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Let $f_n\colon [a,b] \to \mathbb{R}$ be a sequence of continuous functions converging uniformly to a function $f$. Show that if each $f_n$ has a zero then $f$ also has a zero.

Thanks for any help.

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Do you mean all the $\,f_n\,$ have the same zero or just some zero? –  DonAntonio Jun 11 '12 at 4:06
    
@DonAntonio some zero –  Ester Jun 11 '12 at 4:09
    
Thanks. Copper's answer has already cracked up the problem. –  DonAntonio Jun 11 '12 at 4:24
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3 Answers

up vote 3 down vote accepted

Since each $f_n$ has a zero, there is a number $x_n \in [a,b]$ such that $f_n(x_n) = 0$. The set $[a,b]$ is compact, so $x_n$ has a convergent subsequence, call it $x_{n_k}$, and let $x_{n_k} \to x$.

Since $f_n$ converges to $f$ uniformly, and $f_n$ are continuous, then $f$ is continuous.

Now consider $|f_{n_k}(x_{n_k}) - f (x_{n_k})| = | f (x_{n_k})|$. By uniform convergence we have $|f(x_{n_k})| \to 0$, and by continuity of $f$, we have $f(x) = 0$.

Alternative proof:

Another version would be to proceed by contradiction. Suppose $f(x) \neq 0 $ $\forall x \in [a,b]$. Then since $[a,b]$ is compact (and $|f|$ is continuous), there exists $\delta>0$ such that $|f(x)| \geq \delta$, $\forall x \in [a,b]$. By assumption of uniform convergence, we can choose $N$ so that $|f_n(x) - f(x)| < \frac{\delta}{2}$, $\forall n \geq N$, $\forall x$. Since $f_N$ has a zero, we have $f_N(z) = 0$ for some $z \in [a,b]$, then the previous inequality gives $|f(z)| < \frac{\delta}{2}$, which is a contradiction.

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Thanks a lot to both of you . –  Ester Jun 11 '12 at 4:14
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Hint: For each $n$, let $x_n\in[a,b]$ be a zero of $f_n$. By compactness there exists a subsequence $(x_{n_k})_{k=1}^\infty$ convergent to some $x\in[a,b]$. Now what can you say about $f(x)$?

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Let us denote $ x_n$ the $f_n$ zero Since $[a,b]$ is compact there existe a subsequence $x_{\phi(n)}$ of $(x_n)$ such that $x_{\phi(n)}$ converges to $\ell \in [a,b]$ You can try proove that $f(\ell)=0$ by using $f_{\phi(n)} (x_{\phi(n)})=0$ for all $n$ and uniforme convergence ...

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