Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 2F (Topics In Analysis)

False. If uniform convergence held, some polynomial $P_n$ would satisfy ${\sup }_{t\in \mathbb{R}}|P_n(t)-\sin t|<1$. Such a polynomial would be bounded on $\mathbb{R}$, hence constant: a nonconstant polynomial grows without bound in absolute value along a real half-line. For any constant $c$, however,a contradiction.

True. By the Weierstrass approximation theorem, choose a polynomial $Q_n$ with ${\sup }_{|t|\le n}|Q_n(t)-f(t)|<1/n$. For each fixed $t$, this bound applies once $n\ge |t|$, so $Q_n(t)\to f(t)$. This proves pointwise convergence; the same construction even gives uniform convergence on every closed interval.

True. Define $h(u)=g(1/u)$ for $0<u\le 1$ and $h(0)=0$. The hypothesis at infinity makes $h$ continuous on $[0,1]$. The Weierstrass approximation theorem gives polynomials $R_n$ with ${\sup }_{0\le u\le 1}|R_n(u)-h(u)|\to 0$. Since $t\mapsto 1/t$ maps $[1,\infty )$ onto $(0,1]$,