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Fix compact intervals $X, P \subseteq \mathbb{R}$.

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 $f$.

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 $n$.

When can we also say that $x_n^*(p) \to x(p)$ uniformly in $p$? Is there a standard reference for such results?

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How about $f_n(x,p)=(x-p)^2+1/n$? This converges to $f(x,p)=(x-p)^2$, but $f_n(x,p)\neq 0$ for all $x,p$ and $n$. Therefore, your therefore does not seem right, unless I am misinterpreting something. –  Per Alexandersson Apr 24 '12 at 21:01
    
I think " $f_1(x,p)$ " denotes the partial derivative wrto the first variable, x. In other words, he's in the hypotheses of the implicit function thm. –  Pietro Majer Apr 24 '12 at 22:04

1 Answer 1

It's a quick proof by contradiction.

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 $|x_n ^ *(p _ n)-x ^ *(p _ n)|\ge\epsilon$.

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.

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Note that this requires nothing about the derivatives, just uniform convergence of $f_n$ to $f$ on $X \times P$, continuity of $f$ on $X \times P$, and the fact that $f(\cdot,p)$ has a unique zero in $X$ for each $p \in P$. –  Robert Israel Apr 26 '12 at 7:36
    
(and compactness of $X$ and $P$) –  Robert Israel Apr 26 '12 at 7:37
    
Indeed. We may say it's an instance of a kind of general situation: "uniqueness plus compactness gives continuous dependence". –  Pietro Majer Apr 26 '12 at 7:46

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