best Answer
Hint: you know that $\int_{-1}^1 tH(t)dt=1$ where
$H(t)=\begin{cases} 1\quad \quad \quad 0\le x\le 1 &\\ -1\quad -1\le x<0\end{cases}$
but $H$ is not continuous, so try to find a sequence of continuous functions that converge to $H$ on $[-1,1]$. Since the problem with $H$ is the jump at $x=0,$ back off a little and define
$H_n(x)=\begin{cases} 1\quad \quad \quad 1/n\le x\le 1 &\\ ?? \quad -1/n\le x\le 1/n&\\-1 \quad -1\le x<-1/n\end{cases}.$