I think that the professor might mean pointwise bounded but not uniformly bounded? Or is there a way that it makes sense to think of $F$ being bounded? $C([0,1])$ refers to the set of continuous functions on $[0,1]$ by the way.
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Boundedness makes sense only with respect to a norm (or at least metric). Of course pointwise means that we investigate for $x\in[0,1]$ the set $\{f(x)\mid f\in F\}\subseteq \mathbb R$; as this is a subset of $\mathbb R$, boundedness with respect to $|\cdot|$ is implied. Boundedness of $F$ itself must refer to a metric on $C([0,1])$ and one may assume indeed that $\lVert\cdot\rVert_\infty$ is implied. So, yes, it would be less ambiguous to explicitly speak of uniform boundedness.
As for the problem itself: We would like to use $f(x)=\frac1x$, but of course cannot. But how can we almost use it?
Hagen von Eitzen
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So as far as I understand, we need one function $\phi\in C([0,1])$ such that for every $f\in F$, $|f(x)|<\phi(x)$ for all $x\in[0,1]$. But if $\phi\in C([0,1])$ then $\phi$ is bounded, say by $M$, so doesn't this imply that each $f\in F$ is bounded by $M$, and so $F$ is uniformly bounded? Or does $\phi$ not need to be in $C([0,1[)$? – crf Mar 01 '13 at 07:56