= Solution
The <Arzelà-Ascoli theorem> for $C([0,1])$ states that a family has compact closure in the uniform <norm> exactly when it is uniformly bounded and equicontinuous. To prove sufficiency, choose a countable dense set. From any <sequence> of functions, successive subsequence extraction at each dense point and a diagonal selection give convergence at all those points. Given $\varepsilon>0$, <equicontinuity> gives a common $\delta$; cover the interval by finitely many $\delta$-balls centered at dense points. On a sufficiently late tail the functions differ by less than $\varepsilon/3$ at those finitely many centres, and <equicontinuity> makes each of the other two differences less than $\varepsilon/3$. Thus the selected subsequence is uniformly Cauchy. Completeness of $C([0,1])$ gives a continuous uniform limit. In a metric space this proves <compactness> of the closure. Conversely, a compact family has finite uniform $\varepsilon$-nets. The finitely many net functions are bounded and uniformly continuous; approximation by them gives a common bound and <equicontinuity> for the family. This proves both directions.
For $f\in S_N$, a positive interior maximum would have $f'=0$ and $f''\leq0$, contradicting $f''=f>0$. A negative interior minimum likewise contradicts $f''=f<0$. The endpoint bounds therefore give $|f|\leq N$ everywhere.
Now put $w=f'$. Differentiating the equation, using the stipulated third derivative, gives $w''=w+2ww'$. At an interior extremum of $w$, $w'=0$, so $w''=w$. The same maximum/minimum argument and the derivative endpoint bounds imply $|f'|=|w|\leq N$. Consequently $|f(x)-f(y)|\leq N|x-y|$ by the <mean value theorem>. Thus $S_N$ is uniformly bounded and equicontinuous, and the theorem makes it relatively compact, hence \b[totally bounded in $C([0,1])$]. Its closure need not be asserted to remain thrice differentiable.
Back to article page
