The question and its solution are given in the following link:
Finding the norm of an operator.
Here is the solution:
Yes, your bound is correct. Let $X = C[0,1]$ with the $\sup$-norm and let $\{\alpha_k\}_{k = 1}^n \subset [0,1]$ as in the question. Let $f \in X$ be arbitrary with $\Vert f \Vert = 1$, then
$$ \vert Tf \vert = \vert \sum_{k = 1}^n \alpha_k f(x_k) \vert \leq \sum_{k = 1}^n \vert \alpha_k \Vert f \Vert_{\infty} \vert = \sum_{k = 1}^n \vert \alpha_k \vert $$
and hence $$\Vert T \Vert = \sup_{f \in X, \Vert f \Vert = 1} \vert Tf \vert \leq \sum_{k = 1}^n \vert \alpha_k \vert ~~.$$
In fact, this is the norm of the functional, since $\sum_{k = 1}^n \vert \alpha_k \vert$ is attained by the following function $f$. Define
$$ f(x) = \begin{cases} 1 & x = x_k, \alpha_k \geq 0 \\ -1 & x = x_k, \alpha_k < 0 \\ 0 & x \in \{0,1\} \setminus \{x_k\}_{k =1}^n \end{cases}$$
and linear interpolation otherwise. Then this function is continuous on $[0,1]$, it has $\Vert f \Vert = 1$, and we have
$$ \vert Tf \vert = \vert \sum_{k = 1}^n \alpha_k f(x_k) \vert = \sum_{k = 1}^n \vert \alpha_k \vert$$
where in the last step we use that all summands are positive. Hence $\Vert T \Vert \geq \sum_{k = 1}^n \vert \alpha_k \vert$, and thus we have equality.
My question is:
Why in the second line from below in the solution we have:
$$\vert \sum_{k = 1}^n \alpha_k f(x_k) \vert = \sum_{k = 1}^n \vert \alpha_k \vert$$
What $f$ is doing is basically swapping the sign of $\alpha_k$ if it is negative. In other words, $$\alpha_k f(x_k) = |\alpha_k|$$ for each $1\leqslant k\leqslant n$. Thus equality follows immediately since all the summands are positive.