Thread: max

**alexandrabel90** · November 20th 2010, 10:23 AM

in trying to prove that a function on a closed bounded interval attains a max value.

i was reading a proof and the way they proved it was that to show that
let M be the sup of f then show that
f(x) is less than M-1/M_1 hence M-1/M_1 is an upper bound for f, which is smaller than M..hence M is the max..

but im wondering, if that works, then M-1/M_1 should be the max of f right? and not M..

**Plato** · November 20th 2010, 10:34 AM

You have us at distinct disadvantage: you have the book and we don’t.
Moreover, as stated that statement is false.
Consider $f(x) = \left\{ {\begin{array}{rl} {\frac{1} {x},} & {x \ne 0} \\ {1,} & {x = 0} \\ \end{array} } \right.$ .
That function has no maximum on $[0,1].$

Do you mean continuous function?

**alexandrabel90** · November 20th 2010, 10:36 AM

ya sorry. i meant a continuous function on a closed interval.

**Plato** · November 20th 2010, 10:50 AM

Originally Posted by alexandrabel90

ya sorry. i meant a continuous function on a closed interval.

To prove that you need to first show that the function is bounded on the closed finite interval. To do that you need to have the finite open covering theorem. If it bounded the range has a least upper bound. If we suppose that the LUB is not in the range then there is an infinite subset of the interval with no limit point. That is a contradiction.

**alexandrabel90** · November 20th 2010, 10:59 AM

i think i get what you are saying..

but for this method that i have stated above where M is assumed to be the sup of f since f is continuous and hence bounded,

showing that f(x) is less than M-1/M_1 hence M-1/M_1 is an upper bound for f doesnt show that f attains the sup M right?

what the book did was this:
let M be the sup of f. thus f(x)<M
since f(x) is cont, then g(x)= 1/(M-f(x)) is cont
since g(x) is cont, it is bounded and let g(x)< M_1

then
1/(M-f(x)) <M_1
thus f(x)<M-(1/M_1)
thus max of f(x)=M

**Drexel28** · November 21st 2010, 12:49 PM

Originally Posted by alexandrabel90

in trying to prove that a function on a closed bounded interval attains a max value.

i was reading a proof and the way they proved it was that to show that
let M be the sup of f then show that
f(x) is less than M-1/M_1 hence M-1/M_1 is an upper bound for f, which is smaller than M..hence M is the max..

but im wondering, if that works, then M-1/M_1 should be the max of f right? and not M..

You can do this without appealing to the open cover definition of compactness. So, let's start out easy. Let's prove that if $f:[a,b]\to\mathbb{R}$ is continuous then $f([a,b])$ is closed and bounded.

To prove that $f([a,b])$ is bounded, suppose not. Then, there is a sequence $f(x_n)\in[a,b]$ with $f(x_n)>n$ . Recall though that by the Bolzano-Weierstrass Theorem we know that $\{x_n\}$ has a convergent subsequence $\{x_{n_k}\}$ , and in particular $\{x_{n_k}\}$ is Cauchy. Recall though that every continuous function on a closed and bounded interval is uniformly continuous (this is the Heine-Cantor theorem) and that uniformly continuous functions preserve Cauchy sequences, and so $\{f(x_{n_k})\}$ is Cauchy. But, this is clearly impossible since $f(x_{n_k})\geqslant n_k$ . It follows that there exists no such sequence and so $f([a,b])$ is bounded.

So, we know now that $\sup f([a,b])=M$ exists. Suppose then that ]math]f(x)<M[/tex] for all $x\in [a,b]$ . Then, the function $\displaystyle g(x)=\frac{1}{M-f(x)}$ is continuous. Thus, by our last paragraph we know that $g([a,b])$ is bounded above by $C>0$ . So, solving $\displaystyle \frac{1}{M-f(x)}\leqslant C$ for $C$ gives $\displaystyle f(x)\leqslant M-\frac{1}{C}$ for all $x\in [a,b]$ . But, this contradicts that $M=\sup f([a,b])$ . So, $f(x_0)=M$ for some $x_0\in[a,b]$ .

**roninpro** · November 21st 2010, 07:37 PM

Actually, the boundedness part of your proof can be considerably shortened. You can stop right here:

Originally Posted by Drexel28

To prove that $f([a,b])$ is bounded, suppose not. Then, there is a sequence $f(x_n)\in[a,b]$ with $f(x_n)>n$ . Recall though that by the Bolzano-Weierstrass Theorem we know that $\{x_n\}$ has a convergent subsequence $\{x_{n_k}\}$ .

See below for the correction.

**Drexel28** · November 21st 2010, 08:05 PM

Originally Posted by roninpro

Actually, the boundedness part of your proof can be considerably shortened. You can stop right here:

All you have to do is note that any subsequence $\{x_{n_k}\}$ must diverge because it is unbounded (i.e. $x_{n_k}>n_k$ ). But you have found a subsequence that converges - a contradiction.

I don't understand. We know that $\{x_n\}$ has a convergent subsequence, but I don't understand why that, without the rest of my paragraph, implies that $\{f(x_n)\}$ has a convergent subsequence?

