Finding the norm of an operator. Here is the question:
Let $X = C[0,1]$ with $\|.\|_{\max}$ norm, let $\{ \alpha_{k}\}_{k=1}^{n}$ be real numbers, let $\{x_{k}\}_{k=1}^{n} \subset [0,1],$ and define $$ T f = \sum_{k=1}^{n} \alpha_{k} f(x_{k}). $$ Prove that $T$ is a bounded linear functional on $C[0,1]$ and find its norm. 
I managed to prove that $T$ is linear and bounded, where $$\|T\| \leq \sum_{k=1}^{n}|\alpha_{k}|$$\
**My questions are: ** 
1-is my bound correct ?
2- How can I find the norm? I do not know if my function increasing $f$ or decreasing, I do not know how to think to find the norm, any help will be appreciated.   

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. 
