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

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.