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I'm trying to understand a proof for the extreme value theorem: Let $f$ be a continuous real-valued function on a closed interval [a, b]. Then $f$ is a bounded function. Moreover, $f$ assumes its maximum and minimum values on [a, b]; that is, there exist $x_0$, $y_0$ in [a, b] such that $f(x0) ≤ f(x) ≤ f(y0)$ for all $x ∈ [a, b]$.

I understand how to prove that $f$ is bounded, but now I'm trying to understand a proof of showing $f$ assumes its maximum and minimum values. To prove it, let $M = \text{sup}\{f(x):x\in[a,b]\}$ which we know exists because $f$ is bounded. Now, for all $n \in \mathbb{N}$, there exists a sequence $(y_n)$ such that $M - 1/n < f(y_n) \leq M$. Then the limit of $f(y_n) = M$, (but how can I prove this is the limit?) Then, by the Bolzano-Weiersrtass theorem, we can find a subsequence $(f(y_{nk}))$ with limit $y_0 \in [a,b]$. Then since $f$ is continuous, we know that lim$f(y_{nk})=f(y_0)$. Since $(f(y_n))$ is convergent, it has the same limit as its subsequence $(f(y_{nk}))$, $M$. Therefore, lim$f(y_n) = f(y_0) = M$, which is the maximum value and shows that the function assumes its maximum. My only question is how do we know that $(f(y_n))$ converges?

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  • $\begingroup$ Because by choice you have $|M-f(y_n)| \le {1 \over n}$. Pick $\epsilon>0$ and $N$ such that ${1 \over N} < \epsilon$. Then for $n \ge N$ we have $|M-f(y_n)| < \epsilon$. $\endgroup$ – copper.hat Mar 24 '17 at 23:42
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You know it converges because of the squeeze theorem.

For each $n \in \mathbb{N}$, $M-1/n < f(y_n) ≤ M$. Hence by the squeeze theorem, \begin{align*} M = \lim_{n\rightarrow\infty}M-\frac{1}{n} \leq \lim_{n\rightarrow\infty}f(y_n) ≤ \lim_{n\rightarrow\infty}M = M \end{align*}

Therefore $\lim_{n\rightarrow\infty}f(y_n)$ must exist and is equal to $M$

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  • $\begingroup$ That makes sense, why do I need to use a subsequence then? $\endgroup$ – TheStrangeQuark Mar 25 '17 at 2:47
  • $\begingroup$ I have no idea, I wondered about that part too. It looks redundant to me. $\endgroup$ – Harambe Mar 25 '17 at 2:56
  • $\begingroup$ I was thinking it was to show that $y_0$ is in the interval [a,b]. But it still feels redundant. $\endgroup$ – TheStrangeQuark Mar 25 '17 at 3:02
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    $\begingroup$ Ah it makes sense now - we've shown that $f(y_n)$ converges to $M$, but it's not obvious that $y_n$ is convergent. You want to prove that "for some $y_0 \in [a, b], f(y_0) = M$". Without the subsequence part we have no idea how $y_n$ is behaving - we've just shown that $f$ attains some maximum without talking about where it attains its maximum. $\endgroup$ – Harambe Mar 25 '17 at 3:16

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