 Solution

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.