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Let $F:I\times J\to\mathbb{R}$ be a $C^k$ (or analytic) function, with $I,J$ real open intervals. Set $f_\lambda(x):=F(\lambda,x)$ and consider the parametric equation $$f_\lambda(x)=0\,.$$ Assume its solutions are at least $1$ and at most $n$ (not dependent on $\lambda$) and denote them by $$x_1(\lambda)\leq x_2(\lambda)\leq\dots\leq x_n(\lambda)\,.$$ Is it possible to say that $\lambda\mapsto x_1(\lambda)$ is a continuous function? Further conditions on $F$ are required?

Note. For the sake of semplicty one may start with $n=2$. In this case there could be for example a value $\lambda_0$ such that for $\lambda\leq\lambda_0$ the equation $f_\lambda(x)=0$ has 1 solution $x_1(\lambda)$, while for $\lambda>\lambda_0$ it has $2$ different solutions $x_1(\lambda)<x_2(\lambda)$.

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Could you be a bit more explicit in what do you mean by "Assume its solutions are at least 1 and at most $n$ (not dependent on $\lambda$)"? Do you mean that for every $\lambda$ there are at most $n$ solutions? Or there is a finite and fixed number of solutions for every $\lambda$? Thank you! –  Giovanni De Gaetano Apr 18 '13 at 17:06
    
I mean that there exists an $n$ such that for every $\lambda$ there are at most $n$ solutions of the equation. –  qwertyuio Apr 18 '13 at 17:32
    
It's easy to come up with counterexamples if you're allowed to make roots vanish. I think the result should be true in intervals where the roots do not meet each other, even just given continuity of $F$. This seems a more interesting statement to me (: –  Sharkos Apr 19 '13 at 0:38

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The answer to your question is no. It is not possible to claim that $\lambda \mapsto x_1(\lambda)$ is a continuous function. To show it, it is enough to provide a counterexample.

Let us consider $I=J=(0,1)$ for simplicity. Then we define the polynomial function:

$$F(x,\lambda):= -\frac{1}{3}x^3 +\frac{1}{2}x^2 -\frac{2}{9}x +\frac{\lambda}{20}.$$

Observe that, since it is a polynomial, $F$ is analytic, so $\mathcal{C}^\infty$ and so on. I apologize for the "weird" numbers, but I wanted to show exactly what is going on with pictures, and Mathematica essentially forced them on me.

This is the picture of $F(x,0.4)$:

0.4

This is the picture of $F(x,0.6)$:

0.6

And finally this is the picture of $F(x,0.8)$:

enter image description here

You see that, while $\lambda$ is getting bigger, we are shifting the graph of $F$. And it is also clear that, for a specific $\lambda$ in the interval $(0.6,0.8)$, your function $x_1(\lambda)$ will jump to the right. It will assume those values that you were calling $x_3(\lambda)$ until an $\epsilon$ before.

This answer is "sketchy" on purpose, I think that it gives you exactly the idea of the kind of things that can go wrong.

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Thank you very much! Your example is very clear. –  qwertyuio Apr 18 '13 at 18:45
    
So is there some other condition on $F$ or maybe a standard method to study continuity of zeros? –  qwertyuio Apr 18 '13 at 18:47
    
Maybe if one moves to the complex plane.. i.e. looks for solutions $x\in\mathbb{C}$.. –  qwertyuio Apr 18 '13 at 19:03
    
Forcing a fixed number of solutions for any $\lambda$, keeping the interval $I$ open, should give you the result! –  Giovanni De Gaetano Apr 22 '13 at 8:27

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