Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 2F (Topics in Analysis)

For $n\ge 3$, extend the restriction of the continuous function $f$ to $[1/n,1-1/n]$ to a continuous function $f_n$ on $[0,1]$ by making it constant on each of the two remaining intervals. The Weierstrass approximation theorem supplies a polynomial $P_n$ with $\Vert P_n-f_n{\Vert }_{\infty ,[0,1]}<1/n$. Every compact set $K\subset (0,1)$ is eventually contained in $[1/n,1-1/n]$, where $f_n=f$. Hence ${\sup }_K|P_n-f|<1/n$ eventually. This proves uniform convergence on every compact subset, without assuming any endpoint limits of $f$. The first two polynomials can be chosen arbitrarily.

Write $M={\sup }_{(0,1)}|f|<\infty$ and use the same constant extensions $f_n$ as in part (a). They satisfy $\Vert f_n{\Vert }_{\infty }\le M$. Choose polynomials $Q_n$ with $\Vert Q_n-f_n{\Vert }_{\infty }<1/n$ for $n\ge 3$, and set the first two polynomials equal to zero. Then $\boxed{\underset{n}{\sup }\underset{0<x<1}{\sup }|Q_n(x)|\le M+1}$, while the argument of part (a) proves uniform convergence to $f$ on every compact subset of $(0,1)$. The bound concerns the entire interval, rather than just each separate compact subset.