# Finding the norm of this operator

I have this space $$X=C([-1,1],\mathbb{C})$$ the space of continous functions from $$[-1,1] \to \mathbb{C}$$, let $$t_1,\dots,t_n \in [-1,1]$$ and $$c_1,\dots c_n\in \mathbb{C}$$. Consider the operator: $$f: X \to \mathbb{C}$$ as: $$f(x)=\sum_{j=1}^n c_j x(t_j).$$ I have to prove that $$f$$ is a bounded operator from $$X$$ to $$\mathbb{C}$$ and find $$||f||$$. I have managed to prove that it is bounded, but I am having problems finding $$||f||$$. Recall that the norm in $$X$$ is the $$\sup$$ norm. If we denote $$\gamma_j=||c_j||$$: $$\sup_{x\in X}\frac{||f(x)||}{||x||}=\sup_{x\in X}\frac{||\sum_{j=1}^n c_j x(t_j)||}{||x||}=\sup_{x\in X} \sum_{j=1}^n \gamma_j \frac{||x(t_j)||}{||x||}$$ Clearly $$|| f||$$ depends explicitly of $$\gamma_j$$, but I don't know well how to continue and somehow show a more explicit number. Any ideas would help.

Hint 1: You made a mistake, the following equality is not true:

$$\sup_{x\in X}\frac{||\sum_{j=1}^n c_j x(t_j)||}{||x||}\neq \sup_{x\in X} \sum_{j=1}^n \gamma_j \frac{||x(t_j)||}{||x||}$$

Hint 2: For all $$c_j \neq 0$$ denote by $$\omega_i:= \frac{c_j}{||c_j||}$$.

Then $$\|\sum_{j=1}^n c_j x(t_j)\| \leq \sum_{j=1}^n| c_j| | x(t_j)| \leq \sum_{j=1}^n| c_j| \| x\|$$

The first inequality is equality exactly when $$\overline{\omega_j} x(t_i)=|x(t_i)|$$ while the second is equality exactly when $$|x(t_i)|=\|x\|$$

• I think you mean $\omega_i := c_j / |c_j|$? – angryavian Oct 26 at 16:53
• @angryavian yes, ty. Fixed. – N. S. Oct 26 at 16:58
• Thanks, from here I can follow, but still don't get why the equality is wrong, is not just a re-writing using the norm of each $c_j$? – J.Rodriguez Oct 26 at 17:08
• @J.Rodriguez The triangle inequality says $$||\sum_{j=1}^n c_j x(t_j)|| \leq \sum_{j=1}^n|| c_j x(t_j)||$$ and it is easy to construct examples where the inequality is strict. You moved the sum outside the norm, when you do that you don't always get equality. – N. S. Oct 26 at 17:16
• ohh god, how coudn't i see it, thanks. – J.Rodriguez Oct 26 at 17:17