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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 $$\sup_{x\in[0,1]} \|f(x)\| \leq C\|f\|.$$

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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 $f\in X$, $$\sup_{x\in [0,1]}|f(x)|=\sup_{r\in \Bbb Q\cap [0,1]}|f(r)|=\sup_{r\in \Bbb Q\cap [0,1]}|T_r(f)|\leqslant C\lVert f\rVert_X.$$

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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
No $T_r$ is a linear functional. – Davide Giraudo Dec 2 '12 at 15:07

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.

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