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Can someone please verify this?

Let $f_n$ be a sequence of continuous functions on a closed interval $I$ converging uniformly to $f$. Is it true that max $\{f_n(x):x\in I\}$ converges to max $\{f(x):x \in I\}$?

Yes. Let $\alpha = $ max $\{f(x):x \in I\}$

Assume, for the sake of contradiction, that max $\{f_n(x):x \in I\}$ does not converge to $\alpha$.

Then, $$\forall \epsilon > 0, \forall N \in \mathbb{N}, \exists n > N : \left|\operatorname{max} \{ f_n(x)|x \in I\} - \alpha\right| > \epsilon$$

However, this contradicts the fact that $f_n$ converges uniformly to $f$. Note that

$$\forall \epsilon > 0, \forall N \in \mathbb{N}, \exists n > N:|f_n(y) - \alpha| > \epsilon$$

Here, $y$ is the value at which $f$ attains its maximum (it must attain its maximum at some point within $I$, since $I$ is closed). Therefore, it must be the case that max $\{f_n(x):x\in I\}$ converges to max $\{f(x):x \in I\}$

RRL
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user154185
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  • How are you justifying $"\forall \epsilon >0, \forall N \in \mathbb{N}, \exists n>N :|f_n(y)-\alpha|>\epsilon$"? The value $f_n(y)$ is not necessarily further from $\alpha$ than $\max {f_n(x) : x \in I}$. – Kyle Jun 16 '14 at 15:56
  • @user153841 If I changed that to $\forall \epsilon > 0, \forall N \in \mathbb{N}, \exists n > N : |f_n(y) - \alpha| > \epsilon$ or $|f_n(z_n) - \alpha| > \epsilon$, where $y$ is as I stated in the original solution, and $z_n$ is the value at which $f_n$ obtains its maximum, would that be correct? – user154185 Jun 16 '14 at 16:08
  • @user154185 The statement "$\forall \epsilon>0, \forall N \in \mathbb{N}, \exists n >N : |f_n(z_n)-\alpha|>\epsilon$ where $f(z_n)=\max{ f_n(x) : x\in I}$" is equivalent to the hypothesis you're making to obtain a contradiction, namely that $\max {f_n(x) : x\in I}$ doesn't converge to $\alpha$. – Kyle Jun 16 '14 at 16:13

1 Answers1

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I assume $I$ is closed and bounded, so $f_n$ is bounded.

Uniform convegence implies for every $\epsilon>0$ there exists $N \in \mathbb{N}$ such that for every $x \in I$ and $n \geq N$

$$ f(x) - \epsilon < f_n(x) < f(x) + \epsilon.$$

Hence, for every $\epsilon > 0,$

$$f_n(x) < \max(f)+ \epsilon, \\ f(x) - \epsilon < \max(f_n).$$

It follows that,

$$\max(f_n) \leq \max(f)+ \epsilon, \\ \max(f) - \epsilon \leq \max(f_n).$$

Therefore for every $n \geq N,$

$$|\max(f_n)-\max(f)| \leq \epsilon$$

RRL
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