# Proving that a bounded, continuous function has a supremum

Theorem: Let $f$ be a continuous real-valued function on a closed rectangle $R$ in $\Bbb R^2$. Then,

(a) $\quad f$ is bounded on $R$

(b) $\quad$There exist points $c$ and $d$ in $R$ so that $f(c) = \sup_{p\in R} f(p), \quad f(d) = \inf_{p\in R} f(p)$.

(c) $\quad f$ is uniformly continuous on $R$.

My Work

(a) $\quad$ To prove that $f$ is bounded on $R$, we need to show that there is an $U$ so that $|f(p)| \le U$ for all $p \in R$. Suppose this is false. Then for each large integer $n$, there is a $p_n \in R$ such that $|f(p_n)| > n$. By Bolzano-Weierstrass, the bounded sequence $\{p_n\}$ has a convergent subsequence $\{p_{n_k}\}$ that converges to $p_0$ as $k \to \infty$. Since $f$ continuous, $f(p_{n_k}) \to f(p_0)$ as $k \to \infty$. But this is impossible since $f(p_{n_k})$ is unbounded by virtue of $f(p_n)$ being unbounded. So $f$ is bounded on $R$.

(b) $\quad$ Since $S = \{f(x) : x \in R \}$ is nonempty, $S$ has a supremum. Call this supremum $M$. Consider a sequence $\{c_n\}$ and the interval $M - \frac{1}{n} \le f(c_n) \le M$. Since this interval is closed, $f$ must have a convergent subsequence $\{c_{n_k}\}$ that converges as $k \to \infty$. Taking $N \ge 1/\epsilon$, we have $M - (M - 1/n) = 1/n \le \epsilon$ for $n \ge N$. So since $f$ is continuous we must have $f(c_n) \to f(c) = M$.

My Question: I feel pretty sure of my proof of (a), but (b) seems like it lacks a key component. And I am not sure how to prove (c) for any arbitrary $f$, given that the definition of uniform continuity for a continuous $f$ on a set $E \subseteq \Bbb R ^2$ is given $\epsilon > 0$, there exists a $\delta > 0$ such that $$\|q - p \| \le \delta \ \Rightarrow \ |f(q) - f(p) | \le \epsilon \quad \forall \ q, p \in E$$

Edit Revised part (c):

If $f$ is not uniformly continuous on $R$, then, from the definition of uniform continuity, there are two sequences $(x_n)$ and $(y_n)$ in $R$ and an $\epsilon_0>0$ so that for each $n$ we have $$\tag{1}\Vert x_n-y_n\Vert\le 1/n\ \ \ \text{ and }\ \ \ \Vert f(x_n)-f(y_n)\Vert\ge\epsilon_0.$$ Consider a convergent subsequence $(x_{n_k})$ of $(x_n)$. This converges to some $x\in R$. So for some large integer $K \ge 1/\epsilon$, and $k \ge K$, we have $\Vert y_{n_k} - x \Vert \le \epsilon$ so $(y_{n_k})$ converges to $x$ as well.

From this we have that $(x_{n_k}), (y_{n_k}) \to x$, but $\Vert f(x_{n_k}) - f(y_{n_k}) \Vert \ge \epsilon_0$ which implies that $f$ is not continuous at $x$ which contradicts the hypothesis.

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I think the assertion "Since $f$ is bounded on $R$ it acheives a maximum" is exactly what you're trying to prove, isn't it? – yohBS Aug 21 '12 at 14:07
Once you know that any interval $[a,b]$ is compact in $\mathbb{R}$, then (b) is Weierstrass' theorem and (a) is a trivial consequence of (b). – Siminore Aug 21 '12 at 14:31
We have not defined compactness, we are supposed to do these problems simply with continuity and Bolzano-Weierstrass – Moderat Aug 21 '12 at 14:37

For part b), you cannot start by saying "$f$ achieves a maximum...". This is what you are trying to prove. Instead, start by observing that the set $\{f(x)|x\in R\}$ is nonempty and bounded above. It thus has a supremum, $U$. Now (carefully) define your $c_n$ so that $U-1/n\le f(c_n)\le U$ and continue with your argument.

For part c), you could argue as follows:

If $f$ is not uniformly continuous on $R$, then, from the definition of uniform continuity, there are two sequences $(x_n)$ and $(y_n)$ in $R$ and an $\epsilon_0>0$ so that for each $n$ we have $$\tag{1}\Vert x_n-y_n\Vert\le 1/n\ \ \ \text{ and }\ \ \ \Vert f(x_n)-f(y_n)\Vert\ge\epsilon_0.$$ Consider a convergent subsequence $(x_{n_k})$ of $(x_n)$. This converges to some $x\in R$. Show that the corresponding subsequence $(y_{n_k})$ of $(y_n)$ also converges to $x$. Now, show that this and $(1)$ imply that $f$ is not continuous at $x$.

