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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:

  1. 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.
  2. 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?

copper.hat
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Stanislav Hronek
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    If $p$ is a second degree polynomial, and $q(x)=x^3$ then it it not difficult to see that $|(p+{1 \over n} q)-p|_1 \to 0$. – copper.hat Feb 28 '22 at 21:57
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    It is not difficult (think Vandermonde) to exhibit an invertible linear map $L$ between $P_2$ and $\mathbb{R}^3$. Suppose $p_n \to p$ with $p_n \in P_2$, then $Lp_n$ is Cauchy in $\mathbb{R}^3$ and so $Lp_n \to x^$ (in $\mathbb{R}^3$). Then $|p_n-L^{-1}x^| \le |L^{-1}| |Lp_n -x^|$ and so $L^{-1}x^ = p$ and so $p \in P_2$. – copper.hat Feb 28 '22 at 22:33
  • In fact any finite dimensional subspace of a normed linear space is always closed. The answer you have approved does not even address the question in your title. – Kavi Rama Murthy Mar 04 '22 at 04:50

2 Answers2

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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.

Kavi Rama Murthy
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  • I don't understand your argument. It seems to me that your are proving that $P_2$ is a closed subset of $P_2$ itself. Since the space of all polynomial functions is infinite-dimensional, I don't see how is it that you can use the fact that on any finite dimensional normed linear space any linear functional is continuous. – José Carlos Santos Mar 03 '22 at 15:21
  • @JoséCarlosSantos I take a sequence $(p_n)$ in $P_2$ converging to a polynomial $f$ in $L^{1}$ norm. I consider the map $T: P_2 \to \mathbb R$ defined by $T(ax^{2}+bx+c)=a$. This map is continuous for the $L^{1}$ norm by finite dimensionality. It follows that $Tp_n$ is Cauchy sequence of real numbers , hence convergent. Similarly, the other two coefficients of $p_n$'s also converge. It follows that $p_n$ converge some element of $P_2$ and $f$ must be equal to this polynomial. This is all just basic FA and I hope there is no mistake in my argument. – Kavi Rama Murthy Mar 03 '22 at 23:17
  • I understand that. Note that you are using here implicitly the fact that every linear continuous map is uniformly continuous. Besides, your answer (unlike your comment) makes no reference to Cauchy sequences, and I doubt that the OP would think spontaneously about them. – José Carlos Santos Mar 04 '22 at 07:44
  • Thank you both for the answers and for the discussion everything is clear now. – Stanislav Hronek Mar 04 '22 at 18:07
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That space is connected. Therefore, since $P_2$ is neither $\emptyset$ nor the whole space, it cannot be both closed and open.

And it is in fact not open: $0\in P_2$ and$$(\forall n\in\Bbb N):\rho(0,x^n)=\frac1{n+1}.$$So, there are non-quadratic polynomials with degree arbitrarily close to $0$.

As you were told in the comments, $P_2$ is closed, since every finite-dimensional subspace of a normed vector space is closed.

José Carlos Santos
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