= Solution
Write $M=\sup_{(0,1)}|f|<\infty$ and use the same constant extensions $f_n$ as in part (a). They satisfy $\|f_n\|_\infty\le M$. Choose <polynomials> $Q_n$ with $\|Q_n-f_n\|_\infty<1/n$ for $n\ge3$, and set the first two <polynomials> equal to zero. Then $\boxed{\sup_n\sup_{0<x<1}|Q_n(x)|\le M+1}$, 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>.
Back to article page
