Uniform continuity and boundedness - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-21T06:41:40Z http://mathoverflow.net/feeds/question/82604 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82604/uniform-continuity-and-boundedness Uniform continuity and boundedness krje1980 2011-12-04T09:00:36Z 2011-12-04T10:39:49Z <p>Hi.</p> <p>I have come across a proof which I understand almost completely, except for one part:</p> <p>THEOREM: If $f$ is uniformly continuous on a bounded interval $I, [a,b]$ then $f$ is also bounded on $I$.</p> <p>PROOF: Fix an $\epsilon > 0$, for instance $\epsilon = 1$. Since $f$ is uniformly continuous, there is a $\delta > 0$ such that:</p> <p>$|f(x_1) - f(x_2)| &lt; \epsilon = 1$ when $x_1, x_2 \in I$ and $|x_1 - x_2| &lt; \delta$</p> <p>Divide $I$ into $N$ intervals, $I_1, . . ., I_N$, where $N$ is chosen so that $\frac{b-a}{N} &lt; \delta$.</p> <p>Let $z_i$ be the center point of $I_i$. For each $i$ and $x \in I_i$, $|x - z_i| &lt; \delta$, and then we have:</p> <p>$|f(x)| = |f(x) - f(z_i) + f(z_i)| \leq |f(x) - f(z_i)| + |f(z_i)| \leq 1 + |f(z_i)|$. Then for $x \in I_i$,</p> <p>$|f(x)| \leq 1 + max_{1 \leq i \leq N}{|f(z_i)|}$.</p> <p>Let $M = max_{1 \leq i \leq N}{|f(z_i)|}$. Then $|f(x)| \leq 1 + M$</p> <p>QED</p> <p>OK, so the one thing I am a bit unsure of here, is when we write:</p> <p>Let $M = max_{1 \leq i \leq N}{|f(z_i)|}$.</p> <p>How is it that we know for sure that each $|f(z_i)|$ is also bounded? I see how the presence of a maximum value completes the proof, but why is it not possible that we have an $|f(z_i)|$ which is unbounded?</p> <p>If anyone could explain this to me I would greatly appreciate it!</p> <p>Also, for what it's worth, I tried to solve this my own way, but I am not sure if the proof is rigorous enough (it's much simpler!). It goes as follows:</p> <p>PROOF BY CONTRADICTION</p> <p>Suppose $f$ is not bounded on $I$. Then, for each $M > 0$, we have $|f(x)| > M$ for some $x \in I$. However, since $f$ is uniformly continuous, for every $\epsilon > 0$ there exists a $\delta > 0$ such that</p> <p>$|f(x) - f(y)| &lt; \epsilon$ when $x, y \in I$ and $|x - y| &lt; \delta$</p> <p>And it follows from this that:</p> <p>$|f(x)| &lt; \epsilon + f(y)$</p> <p>Which is a contradiction if $|f(x)|$ is greater than any $M > 0$.</p> <p>QED</p> <p>If anyone also can let me know if my proof is OK, I would also be very grateful!</p> http://mathoverflow.net/questions/82604/uniform-continuity-and-boundedness/82606#82606 Answer by Andrei MF for Uniform continuity and boundedness Andrei MF 2011-12-04T09:34:52Z 2011-12-04T09:34:52Z <p>The theorem you mention is kind of strange. You don't need to assume uniform continuity, it is enough to suppose that your function $f$ is continuous: every continuous function on a compact subset of $\mathbb R$ is automatically uniformly continuous. Then, what you are trying to prove is that continuity on a compact $\Rightarrow$ boundedness (so called, extreme value theorem, see <a href="http://en.wikipedia.org/wiki/Extreme_value_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Extreme_value_theorem</a> where a the standard proof is outlined).</p> http://mathoverflow.net/questions/82604/uniform-continuity-and-boundedness/82609#82609 Answer by Toby Bartels for Uniform continuity and boundedness Toby Bartels 2011-12-04T10:39:49Z 2011-12-04T10:39:49Z <p>I'm afraid that I don't like your proposed proof. You derive a bound on $f(x)$, namely $\epsilon + f(y)$, but this is not fixed. Although you may choose any positive $\epsilon$ you wish (which then gives you $\delta$), $y$ must be chosen to be within $\delta$ of $x$. So as you vary $M$, you vary $x$ (to keep $f(x) > M$), but then (to keep it close enough to $x$) you vary $y$, and it seems possible that $f(y)$ would grow fast enough that $\epsilon + f(y) > M$ would be maintained.</p>