Let $ f,g : [0,1] -> \mathbb{R}$ are continuous one-one function such that $Range (f)\supseteq Range (g)$. Then
(a) There exist a sequence of polynomials $P_n$ such that $f(P_n)$, converges uniformly to $g$ on $[0,1]$.
(b) There exist a sequence of polynomials $P_n$ such that $g(P_n)$ converges uniformly to $ƒ$ on $[0,1]$
(c.) There does not exist sequence of polynomials $P_n$ such that $f(P_n)$ converges uniformly to $g$ on [0,1]
(d) There does not exist sequence of polynomials $P_n$ such that $g(P_n)$ converges uniformly to $ƒ$ on $[0,1]$
I tried using Weirstrass approximation theorem which states " There exists a sequence of polynomials converging to a continuous functions $ f: [a,b]-> \mathbb{R} $ "
So $P_n$ converges uniformly to both f and g. But I don't have any idea about uniform convergence after composition of functions.I need help with all options
First, note that there is no guarantee that we can find a sequence such that $g(P_n) \rightarrow f$: for example, take $f(x)=x$ and $g(x)=0$.
Now, to show that the other statement holds, we want to find a continuous function $h \colon [0,1] \rightarrow \mathbb{R}$ such that $f \circ h = g$, after which we may apply the Weierstrass approximation theorem. Since $f$ is a homeomorphism onto its image (this is because $[0,1]$ is compact and $\mathbb{R}$ is Hausdorff), there is a well-defined continuous function $f^{-1} \colon Range(f) \rightarrow [0,1]$ such that $f \circ f^{-1} = id_{Range(f)}$.
Now, since $Range(f) \supseteq Range(g)$, the composition $h= f^{-1} \circ g$ makes sense and is the composition of continuous functions, hence continuous. Furthermore, $f \circ h = f \circ f^{-1} \circ g = g$.