If not stated otherwise, we assume that the metric spaces \mathbb{R} and \mathbb{Q} are quipped with the Euclidean metric d(x,y)=|x-y|.
Exercise: If K\subset\mathbb{R} and f:K\rightarrow\mathbb{R} is continuous. Show that either f(x)=0 for some x\in K or there exists n\in\mathbb{N} such that |f(x)|>1/n for all x\in K.
Proof. Since f is continuos and K is compact, f(K) is a compact subset of \mathbb{R}, hence is closed. If 0\notin f(K), then there exists n\in\mathbb{N} such that [-1/n, 1/n]\cap f(K)=\emptyset, which gives us that |f(x)|>1/n for all x\in K.
Exercise: Let I=[0,1] be the closed unit interval. Suppose f is a continuous mapping from I to I. Prove that there exists at least one x\in I such that f(x)=x. Hint: Consider the function g(x)=f(x)-x.
Proof. Since g(0)=f(0)-0\geq 0 and g(1)=f(1)-1\leq 0, by the intermediate value theorem, there exists at least one x\in I such that g(x)=0, i.e., f(x)=x.
Exercise: Give an example (with proof) of two real-valued functions f and g which are uniformly continuous, but whose product f\cdot g is not uniformly continuous.
Proof. (An example without proof) Let f(x)=g(x)=x for all x\in\mathbb{R}.
Exercise: Let f:E\rightarrow\mathbb{R} be a uniformly continuous function on the open interval E=(a,b)\subset\mathbb{R}. Show that f can be extended to a uniformly continuos function on the closed interval [a,b]. More precisely, show that there is a uniformly continuous function g:[a,b]\rightarrow\mathbb{R}such that f(x)=g(x) for all x\in (a,b). Hint: show first that \lim_{x\rightarrow a}f(x) and \lim_{x\rightarrow b}f(x) exist.
Proof. (Sketch) Following the hint, we first prove that the limits exist.
Claim: For any sequence \{a_n\} that converges to a, the sequence \{f(a_n)\} is actually a Cauchy sequence in \{\mathbb{R}\}, thus converges.
Proof of claim: to show that \{f(a_n)\} is a Cauchy sequence, it’s sufficient to show that for every \epsilon>0, there is N such that if m,n\geq N then |f(a_m)-f(a_n)|<\epsilon. Pick your favorite \epsilon first. As f is uniformly continuous, there is \delta> 0 such that for any x,y\in(a,b) if |x-y|\delta, |f(x)-f(y)|<\epsilon. As \{a_n\} converges to a, there is N such that for all n\geq N, |a_n-a|<\delta/2. Thus for any m,n\geq N, |a_m-a_n|\leq|a_m-a|+|a_n-a|<\delta, whence |f(a_m)-f(a_n)|<\epsilon.
Now, we’ve shown that for every sequence \{a_n\} that converges to a, the sequence \{f(a_n)\} converges. But how do you know for two different sequences \{a_n\} and \{b_n\} that converge to a, \{f(a_n)\} and \{f(b_n)\} have the same limit?
There is a slick way to resolve this. We can mingle \{a_n\} and \{b_n\} together. Specifically, consider a new sequence \{a_1, b_1, a_2, b_2, \ldots\}, denoted as c_n. Now c_n converges to a, so \{f(c_n)\} converges. Meanwhile, we know that \{f(a_n)\} and \{f(b_n)\} are two subsequences of \{f(c_n)\}, so they share the same limit.
Now we can extend f to g by feeding in the limits at a and b. Since g is a continuous function on a closed interval [a,b], its uniformly continuity follows immediately.
Exercise: Let a< b be real numbers. A function f:(a,b)\rightarrow\mathbb{R} is said to be convex if
f(\lambda x+(1-\lambda)y)\leq f(x)+(1-\lambda)f(y) \text{ for all } x,y\in(a,b) \text{ and all } \lambda\in (0,1).
Prove that every convex function is continuos. Hint: prove first that if a<s<t<u<b, then
\frac{f(t)-f(s)}{t-s}\leq\frac{f(u)-f(s)}{u-s}\leq\frac{f(u)-f(t)}{u-t}
Proof. First, we prove the first inequality in the hint. Notice that
\frac{f(t)-f(s)}{t-s}\leq\frac{f(u)-f(s)}{u-s}
is equivalent to
f(t)\leq\frac{t-s}{u-s}f(u)+\frac{u-t}{u-s}f(s).
But notice that if we take x=u, y=s and \lambda=\frac{t-s}{u-s} and plug them into the convexity inequality, the above inequality holds immediately. Similarly, we can prove the second inequality in the hint.
For a<x<y<b, let k(x,y)=\frac{f(x)-f(y)}{x-y}. By the inequality chain in the hint, we know that if a<s<t<u<b, k(s,t)\leq k(s,u)\leq k(t,u). Now for any x\in(a,b) and \epsilon>0, pick \delta\in(0,\epsilon) such that a<x-\delta<x+\delta<b and x_1\in(a,x-\delta), x_2\in(x+\delta,b). For any y\in(x-\delta,x+\delta), by the chain inequality in the hint, k(x,y) is bounded below by k(x_1,x) and is bounded above k(x,x_2). Thus there is B such that |k(x,y)|<B for all y\in(x-\delta,x+\delta). Whence, for all y\in(x-\delta/B, x+\delta/B), |f(x)-f(y)|<B|x-y|<\delta<epsilon. This shows that f is continuous.