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Let $f:\mathbb{R}^{2n+1}\to\mathbb{R}$ be defined through: $f\left(x_0,...,x_{2n}\right)$ is the greatest root of the polynomial $p(t)=\sum_{k=0}^{2n}x_kt^k$. Is $f$ continuous? If so, what is the greatest $r$ such that $f\in C_r\left(\mathbb{R}^{2n+1}\right)$? It seems a bit difficult to directly prove the continuity, but I wasn't able to use tools like the implicit function theorem either. How to tackle the problem?

Redundant Aunt
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  • The function $f:\mathbb R\to \mathbb R$ is undefined. The function $f:\mathbb R^3\to\mathbb R$ has no values whenever $x_1^2-4x_0x_2 < 0$. I think you need to restrict the domain of $f$. – kennytm Jun 17 '16 at 00:08
  • For $x_2 \ne 0$, we're talking about the root of a cubic...which always has at least one real root, right, kennfyfm? Or am I missing something. But for $x_2 = x_1 = 0$ and $x_0 \ne 0$, the function $f$ is certainly undefined. – John Hughes Jun 17 '16 at 00:14

1 Answers1

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consider the polynomial $$ g_s(x) = (x^2 + s) (x + 1) $$ For small negative $s$, this has two roots near zero, and another at $-1$.

As $s$ increases to become positive, the roots near zero disappear. So your function $f$, applied to $g_s$, as $s$ increases through $0$, has a jump in value from $0$ to $-1$. Evidently, it's not a continuous function of $s$.

Since the polynomial coefficients of $g$ clearly ARE continuous functions of $s$, if your function $f$ were continuous, so would be $s \mapsto f(g_s)$. But it's not, so $f$ isn't continuous.

This answers your question in the negative for $n = 1$, hence for every $n$ (by ignoring the other variables).

John Hughes
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  • How about if you consider the complex roots? Are they then continuous? – marty cohen Jun 17 '16 at 00:31
  • @martycohen We don't have an order in complex number, so there is no greateset root. – Emre Jun 17 '16 at 01:01
  • The most you might be able to say (for one var) is something like this: "if $g_s$ is a (generic?) continuous family of degree-$n$ polynomials in one variable depending on a paramter $s$, then there are $n$ continuous functions $s\mapsto r_1(s), \ldots, s\mapsto r_n(s)$ such that $g_s(r_i(s)) = 0$ if $g_s(x) = 0$, then $x = r_i(s)$ for some $i$, and (something about the functions $r_i$ all being distinct). " I don't know the right statement to make this true, but in general, multiple-roots will be your big bugaboo. – John Hughes Jun 17 '16 at 01:23
  • @Emre: If we just look at the roots, it is known that the roots of a polynomial are a continuous function of the coefficients. Look it up. – marty cohen Jun 17 '16 at 02:58
  • Marty is right, and here (http://math.stackexchange.com/questions/63196/continuity-of-the-roots-of-a-polynomial-in-terms-of-its-coefficients) is a nice reference to a couple of proofs of the very sort of thing I was trying to say. – John Hughes Jun 17 '16 at 15:04