**roninpro** · November 21st 2010, 09:24 PM

Sorry, I missed the function $f$ in the sequences. Let me rewrite the argument correctly and fully.

We found a sequence $\{f(x_n)\}$ such that $f(x_n)>n$ for all $n$ ; the sequence diverges to infinity. Note that if we take any subsequence of this, $\{f(x_{n_k})\}$ , it too will diverge, since $f(x_{n_k})>n_k$ for all $k$ .

Now from the Bolzano-Weierstrass theorem, the sequence $\{x_n\}\subset [0,1]$ has a convergent subsequence, call it $\{x_{n_j}\}\to x$ . Since $f$ is continuous, $\{f(x_{n_j})\}\to f(x)$ . In other words, we found that the subsequence $\{f(x_{n_j})\}$ converges, which contradicts our statement above.

**Drexel28** · November 21st 2010, 09:32 PM

Originally Posted by roninpro

Sorry, I missed the function $f$ in the sequences. Let me rewrite the argument correctly and fully.

We found a sequence $\{f(x_n)\}$ such that $f(x_n)>n$ for all $n$ ; the sequence diverges to infinity. Note that if we take any subsequence of this, $\{f(x_{n_k})\}$ , it too will diverge, since $f(x_{n_k})>n_k$ for all $k$ .

Now from the Bolzano-Weierstrass theorem, the sequence $\{x_n\}\subset [0,1]$ has a convergent subsequence, call it $\{x_{n_j}\}\to x$ . Since $f$ is continuous, $\{f(x_{n_j})\}\to f(x)$ . In other words, we found that the subsequence $\{f(x_{n_j})\}$ converges, which contradicts our statement above.

I think there is something missing from your argument. Do you agree that $\frac{1}{x}$ is continuous on $(0,1)$ and $\frac{1}{n}\to 0$ etc.

You need to mention that $[0,1]$ is closed and thus since the real numbers are first countable that $x_n\to x\implies x\in\overline{[0,1]}=[0,1]$ so that $f$ is defined at $x$ . Otherwise, your argument is incorrect. Then, with this I don't feel as though your argument is any shorter than mine.

**roninpro** · November 22nd 2010, 08:56 AM

You are right. I implicitly used closure here: $\{x_{n_j}\}\to x\in [a,b]$ . (I specifically would like to point out that you can prove this simply by using the fact that $[a,b]$ is a "closed" interval. First-countability or the topological concept of closure is not necessary.) This allows me to make the continuity claim, that $\{f(x_{n_j})\}\to f(x)$ . Other than this, I do not see why the argument is not sound.

My main issue with your proof was that it calls on some "heavy" tools: uniform continuity and Cauchy sequences. The proving their associated results actually takes some work, and therefore leads to a longer proof.

**Drexel28** · November 22nd 2010, 09:09 AM

Originally Posted by roninpro

You are right. I implicitly used closure here: $\{x_{n_j}\}\to x\in [a,b]$ . (I specifically would like to point out that you can prove this simply by using the fact that $[a,b]$ is a "closed" interval. First-countability or the topological concept of closure is not necessary.) This allows me to make the continuity claim, that $\{f(x_{n_j})\}\to f(x)$ . Other than this, I do not see why the argument is not sound.

My main issue with your proof was that it calls on some "heavy" tools: uniform continuity and Cauchy sequences. The proving their associated results actually takes some work, and therefore leads to a longer proof.

Trust me, friend. I understand what's going on here. I just feel as though uniform continuity and Cauchy sequences are A) not "heavy" and B) provide an easier proof. But, I guess when it comes down to it it's a matter of opinion.

**roninpro** · November 22nd 2010, 09:24 AM

I meant no insult to your competency. I'm just trying to be very clear about where I am attempting to shorten the proof.

I think that the issue boils down to what we mean by "shorter proof". If we mean less words, then I can agree that after all of the details are written, our two approaches are more or less the same. However, when I said "shorter", I meant that it uses less tools and results; in other words, it takes less to make it self-contained. (I think that it is better of have a more elementary proof - though, this could be a matter of opinion.) Your method uses sequences, Cauchy sequences, continuity, and uniform continuity and some results. I'm sure that you would agree that this is much more than my approach, which uses sequences, continuity and some results.

**Drexel28** · November 22nd 2010, 09:28 AM

Originally Posted by roninpro

I meant no insult to your competency. I'm just trying to be very clear about where I am attempting to shorten the proof.

I think that the issue boils down to what we mean by "shorter proof". If we mean less words, then I can agree that after all of the details are written, our two approaches are more or less the same. However, when I said "shorter", I meant that it uses less tools and results; in other words, it takes less to make it self-contained. (I think that it is better of have a more elementary proof - though, this could be a matter of opinion.) Your method uses sequences, Cauchy sequences, continuity, and uniform continuity and some results. I'm sure that you would agree that this is much more than my approach, which uses sequences, continuity and some results.

No offense taken! I don't know why we're discussing this. I'm just saying that if you have the tools use them.

**roninpro** · November 22nd 2010, 09:35 AM

Fair enough. I guess we'll leave it at that.

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