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Let $||\cdot||$ be such complete norm on $C[0,1]$, that for every $q \in \mathbb{Q}\cap[0,1]$ there exists such $M_q \geqslant 0$, that $|f(q)| \leqslant M_q ||f||$ for every $f \in C[0, 1]$. Show that there exists constant $M \geqslant 0$ such that $||\cdot||_\sup \leqslant M||\cdot||$.

We may assume that $M_q = \sup_{||f|| = 1}|f(q)|$. We can extend this definition to all numbers $x \in [0, 1],$ defining $M_x = \sup_{||f||=1}|f(x)| \in \mathbb{R} \cup \{+\infty\}$. We want to show that in fact $M_x < +\infty$ for all $x \in [0, 1]$ and the mapping $([0, 1] \ni x\mapsto M_x \in \mathbb{R})$ is bounded. So it suffices to prove that it is continuous. But is it really the case? And how can I use the assumption that the norm is complete?

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You need the completeness assumption because this is an exercise on the Uniform Boundedness Principle. Consider the collection of functionals $\phi_q(f) = f(q)$, $q\in\mathbb{Q}$. Each of them is bounded in the norm $\|\cdot \|$. Furthermore, for every $f\in C[0,1]$ the set $\{\phi_q(f):q\in\mathbb{Q}\}$ is bounded by $\sup|f|$. By the Uniform Boundedness Principle, there exists a constant $M$ such that $\|\phi_q\|\le M$ for all $q\in \mathbb{Q}$. (The norm here is the norm induced by $\|\cdot\|$.)

The rest goes by the density of rationals: for every $f\in C[0,1]$, $$\sup|f| = \sup_{q\in\mathbb{Q}} |\phi_q(f)| \le M\|f\|$$ as claimed.