Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(X, \|.\|)$ be a Banach space such that

  • $X \subset C([0,1]) $
  • For every $r\in \mathbb{Q}\cap[0,1], f\rightarrow f(r)$ defines a bounded linear functional on $X$.

Prove that there exists a $C>0$ such that for all $f\in X$ we have $$\sup_{x\in[0,1]} \|f(x)\| \leq C\|f\|.$$

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

Let for $r\in \Bbb Q\cap [0,1]$, $T_r\colon X\to \Bbb R$ given by $T_r(f):=f(r)$. As $f$ is bounded, $|T_r(f)|\leqslant \sup_{x\in [0,1]}|f(x)|$. By the principle of uniform boundedness, $C:=\sup_{r\in\Bbb Q\cap [0,1]}\lVert T_r\rVert<\infty$. So for $f\in X$, $$\sup_{x\in [0,1]}|f(x)|=\sup_{r\in \Bbb Q\cap [0,1]}|f(r)|=\sup_{r\in \Bbb Q\cap [0,1]}|T_r(f)|\leqslant C\lVert f\rVert_X.$$

share|improve this answer
    
What is $T_r$? dont you need to use that $\mathbb{Q}$ is dense? –  Johan Dec 2 '12 at 13:36
    
I defined it at the first line. Yes, I used density of $\Bbb Q\cap [0,1]$ in $[0,1]$ and continuity of $f$ to get the first equality in the displayed equation. –  Davide Giraudo Dec 2 '12 at 13:37
    
So $T_r : X^* \rightarrow X$? –  Johan Dec 2 '12 at 14:21
1  
No $T_r$ is a linear functional. –  Davide Giraudo Dec 2 '12 at 15:07
add comment

Call the bounded linear functionals $\Lambda_r$.

Apply the Banach-Steinhaus theorem. We either have $E \subset X$ (a dense $G_\delta$) for which

$$ \sup_{r \in \mathbb{Q} \cap [0, 1]} \left|f(r)\right| = \infty $$

for all $f \in E$. Or there exists $M < \infty$ such that

$$ \|\Lambda_r\| \le M $$

for all $r \in \mathbb{Q} \cap [0, 1]$. Since $X$ is a subset of $C([0, 1])$, all of its functions are bounded and the first case is impossible. Hence

$$ \left|f(r)\right| \le M \|f\| $$

for all $r \in \mathbb{Q} \cap [0, 1]$ and for all $f \in X$. Since the set $\mathbb{Q} \cap [0, 1]$ is dense in $[0, 1]$ and all functions in $X$ are continuous, the desired result follows.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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