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Let $R_n$ denote the set of all monic real polynomials of degree $n$ all of whose roots are real. Then $R_n$ is a closed subset of the $n+1$-dimensional space ${\mathbb R}_n[X]$. For $P\in R_n$, denote by $\theta_1(P) \leq \theta_2(P) \leq \ldots \leq \theta_n(P)$ the roots of $P$ in increasing order. Is the map $(\theta_1,\theta_2, \ldots,\theta_n): R_n \to {\mathbb R}^n$ continuous? This is true for $n=2$.

I know about more usual "continuity of roots" properties (see for example here ), but I don’t see clearly how they might be useful here. Most versions only deal with polynomial without repeated roots.

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Ewan Delanoy
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The answer is yes. I've not worked through the technical difficulties in making a direct proof, so I'll cheat and suggest that you construct the companion matrix, and use the fact that you can write the eigenvalues as continuous functions. (I don't have a better reference off hand)

All that remains is to show that if you have $n$ continuous functions $f_i(x)$, that if you let the list $g_i(x)$ be the reordering of the values of the $f_i(x)$ from least to greatest, that the $g_i$ are also continuous.

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  • The reordering issue is easy : the functions $\min$ and $\max$ are continuous ${\mathbb R}^2 \to {\mathbb R}$, because $\min(x,y)=\frac{x+y-|x-y|}{2}$ and $\max(x,y)=\frac{x+y+|x-y|}{2}$ – Ewan Delanoy Aug 15 '14 at 16:14