Given the metric space of polynomials over the unit interval $[0,1]$ with the metric
$$
\rho(P,Q) = \int\limits_{0}^{1} \vert P(x) -Q(x) \vert {\rm d}x
$$
and given a subset $P_2= \{ a+bx+cx^2, x\in [0,1], a,b,c \in \mathbb{R}\}$ of degree 2 polynomials.
The question is this subset open or closed or both?
My guess would be that both, for the following reasons:
- I cannot think of any polynomial that would be in the boundary of this set. Finding any polynomial that is not inside this set and has a zero distance to this set would be in contradiction with the fact that this is a metric space. I mean that we would not satisfy this rule: $\rho(x,y)=0\quad \text{iff}\quad x=y$. Meaning that this set equals its closure and is therefore closed.
- When I think about the complement of this set (all generic polynomials such that $a=b=c=0$) I also tend to think this set is closed because of the same argument as above again using that the boundary of this set is empty.
Are there some faults in my logic or something that I missed? This is all quite vague and intuitive so any idea how I would properly prove this?
Best Answer
The set is closed. On any finite dimensional normed linear space any linear functional is continuous. Consider the maps $a+bx+cx^{2} \to a, a+bx+cx^{2} \to b$ and $a+bx+cx^{2} \to c$. These are continuous (for any norm, in particular the $L^{1}$ norm). Hence convergence of $a_n+b_nx+c_nx^{2}$ to some function $f$ in $L^{1}$ norm implies convergence of the coefficient sequences, so the limit function $f$ is necessarily a polynomial of degree at most $2$. [If a linear map $T$ is continuous then $\|Tp_n-Tp_m\| \leq \|T\|\|p_n-p_m\| \to 0$ for any Cauchy sequence $(p_n)$. This gives convergence of he coefficients since any Cauchy sequence of real numbers is convergent].
It cannot be open because the space $P_2$ is connected.
In fact any finite dimensional subspace of a normed linear space is always closed.