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I am struggling to understand a step From Angel De La Fuente's book.

Thereom 6.22: enter image description here The yellow bits are giving me some grief.

The first highlight - is this just by assumption of $f$'s unboundedness? I can always find some $n \in \mathbb{N}$ such that for some $x_n \in [a,b]$, $f(x_n) > n$ ?

Is the second highlighted bit seems to be applying the same logic?

Thanks

codenoob
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  • The interval would need to be closed and bounded. – copper.hat Oct 18 '19 at 01:28
  • The first point is using the unboundedness as you suggested to construct a sequence $(x_n)n$ with $f(x_n) \geq n$. The second point is just using the fact that $(x{n_k})k$ is a subsequence of $(x_n)$ and there fore $f(x{n_k}) \geq n_k$. But note that $n_k \geq k$ (the strict inequality might be a typo). – parsiad Oct 18 '19 at 01:28
  • They are assuming that $f$ is unbounded, so either $f$ is unbounded above or below. They assume above for definiteness. The second is different. Since the $x_n \in [a,b]$, Bolzano-Weierstrass says there is a convergent subsequence $x_{n_k}$. – copper.hat Oct 18 '19 at 01:32
  • Since $[a, b]$ is a closed, bounded set of reals, it is compact and also sequentially compact. Hence, every sequence in this set admits a convergent subsequence. See https://math.stackexchange.com/questions/3381175/a-closed-bounded-set-of-a-complete-metric-space-can-fail-to-be-compact – rims Oct 18 '19 at 01:45
  • not that you could always find some $n$: It is that for every $n$ you could find $x_n$ with $f(x_n)>n$, assuming $f$ is not bounded above. – Mirko Oct 18 '19 at 02:33

2 Answers2

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In answer to your first question, yes, this is because of the unboundedness assumption. If no such $x_n$ existed for some $n \in \Bbb{N}$, then this would imply that $f(x) < n$ for all $x \in [a, b]$, which is to say $f$ is bounded above by $n$.

In answer to your second question, this follows from the first highlight. You know that $f(x_n) \ge n$ for all $n$, so $f(x_{n_k}) \ge n_k$ for all $k$. Now, here the proof makes a small, inconsequential error: it really should claim $n_k \ge k$, rather than $n_k > k$. This is due to the nature of subsequences; the sequence $n_k$ must be a sequence of natural numbers that is strictly increasing, and a simple induction proof shows $n_k \ge k$ for all $k$.

Theo Bendit
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    Ah, ofcourse! So, just to make sure I understand.

    If I have the following sequence:

    $(x_1, x_2, x_3, x_4, ... )$ where $ n = 1, 2, 3, ...$ a subsequence $x_{n_k}$ could be every second $x$, eg. $(x_2, x_4, ...)$ so obviously, if $k=1$, then the subsequence $n_k=2$ - is that correct/

     – codenoob Oct 18 '19 at 01:44
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    yes, that's right. – rims Oct 18 '19 at 01:49
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Let $x_n$ be a sequence such that $|f(x_n)| \to \sup_{x \in [a,b]} |f(x)|$.

Bolzano Weierstrass gives a subsequence $x_{n_k}$ that converges to some $x^* \in [a,b]$.

Since $|f|$ is continuous, we have $|f(x_{n_k})| \to |f(x^*)|$ and hence $|f(x^*)| = \sup_{x \in [a,b]} |f(x)|$.

In particular, $f$ is bounded.

copper.hat
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