most recent 30 from http://mathoverflow.net 2013-05-24T23:07:38Z http://mathoverflow.net/feeds/question/95049 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95049/f-nx-p-converge-uniformly-to-nice-fx-p-do-zeros-of-f-n-p-converge-uniform Ben Golub 2012-04-24T17:45:27Z 2012-05-10T07:22:00Z

<p>Fix compact intervals $X, P \subseteq \mathbb{R}$.</p> <p>Let $f_n : X \times P \to \mathbb{R}$ be a sequence of $C^2$ functions converging uniformly to a $C^2$ function $f$. The first and second derivatives of $f_n$ also converge uniformly to those of $f$.</p> <p>For any $p \in P$, the function $f(\cdot,p) : X \to \mathbb{R}$ has a unique zero $x^*(p)$, and $f_1(x^*(p),p) \neq 0$. Therefore there is a sequence $x_n^* (p) \to x^*(p)$ such that $f_n(x_n^*(p),p)=0$ for large $n$.</p> <p>When can we also say that $x_n^*(p) \to x(p)$ <em>uniformly</em> in $p$? Is there a standard reference for such results?</p>

http://mathoverflow.net/questions/95049/f-nx-p-converge-uniformly-to-nice-fx-p-do-zeros-of-f-n-p-converge-uniform/95058#95058 Pietro Majer 2012-04-24T18:56:24Z 2012-04-26T06:41:50Z

<p>It's a quick proof by contradiction.</p> <p>Assume that $ x _ n ^ * $ does not converge uniformly to $x^*$. Then, there exists $\epsilon > 0$ and, for all $n\in \mathbb{N}$, a point $p_n\in P$ such that $|x_n ^ *(p _ n)-x ^ *(p _ n)|\ge\epsilon$. </p> <p>A subsequence $\big( p_{n_k}\ ,\ x_ {n_k} ^ *(p _{n_k})\big)$ converges to some $( p, y ^ * ) \in P \times X$, a zero of $f$ different from $(p,x^*(p))$, contradiction. </p>