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?

  • $\begingroup$ 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 $f(c) = 0$? $\endgroup$ – Daniel Fischer Nov 14 '16 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 Tamaroff Nov 14 '16 at 18:32
  • $\begingroup$ You're close. Use the fact that $f$ is continuous. $\endgroup$ – zhw. Nov 14 '16 at 18:32
  • 1
    $\begingroup$ 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 $f$. $\endgroup$ – Daniel Fischer Nov 15 '16 at 11:51
  • 1
    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Nov 15 '16 at 13:47

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

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

  • $\begingroup$ 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 '16 at 5:27
  • $\begingroup$ Made the small correction and minor additions. $\endgroup$ – Fimpellizieri Nov 15 '16 at 5:37
  • $\begingroup$ I didn't see role of $m$ in the stronger property you have mentioned. :) $\endgroup$ – Error 404 Nov 15 '16 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. $\endgroup$ – Fimpellizieri Nov 15 '16 at 5:58
  • $\begingroup$ Thank you so much for your nice explanation! :) $\endgroup$ – Error 404 Nov 15 '16 at 6:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.