$F = \{f\in C^1([0,1])| \hspace{2mm} \|f\|\leq M, \|f'\|\leq N\}$. Showing it is precompact and not closed.

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I have an example in my book: Let $C([0,1])$ denote the space of all continuous functions $f$ on $[0.1]$ with continuous derivative $f'$. For constants $M>0$ and $N>0$, we define the subset $F$ of $C([0,1])$ by
$$F = \{f\in C^1([0,1])\mid \|f\|\leq M, \|f'\|\leq N\}.$$

Where $\|\cdot\|$ denotes the $\sup$-norm.The book claims that $F$ is precompact in $C(K)$ and it is not closed because the uniform limit of continuously differentiable functions need not be differentiable.

My questions are: 1) How can I prove this set is precompact? I know I have to show it is bounded and equicontinuous.

2) Is there an example of such a function that can show this is not closed?

This is not a homework exercise. I am trying to fill in the blanks that my text book has not filled in so I can have a better understanding.

Thank you in advance for any help, advice and comments.

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1) You have a uniform bound on the derivatives, so equicontinuity follows easily from the mean value theorem. As for boundedness, it should be pretty clear from the definition of $F$.

2) Consider sequences of rapidly oscillating functions, e.g. $f_n (x) = g(n) \sin (h(n) x)$. By choosing $g,h$ appropriately you should be able to make $f_n$ converge uniformly to the zero function while $\Vert f_n ' \Vert$ stays constant.