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Let $f(x,y,t):[-1,1]^3\to \mathbb{R}$ be a real-analytic function. Assume that for any fixed $x,y$, $f(x,y;t)$ is not a constant function $[-1,1]\to \mathbb{R}$. Since the zeros of a non-constant real-analytic function of one variable are isolated, we denote the number of the zeros of $f(x,y;t)$ on $[-1,1]$ by $N(x,y)$. Then do we have the uniform bound $$\sup_{(x,y)\in[-1,1]^2}N(x,y)\le C$$ where $C$ is a constant?

I can not find a counterexample such that for some sequence $(x_n,y_n)\in[-1,1]^2$, $N(x_n,y_n)\to \infty$, as $n\to \infty$.

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Let's do this for two variables $x,t$ rather than three for ease of notation. Also, I assume that by real analytic on a compact set, you mean real analytic on some open neighborhood of this set.

Then your claim follows because $N(x)$, with the zeros counted according to multiplicity, is upper semicontinuous, so if $N$ were unbounded, then $N(x)=\infty$ somewhere, contrary to your assumption.

To see that $N(x)\ge\limsup N(x_n)$ if $x_n\to x$, suppose you have consistently at least $k$ zeros $a_n(1),\ldots, a_n(k)$ for each $x_n$. On a suitable subsequence $a_n(j)\to a(j)$, which already gives us zeros at $x$. The only issue is that some of these limit points could agree, but then just look at the derivative(s) to see that they pick up enough multiplicity to compensate for this. Finally, if the original zeros already had multiplicity $>1$, this won't get lost in the limit.

This argument actually doesn't use analyticity; it works the same way for smooth functions.

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  • $\begingroup$ Awesome! Could you add some explanations on how to see "...they pick up enough multiplicity to compensate for this. Finally, if the original zeros already had multiplicity >1, this won't get lost in the limit" ? It seems not very clear to me. Thx! $\endgroup$
    – Right
    Commented May 27, 2017 at 18:29
  • $\begingroup$ @Right: (1) Let's say $a_n(1), a_n(2)\to a$. Then $f'(x_n,t_n)=0$ for some $t_n$ between $a_n(1)$ and $a_n(2)$ by Rolle's theorem (and ' means $t$ derivative), so $f'(x,a)=0$ also by passing to the limit, and this means that $a$ is a zero of multiplicity (at least) $2$. (2) (this is the same really) If the approximating zeros have higher multiplicity, then $f(x_n,a_n)= \ldots = f^{(k)}(x_n,a_n)=0$, which again gives that $f(x,a)=\ldots = f^{(k)}(x,a)=0$ also by passing to the limit. $\endgroup$
    – Christian Remling
    Commented May 27, 2017 at 19:13
  • $\begingroup$ @Right: Alternatively, you can also establish the semicontinuity of $N(x)$ by fixing a smooth function with $\le k$ zeros and then show that a sufficiently small perturbation (meaning function and derivatives are uniformly close) will also have $\le k$ zeros. $\endgroup$
    – Christian Remling
    Commented May 27, 2017 at 19:15
  • $\begingroup$ Nice answer! If the assumption that "for any fixed $x$, $f(x;t)$ is not a constant function" is removed, do you think $N(x)$ is still bounded on the set $\{x\in[-1,1]: f(x;t)\ {\rm is\ not\ a\ constant\ function}\}$ (which may not be compact)? I can not find a counterexample.Thanks for your patience:) $\endgroup$
    – Right
    Commented May 28, 2017 at 1:42
  • $\begingroup$ @Right: Not sure about this off the top of my head. The corresponding claim about smooth functions is of course false, so this would require a different argument (if true). $\endgroup$
    – Christian Remling
    Commented May 28, 2017 at 2:48

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