When do functions near F have zeros near a zero of F? - MathOverflow

When do functions near F have zeros near a zero of F?

2010-09-29T18:53:33Z

Consider a sequence of functions $F_n : \mathbb{R}^d \to \mathbb{R}^d$, a function $F: \mathbb{R}^d \to \mathbb{R}^d$, and an $\mathbf{x} \in \mathbb{R}^d$ so that $F(\mathbf{x}) = \mathbf{0}$. In what sense should the $F_n$ converge to $F$, and what additional conditions should be placed on them, to ensure that (for large enough $n$) we can find a sequence of $\mathbf{x}_n$ with $F_n(\mathbf{x}_n)=0$ and $\mathbf{x}_n$ converging to $\mathbf{x}$?

Apologies for the elementary question. I know this must be a very standard result but I can't seem to find it. Can anyone point me to the right reference?

Answer by André Henriques for When do functions near F have zeros near a zero of F?

2010-09-29T19:19:19Z

You probably want the function to be continuousely differentiable, and the convergence to be in the sense of the C¹-norm (uniform convergence + uniform convergence of all the partial derivatives). You then also want the total derivative of F at the point x to be an invertible matrix.

Ah! Here's a way of getting to the same conclusion with much weaker assumptions:

If F is a continuous function and its derivative at the point x exists and is invertible, then it's enough to assume that the functions F_n are continuous and that they converge in the C⁰ norm (uniform convergence).
The reason is that F, when viewed as a map from a little sphere around x to the punctured space ℝ^d-{0} has degree one. So any other map that is sufficiently close to it will also have degree one. A degree one map cannot extend to the disc bounding the sphere. So the function F_n must have a zero somewhere in that disc. Take smaller and smaller discs to finish the argument.

Answer by Deane Yang for When do functions near F have zeros near a zero of F?

2010-09-29T19:23:17Z

Here's one possible answer: First, consider the case $d = 1$ and think about the graph of $F$. If you assume $F$ to be continuous (anything worse would be much more difficult), then there are roughly speaking two ways to have a zero, one because the graph crosses the $x$-axis (like $F(x) = x$ and one because the graph touches the $x$-axis but does not cross (like $F(x) = x^2$). It is easy to see that the former is more "stable" in the sense that any small continuous change in $F$ will not cause the zero to disappear, whereas in the latter case it is easy to make the zero disappear by a small change in $F$. So the former case is what you want. You want the graph of $F$ to cross and not just touch the $x$-axis at the solution.

The term for this is "transversality", you can look for more information about this. If the graph of $F$ (in any dimension) is transversal to the graph of the zero function at the solution, then any small continuous perturbation of $F$ will still have a zero.

If $F$ is sufficiently smooth, a simple sufficient condition for this is that the derivative of $F$ is invertible at the solution. This follows by the inverse function theorem, which tells you that $F$ can be made to look like the function $y = x$.

Answer by Fiktor for When do functions near F have zeros near a zero of F?

2010-09-29T19:26:11Z

If function $F$ is differentiable at the point $x$, its Jacoby matrix in this point has nonzero determinant, $F_n$ and $F$ are continuous in some neighborhood of $x$ and $F_n\to F$ pointwise, then there exists a sequence of $x_n$, s.t. $F(x_n)=0$ and $x_n\to x$.

Answer by Pietro Majer for When do functions near F have zeros near a zero of F?

2010-09-29T19:29:01Z

Since you are requiring, among other things, some stability for the existence of solution of $F(x)=0$ under perturbation, it seems quite natural to think in terms of topological degree.

Assume that there is an open bounded nbd of $x,$ $\Omega\subset\mathbb{R}^d,$ such that $x$ is the only solution of $F=0$ in $\bar\Omega.$ In particular this implies that $\deg(F,\Omega,0)$ is well defined.
Assume further $\deg(F,\Omega,0)\ne 0.$ (several different facts may ensure this: $F$ is an odd map, or it is injective; or you can compute its degree etc).
Assume that $F_n$ converge to $F$ uniformly on $\bar\Omega$ (several diffferent facts may ensure this too: for instance, $F_n$ converges pointwise to $F, $ and it is uniformly equicontinuous on $\bar\Omega$, etc).

Then, $\deg(F_n,\Omega,0)$ is eventually defined and different to $0$. So there exists $x_n\in\Omega$ solving $F_n(x_n)=0$ (not necessarily unique). By the compactness of $\bar\Omega$ together with the uniform convergence of $F_n$ to $F$ on $\bar\Omega$, we can conclude $x_n\to x$ as we wish.

PS: all maps are assumed to be continuous, of course. $$*$$

Further remark. Generally speaking, even in more general contexts than $\mathbb{R}^d$, any existence result for solutions of $F(x)=0$ (via degree theory, topologic and metric fixed point theorems, minimization, critical point methods,... &c) will give you a perturbation result as the one you are saying, provided you have some uniform a priori bounds for the solutions of $F_n=0,$ and some compactness. Also, if the unperturbed equation $F(x)=0$ is thought as a trivial case, and what you are mainly interested in is the perturbed equation $F_n=0,$ then bifurcation theory is what you want. Lastly, one more very elementary example, for all.

Perturbation for the contraction principle. Assume $T_n$ is a pointwise convergent sequence of contractions, with (uniform) Lipschitz constant $k<1.$ Then the limit map $T$ is a $k$-contraction too, and the sequence $x_n$ of the fixed points of $T_n$ converges to the fixed point of $T$. Indeed, it's immediate to check that $\|x-x_n\|\le\frac{1}{1-k}\|T(x)-T_n(x)\|.$