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I am unsure what you mean by "carefully define your $c_n$." I know that I cannot simply say "let $c_n$ be such that $U - 1/n \le f(c_n) \le U$" but I am unsure how else to go about constructing such a $c_n$. – Moderat Aug 21 '12 at 14:32
@jmi4 It was your phrasing you used. "consider the sequence $\{c_n\}$... " should be changed to "consider a sequence $\{c_n\}$...". I also think a bit of justification is needed here. Not much is needed though, you could just say "from the definition of the supremum of a set, for each positive integer $n$, there is a $c_n$ in $R$ with $U-1/n\le f(c_n)\le U$". – David Mitra Aug 21 '12 at 14:41
how does part (b) look now that it has been revised? – Moderat Aug 21 '12 at 14:56
@jmi4 Mostly ok; but it seems poorly phrased to me. To show that $(y_{n_k})$ converges to $x$, you should first write $|y_{n_k}-x|\le|y_{n_k}-x_{n_k}|+|x_{n_k}-x|\le {1\over n_k}+|x_{n_k}-x|$. Now given $\epsilon>0$, choose $k$ so that the right hand side of the preceding inequality is less than $\epsilon$. – David Mitra Aug 21 '12 at 14:58
@jmi4 In general, to prove something converges, you have to 1) fix a value of $\epsilon>0$. 2) explicitly state what your $N$ or $\delta$ or whatever is. 3) show that your choice made in 2) "works". – David Mitra Aug 21 '12 at 15:00

The answer to a) looks good. For b), how do you know that there is some $c\in R$ where $f$ achieves its maximum? Isn't this what you are supposed to prove?
All you know so far is that $f$ is bounded.

But a proof similar to your argument for a) works. Since $f$ is bounded, there is a supremum $S$ of the set $\{f(x):x\in R\}$. Choose a sequence $(p_n)$ of points in $R$ such that $S-f(p_n)<1/n$.
As before, since $R$ is compact, the sequence $(p_n)$ has a convergent subsequence $(p_{n_k})$. Now show that $p=\lim_{k\to\infty}p_{n_k}$ satisfies $f(p)=S$.

Alternatively, use the open cover formulation of compactness (every open cover of $R$ has a finite subcover) to show that continuous images of compact sets are compact. Then you get that $f[R]$ is compact. By Heine-Borel, $f[R]$ is closed and bounded and therefore has a minimum and a maximum.

For the uniform continuity part I prefer to use compactness of $R$ in the formulation with open covers.
Given $\varepsilon>0$, for each $x\in R$ find $\delta_x$ such that for each $y\in R$ with $|x-y|<\delta_x$ we have $|f(y)-f(x)|<\varepsilon/2$. This is possible since $f$ is continuous.

Now the collection of all open balls $B(x,\delta_x)$ (open ball of radius $\delta_x$ around $x$) is an open cover of $R$. This cover has a finite subcover. Let $x_1,\dots,x_n$ be the centers of the open balls in the finite subcover. Now $\delta=\min(\delta_{x_1},\dots,\delta_{x_n})$ works for uniform continuity.

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Unfortunately, we have not used the fundamentals of open balls and coverings so that machinery is unavailable to me... – Moderat Aug 21 '12 at 14:47
Sorry, but I see that you now have a proof that only uses sequences. So this works, too. I like the open covers approach, as this works in more general situations. – Stefan Geschke Aug 21 '12 at 18:11

I'm posting this as an answer to $(1)$: take advantage of the length that is allowed for an answer and $(2)$: to refrain from further cluttering the question.

Let part $(a)$ stand.

$(c)$: Given $f$ is continuous on $R \subset \Bbb R^2$, prove that $f$ is uniformly continuous on $R$.

Proof: By the definition of continuity, for $x, y \in R$, given $\epsilon > 0$, there exists a $\delta > 0$ s.t. $$\tag{1} \Vert x - y \Vert \le \delta \Rightarrow \Vert f(x) - f(y) \Vert \le \epsilon$$ Now, suppose that $f$ is not uniformly continuous on $R$ despite being continuous on $R$. Then there exists some $\epsilon_0 > 0$ for which no $\delta$ can be found to satisfy $(1)$. In particular, for $\delta_n = {1 \over n}, n \in \Bbb N$, there are sequences $(x_n), (y_n)$ so that for each $n \in \Bbb N$ $$\|x_n - y_n \| \le \delta_n \Rightarrow \|f(x_n) - f(y_n) \| \ge \epsilon_0$$ Since $(x_n), (y_n)$ are bounded, by Bolzano-Weierstrass they each have subsequences $(x_{n_k}), (y_{n_k})$ that respectively converge to some point $x, y \in R$. We want to show that $x = y$, that is, $y_{n_k} \to x$.

To see this, consider the following: $$\|y_{n_k} - x \| \le \|y_{n_k} - x_{n_k}\| + \|x_{n_k} - x\| \le {1\over n_k} + \| x_{n_k} -x\|$$ Now, given $\epsilon > 0$, since $x_{n_k} \to x$, we can take $K_1$ to be some large integer such that for $k > K_1$, $\|x_{n_k} - x \| \le {\epsilon \over 2}$. Additionally, we can pick $K_2$ so that for $k > K_2$, $n_k \ge {2 \over \epsilon}$. Thus, for $k \ge K_3 = \max\{K_1, K_2\}$ we have $${1 \over n_k} + \|x_{n_k} - x \| \le {\epsilon \over 2} + {\epsilon \over 2} = \epsilon$$ Therefore, we have that $(y_{n_k}) \to x$. However, even though $(x_{n_k}), (y_{n_k}) \to x$, we cannot have $(f(x_{n_k})), (f(y_{n_k}))$ converge to the same number because $\|f(x_n) - f(y_n) \| \ge \epsilon_0$ for all $n$. However, this contradicts the hypothesis that $f$ was continuous, so we are done.

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