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So every $f_n : [a,b] \to \Bbb R$ is a sequence of continuous functions and $f_n \to f$ uniformly. If each $f_n$ has a zero, then we have to show that $f$ also has a zero.

To prove this first I list out the definitions I am gonna use:

  1. Uniform continuity of every $f_n$. (since $[a,b]$ is given compact.)

$\forall \epsilon \gt 0, \exists \delta_n \gt 0$ such that $\forall x,y \in [a,b]$ & $|x-y| \lt \delta_n$ we have $|f_n(x)-f_n(y)| \lt \epsilon$ $\forall n \in \Bbb N.$

  1. Uniform convergence of $f_n$.

$\forall \epsilon \gt 0, \exists N \in \Bbb N$ such that if $n \ge N$ then $|f_n(x)-f(x)| \lt \epsilon$ $\forall x \in [a,b]$.

Proof: Let $c_n$ be zero of $f_n$. Then $f_n(c_n)=0 \; \forall n\in \Bbb N.$

For a given $\epsilon \gt 0 \; \exists N \in \Bbb N$ such that if $n \ge N$ then $|f_n(c_n)-f(c_n)| \lt \epsilon \implies |0-f(c_n)| \lt \epsilon.$ (Using definition 2)

This means that $f(c_n)$ is sequence of real numbers converging to zero. Hence there exists $c \in [a,b]$ such that $f(c)=0.$

Now I am particularly suspicious at the last step! Somehow I am not content with it. Is it correct? Also I haven't used definition 1. Could it be key here in the last step?

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  • $\begingroup$ You are missing an argument in that last step. You're probably thinking of the right argument, but you need to explicitly make it. Why do you conclude that there exists a $c \in [a,b]$ with $f(c) = 0$? $\endgroup$ Nov 14, 2016 at 18:32
  • $\begingroup$ Your first condition is mixing up quantifiers. You're not given the family is uniformly equicontinuous, that is, you cannot guarantee the same $\delta$ works for every $f_n$ and every $\varepsilon$. You just know that each $f_n$ is uniformly continuous on its own. $\endgroup$
    – Pedro
    Nov 14, 2016 at 18:32
  • $\begingroup$ You're close. Use the fact that $f$ is continuous. $\endgroup$
    – zhw.
    Nov 14, 2016 at 18:32
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    $\begingroup$ Yes, you need to use the fact that $(c_n)$ has a convergent (in $[a,b]$) subsequence . Then the continuity of $f$ and the uniform convergence show $f(c) = 0$ where $c$ is the limit of the subsequence. An alternative proof: $$\inf_{x\in [a,b]} f(x) = \lim_{n\to\infty} \inf_{x\in [a,b]} f_n(x) \leqslant 0,\quad \sup_{x\in [a,b]} f(x) = \lim_{n\to\infty} \sup_{x\in [a,b]} f(x) \geqslant 0,$$ and now the intermediate value theorem yields the existence of a zero of $f$. $\endgroup$ Nov 15, 2016 at 11:51
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    $\begingroup$ No, I didn't mean a limes inferior. Let $m = \inf \{ f(x) : x \in [a,b]\}$ and $m_n = \inf \{ f_n(x) : x \in [a,b]\}$. Then my assertion is $m = \lim\limits_{n\to\infty} m_n$. And similarly for the suprema. The uniform convergence is of course important to have that. $\endgroup$ Nov 15, 2016 at 13:47

1 Answer 1

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Since $f_n \to f$ uniformly and the $f_n$ are continuous, so too is $f$ continuous.

Now, let's try and adapt your partial solution. For each $n \in \mathbb{N}$, let $c_n \in [a,b]$ be a zero of $f_n$ (which exists by hypothesis). Since $[a,b]$ is compact, there must be some convergent subsequence $c_{n_k}$ of $c_n$; call $c \in [a,b]$ its limit. Then:

$$|f(c)-f_{n_k}(c_{n_k})| \leq |f(c) - f(c_{n_k})| + |f(c_{n_k}) - f_{n_k}(c_{n_k})|$$

As $k \to \infty$, the first can be made as small as desired, because $f$ is continuous (and $c_{n_k} \to c$); and so too with the second term, because the convergence $f_n \to f$ is uniform.

This means $f_{n_k}(c_{n_k}) \to f(c)$, but since $f_{n_k}(c_{n_k}) = 0$ for all $k$, it follows that $f(c)=0$. In other words, we've shown a slightly stronger property:

For each $n \in \mathbb{N}$, let $Z_n=\{x \in [a,b]\,|\, f_n(x) = 0$}. Consider $$Z = \{x \in [a,b]\,|\,\forall \text{ neighborhood $V$ of $x$},\,\forall k\in\mathbb{N},\, \exists m\geq k, \, V\cap Z_m \neq \emptyset\} $$ Then $f(Z)=\{0\}$.

A 'cute' way to characterize $Z$ is by

$$Z = \bigcap_{n\geq 1}{\left(\overline{\bigcup_{k\geq n}Z_k}\right)}$$

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  • $\begingroup$ It's $c_{n_k}$ that converges to $c$. We don't know whether $c_n \to c$ or not. Also small correction, our domain is $[a,b]$. :) $\endgroup$
    – Error 404
    Nov 15, 2016 at 5:27
  • $\begingroup$ Made the small correction and minor additions. $\endgroup$ Nov 15, 2016 at 5:37
  • $\begingroup$ I didn't see role of $m$ in the stronger property you have mentioned. :) $\endgroup$
    – Error 404
    Nov 15, 2016 at 5:56
  • $\begingroup$ Oops, fixed. Sorry, I'm way past sleep time. Basically means there are zeroes of $f_n$'s arbitrarily close to $x$ and with arbitrarily high indices. $\endgroup$ Nov 15, 2016 at 5:58
  • $\begingroup$ Thank you so much for your nice explanation! :) $\endgroup$
    – Error 404
    Nov 15, 2016 at 6:20

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