Calculating operator norm of operator $(Ax)(k) = \int_{0}^{1} x(t)\operatorname{arcctg}\left((-1)^kt\right) \,\mathrm{d}t$

Question

I have no idea how to solve this:

$$A: CL_1[0,1] \rightarrow \ell_\infty$$

$$(Ax)(k) = \int_{0}^{1} x(t)\operatorname{arcctg}\left((-1)^kt\right) \,\mathrm{d}t,\quad\forall x \in CL_1[0,1],\quad\forall k \in \mathbb{N}.$$

I need to calculate the norm of A

UPD: By $CL_1$ i mean the space of continuous functions on the [a,b] with the norm $||f|| = \max_{x \in [a, b]} |f(x)|$

As I understood,

$$\forall x \in \ell_\infty\quad\lVert x\rVert_\infty = \sup_n\lvert x_n\rvert$$

$$\forall x \in CL_1[0,1] \rightarrow \lVert x\rVert_c = \max_{t\in[0,1]}|x(t)| < + \infty$$

$$\implies \lvert(Ax)(k)\rvert \leq 1 \cdot \|x\|_c,\quad\forall k \in \mathbb{N}.$$

But I don't know what to do globally.

Answer 1

$\newcommand{\dt}{\mathrm dt}$

I'll use a different notation: $$A:(C([0,1]), \| f\|_{\infty})\to (\ell^{\infty}, \| (x_k)_{k\in \Bbb N}\|_{\infty}) $$ where $((Af)_k)_{k\in \Bbb N}$ is defined as $(Af)_k=\int_0^1f(t)\operatorname{arccot}((-1)^kt)\dt$.

Note that: $$\begin{split} \|((Af)_k)_{k\in \Bbb N} \| _{\infty}&=\sup_{k\in \Bbb N} \left|\int_0^1f(t)\operatorname{arccot}((-1)^kt)\dt\right|\\ &\le \sup_{k\in \Bbb N} \int_0^1|f(t)||\operatorname{arccot}((-1)^kt)|\dt \\ &\le\|f\|_{\infty}\sup_{k\in \Bbb N} \int_0^1|\operatorname{arccot}((-1)^kt)|\dt \end{split}$$ Can you take it from here?