continuity in normed spaces

Question

Please I want the proof of continuity of the map $f: \Bbb Rn[X] \longrightarrow\Bbb R^n$ we associate to each space a norm (euclidean) where Rn[X] is identified to the to functions like $g(x)= x^n +an*x^{n-1}...$ and can be written like $(x-zn)*(x-z(n-1))...$ in Rn[x] where zi are roots sending each monic polynomial like this to its set of ordered roots?

Answer 1

Sorry, I have a better way. I'm still assuming the roots are distinct. First, a lemma:

Lemma: Suppose $p_m \in P_n(\Bbb{R})$ is a sequence of polynomials of degree $n$ or less, which converges to $p \in P_n(\Bbb{R})$ in the norm topology. Further, let $I$ be a compact interval. Then $p(I)$ and $p_m(I)$ are compact intervals for any $m$, and \begin{align*} \sup p_m(I) &\to \sup p(I) \\ \inf p_m(I) &\to \inf p(I). \end{align*}

Proof. Since $P_n(\Bbb{R})$ is a finite-dimensional space, every norm will generate the same topology.

Consider first the case where $I$ is a singleton. Then, the result is equivalent to the evaluation maps $p \mapsto p(x_0)$ being continuous, which they are, since they are linear maps between finite-dimensional spaces.

Otherwise $I$ is an infinite set. Consider the norm $\| \cdot \|_I$ defined by $$\|p\|_I := \sup_{x \in I} |p(x)|.$$ This is clearly positive semi-definite and sublinear. It will additionally be positive-definite, since if $\|p\|_I = 0$, then $p|_I = 0$, and hence $p$ has infinitely many roots, and is the $0$ polynomial.

So, if $p_m \to p$, then $\|p_m - p\|_I \to 0$. Thus, for all $\varepsilon > 0$, there exists some $M$ such that \begin{align*} m > M &\implies \|p_m - p\|_I < \varepsilon \\ &\implies \forall x \in I, |p_m(x) - p(x)| < \varepsilon \\ &\implies \forall x \in I, p(x) - \varepsilon < p_m(x) < p(x) + \varepsilon \\ &\implies \sup_{x \in I} p(x) - \varepsilon \le \sup_{x \in I} p_m(x) \le \sup_{x \in I} p(x) + \varepsilon \\ &\implies |\sup p_m(I) - \sup p(I)| \le \varepsilon. \end{align*} We can similarly (or symmetrically, replacing polynomials by their negatives) show that $\inf p_m(I) \to \inf p(I)$. QED.

Now, let $X_n$ be the set of monic real polynomials of degree $n$ with $n$ distinct real roots. Note that, if $x$ is a root of some $p \in X_n$, then $p'(x) \neq 0$, otherwise we'd have a repeated root. Therefore, $p$ is strictly monotone (increasing or decreasing) on some neighbourhood of each of its roots.

Suppose that $p_m \in X_n$ converges to $p \in X_n$. Let $z_1, \ldots, z_n$ be the roots of $p$, in increasing order. Let $\varepsilon > 0$ be sufficiently small that $I_j:= [z_j - \varepsilon, z_j + \varepsilon]$ contains no roots of $p'$, and hence $p$ is strictly monotone in the interval, for $j = 1, \ldots, n$. Note that, for all $j$, $\inf p(I_j) < 0$ and $\sup p(I_j) > 0$. Note also that the intervals $I_j$ are pairwise disjoint.

Using the lemma, there exists some $M$ such that, for all $m \ge M$ and $j = 1 \ldots, n$, $\sup p_m(I_j) > 0$ and $\inf p_m(I_j) < 0$. In particular, the intermediate value theorem implies that $p_m$ has a root in the interval $I_j$. Since there are $n$ pairwise disjoint such intervals, this produces all $n$ roots of $p_m$. The order of these roots are preserved too, since the intervals $I_j$ are ordered too.

Therefore, $f(p_m) \to f(p)$ as $m \to \infty$.