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Suppose we have two functions $f,g:\mathbb{R}^n\rightarrow \mathbb{R}$. Assume the existence of unique roots to both function, i.e. $x^f,x^g\in \mathbb{R}^n$ such that $f(x^f)=0=g(x^g)$. Define the homotopy $$H(x,t)= tf(x)+(1-t)g(x) .$$ Are there any known conditions such that $H(\cdot,t)$ has a unique root for every $t$? This is not my field. I have found this example which seems to suggest that some non-trivial conditions have to be made. I am fine with any regularity assumptions (continuity, smoothness, ...) but judging by the example with polynomials, this is not enough. I am also fine with assuming that $f,g$ take values in a bounded interval (which includes 0) if that helps.

Gorgo
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  • You are aware the the locus of the roots is in general in this setup a hypersurface? Are you also asking about the homotopy of the hypersurfaces? – Lutz Lehmann Apr 14 '19 at 08:57
  • No I am not aware of this. What is a locus of the roots? I'm interested in the case when $H(\cdot ,t)$ has a \textit{unique} root for every $t$. Sorry, that was not clear in my question. I just edited it. – Gorgo Apr 14 '19 at 09:26
  • Then you need scalar functions that are monotonous. Or you need something singular like a sum of squares where the corresponding system where the system of the single terms only has one single (common) root. This situation is not stable under perturbation, and a homotopy is such a perturbation. Most likely there will be no roots in-between. – Lutz Lehmann Apr 14 '19 at 09:39
  • By scalar functions do you mean functions with scalar arguments or with scalar values or both? The second condition you give sounds more interesting although it is not perfectly clear to me. Do you happen to have a reference? – Gorgo Apr 14 '19 at 16:38
  • Both. Everything one-dimensional. // A multi-dimensional condition for a single root is convexity of a non-negative function, the sum-of-squares construction can easily achieve that. A convex combination of convex functions, like in the homotopy, is again convex, however there will be no roots in-between as it takes values strictly between $f(x)$ and $g(x)$, thus strictly positive. – Lutz Lehmann Apr 14 '19 at 16:43
  • Convexity is interesting. So you are saying even if $f$ and $g$ are strictly convex and have a unique root, then there will be no roots between them? That doesn't seem right, for example if $f=g$ or $f(x)=x^2-4$ and $g(x)=x^2-1$. – Gorgo Apr 14 '19 at 18:32
  • These are not non-negative. Try $f(x)=(x-2)^2$ and $g(x)=(x-1)^2$. However, double roots might fail your condition of single roots. At a simple root, the derivative or gradient has to be non-zero, and then we are back at the first comment, the set ${x:f(x)=0}$ being a co-dimension one hypersurface. – Lutz Lehmann Apr 14 '19 at 20:28

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