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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.

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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 $$|x(t_i)|=\|x\|$$

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  • $\begingroup$ I think you mean $\omega_i := c_j / |c_j|$? $\endgroup$ – angryavian Oct 26 at 16:53
  • $\begingroup$ @angryavian yes, ty. Fixed. $\endgroup$ – N. S. Oct 26 at 16:58
  • $\begingroup$ 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 $c_j$? $\endgroup$ – J.Rodriguez Oct 26 at 17:08
  • $\begingroup$ @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. $\endgroup$ – N. S. Oct 26 at 17:16
  • $\begingroup$ ohh god, how coudn't i see it, thanks. $\endgroup$ – J.Rodriguez Oct 26 at 17:17

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