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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?

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I don't think you wrote the question you meant... – Nate Eldredge Sep 29 '10 at 19:02
Thanks for catching that and sorry about the typo. – Ben Golub Sep 29 '10 at 19:06
The case $d=1$, $F(x)=x^2$, $F_n(x) = x^2 + (e^{-x^2})/n$ makes me think that more a priori knowledge is needed about your particular problem. – Yemon Choi Sep 29 '10 at 19:17
Functions don't have solutions. Equations have solutions; functions have zeros. Can you please rewrite your question? – Gerry Myerson Sep 30 '10 at 5:00
I wasn't really comfortable with "zero" of a multivariate function, but if you think that's better, I've changed it. – Ben Golub Sep 30 '10 at 5:04

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 $y = x$.

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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 Fn 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 ℝd-{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 Fn must have a zero somewhere in that disc. Take smaller and smaller discs to finish the argument.

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Thank you-- it is really remarkable that uniform convergence is enough... – Ben Golub Sep 30 '10 at 5:26

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 $\|x-x_n\|\le\frac{1}{1-k}\|T(x)-T_n(x)\|.$

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Thanks! This is exactly what we needed. I really appreciate your help. – Ben Golub Sep 30 '10 at 5:04

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 $x_n\to x$.

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