Bounded functionals on Banach spaces. 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
$$\sup_{x\in[0,1]} \|f(x)\| \leq C\|f\|.$$
 A: 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 $f\in X$, 
$$\sup_{x\in [0,1]}|f(x)|=\sup_{r\in \Bbb Q\cap [0,1]}|f(r)|=\sup_{r\in \Bbb Q\cap [0,1]}|T_r(f)|\leqslant C\lVert f\rVert_X.$$
A: 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.
