Uniformly continuous and bounded

[math2006]

Let $$f$$ be a real uniformly continuous function on the bounded
set $$A$$ in $$\mathbb{R}^1$$. Prove that $$f$$ is bounded
on $$A$$.

Since f is uniformly continuous, take $$\epsilon = m, \exists \delta > 0$$
such that
$$|f(x)-f(p)| < \epsilon$$
whenever $$|x-p|<\delta$$ and $$x,p \in A$$
Now we have
$$|f(x)| < m + |f(p)|$$

Obviously i should show $$b + |f(p)|$$ is bounded, but no idea how.
Could someone help me? thx

 

[AKG]

Obviously i should show $$b + |f(p)|$$ is bounded, but no idea how.

What is b? And why would you want to show that b + |f(p)| is bounded? You mean bounded as a function of p? Well that's no easier than showing directly that f is bounded, so this doesn't seem to be a worthwhile approach.

The idea is simple. Fix $\epsilon > 0$. There exists $\delta > 0$ such that, well, you know. Since A is bounded, you can choose a FINITE number of points x₁, ..., x_n in A such that the intervals of radius $\delta$ about the x_i cover A. The total variation of f on these intervals is at most $2\epsilon$, so the total variation of f over all of A is at most $2n\epsilon$.

Oh, and if this was homework, it should have been in the homework section.

 

[tacman]

To prove that f is bounded on A, we need to show that there exists a real number M such that |f(x)| < M for all x \in A.

Since A is a bounded set, there exists a real number M' such that |x| < M' for all x \in A.

Now, let M = m + M'. We can see that for any x \in A, we have
|f(x)| < m + |f(p)| < m + M' = M
where p is some point in A.

Therefore, we have shown that |f(x)| < M for all x \in A, which means that f is bounded on A. This is because for any x \in A, we can always find a point p \in A such that |f(p)| < M'. Then, using the fact that f is uniformly continuous, we can find a \delta > 0 such that |f(x)| < m + |f(p)| < m + M' = M whenever |x-p| < \delta.

Hence, we have proven that f is bounded on A.