Question

Hi.

I have come across a proof which I understand almost completely, except for one part:

THEOREM: If $f$ is uniformly continuous on a bounded interval $I, [a,b]$ then $f$ is also bounded on $I$.

PROOF: Fix an $\epsilon > 0$, for instance $\epsilon = 1$. Since $f$ is uniformly continuous, there is a $\delta > 0$ such that:

$|f(x_1) - f(x_2)| < \epsilon = 1$ when $x_1, x_2 \in I$ and $|x_1 - x_2| < \delta$

Divide $I$ into $N$ intervals, $I_1, . . ., I_N$, where $N$ is chosen so that $\frac{b-a}{N} < \delta$.

Let $z_i$ be the center point of $I_i$. For each $i$ and $x \in I_i$, $|x - z_i| < \delta$, and then we have:

$|f(x)| = |f(x) - f(z_i) + f(z_i)| \leq |f(x) - f(z_i)| + |f(z_i)| \leq 1 + |f(z_i)|$. Then for $x \in I_i$,

$|f(x)| \leq 1 + max_{1 \leq i \leq N}{|f(z_i)|}$.

Let $M = max_{1 \leq i \leq N}{|f(z_i)|}$. Then $|f(x)| \leq 1 + M$

QED

OK, so the one thing I am a bit unsure of here, is when we write:

Let $M = max_{1 \leq i \leq N}{|f(z_i)|}$.

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?

If anyone could explain this to me I would greatly appreciate it!

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:

PROOF BY CONTRADICTION

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

$|f(x) - f(y)| < \epsilon$ when $x, y \in I$ and $|x - y| < \delta$

And it follows from this that:

$|f(x)| < \epsilon + f(y)$

Which is a contradiction if $|f(x)|$ is greater than any $M > 0$.

QED

If anyone also can let me know if my proof is OK, I would also be very grateful!

Answer 1

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 http://en.wikipedia.org/wiki/Extreme_value_theorem where a the standard proof is outlined).

Answer 2

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.