Assume f is uniformly continuous on a bounded set S in Rⁿ. Prove that f must be bounded on S.

Let f: S ® T ; (S,d_s),(T,d_t) are metric spaces.

f is uniformly continuous on S means " e > 0 there is a d_e> 0 such that if x,p are in S then:

The proof is fairly good. The problem with it is the fact that he didn't use the boundedness of S, and he (somewhat) missed talking about d in his last two arguments.

We must use the fact that S is bounded. The reason is that if S was not bounded, and we assume (for contradiction) that f is not bounded, we can not be sure that we can find a situation where (*) and (**) are satisfied.

For instance, take S to be R. Let f(x) = x. This function is uniformly continuous, even though it is unbounded.

Also, the last two statements are not stated very clearly.

A better (but not perfect) proof (using his answer as a template) would look like this:

Assume f is uniformly continuous on a bounded set S in Rⁿ. Prove that f must be bounded on S.

Let f: S ® T, where (S,d_s) and (T,d_t) are metric spaces.

We have that f is uniformly continuous on S. That is, " e > 0 there exists a d_e > 0 such that if x, p are in S then: d_t(f(x),f(p)) < e whenever d_s(x,p) < d

We must show that f is bounded on S.

Assume (for contradiction) that f is not bounded on S. Since S is bounded, there exists a d such that d_s(x,p) < d for all p in S (!). Since f is not bounded, " x in S, there exists a p in S such that given some e₁ > 0, d_s(x,p) < d but d_t(f(x), f(p)) > e₁. (*)

Choose some x in S. Then " p in S (Note that d_s(x,p) < d, see (!)) we have, since f is uniformly continuous, that for e₁ > 0, there exists some d_e < d such that if d_s(x,p) < d_e then d_t(f(x),f(p)) < e₁ (**).

Thus, x and p satisfy (*), since f is unbounded. But, x and p also satisfy (**), since f is uniformly continuous and S is bounded. Then we have that e₁ < d_t(f(x),f(p)) > e₁. Contradiction.

Hence f must be bounded on S.

Questions for Robert.

1. What is exactly a bounded function ? (definition please)

2. Give an example of a continuous function f:S ® R^k, where S is a bounded subset of Rⁿ, such that f is not bounded on S.

1. A function f: S ® R^k is bounded on S if there is a positive number M such that ||f(x)|| <= M for all x in S.

That is, the image f(S) is a bounded set.

2. f: (0,1) ® R defined by f(x)=1/x.

Note that f is continuous everywhere that it is defined, and it is defined on (0,1).

(0,1) is bounded by 0 to the left, and 1 to the right.

f((0,1)) = (0,+¥) which obvoiusly is not bounded to the right.

Closing comments:

The definition of bounded function includes that the range is R^k (not just any metric space). Both Robert and Justin use another definition. There are some gaps in the first proof; the second one is sufficiently essentially correct. Problem closed.