In the elementary case in which f is a M-Lipschitz real function(M>0), the constants can be chosen alpha=M and beta=/f(0)/.

I have a solution for the second question of the problem.

The problem (second question) assumes that a real function f from R to R is continuous, monotone, and bounded on R, and asks to show that f is uniformly continuous on R.

Using that f is monotone and bounded, we will first show that f has finite limits at minus infinity and infinity. Then, using that f is continuous on R and has finite limits at minus infinity and infinity, we will show that f is uniformly continuous on R.

We assume that f is increasing. This is no loss of generality, since if f is decreasing, then –f is increasing, and if the limits of –f exist (as we will show), then the limits of f are the opposites of those of –f; thus, they exist, as well.

Since f is bounded, the image f(R)={f(x):x in R} of f is bounded; thus, it is bounded below and above. Clearly, f(R) is nonempty. As a result, both inff(R) and supf(R) exist in R. We will show that the limit of f at minus infinity is equal to inff(R) and the limit of f at infinity is equal to supf(R).

For every epsilon>0, we have inff(R)< inff(R)+epsilon, and thus inff(R)+epsilon is not a lower bound of f(R). Hence, there exists x0 in R such that f(x0)< inff(R)+epsilon. Since f is increasing, we have f(x) supf(R)-epsilon, for every x>x1. Also, f(x)

Thanks Spiros for your valuable comment. Regarding last part of your comment, such domains A are interesting when A=(-oo, a]\/[b,oo) with a

Dear Dinu and all,

I nice solution of the first question of the problem, using the epsilon-delta definition of uniform continuity, can be found here https://math.stackexchange.com/questions/1014598/if-f-is-a-uniformly-continous-function-then-fx-leq-axb (Alfred Chern's answer).

I tried unsuccessfully, using a sequential argument, to show that if a function f from R to R is uniformly continuous on R, then there exists c>0 such that the function f/x is bounded for |x|>c. If this is true, then |f(x)|c and for some a>=0. Also, since f is continuous, we have |f(x)|

Dear Spiros,

The proof from stackexchange is valid and your idea is almost the same, so yes, I think statement "there exists c>0 such that the function f/x is bounded for |x|>c" is also valid.

I try below to show you are right.

From f uniformly continuous, using the definition, taking epsilon=1, we obtain

there exist c>0 such that /f(x)-f(y)/

It may be useful to note that the converse statement of that in question 1 of the problem does not hold; i.e., if f:R->R is continuous on R and |f(x)|

I'm thinking we could try a proof by contradiction for question 1! Using sequences.

Suppose that for all positive reals a,b there exists a real x(a,b) with

/f(x(a,b))/ > a/x(a,b)/+b. Now consider for all n positive integer some suitable a=a(n), b=b(n) >0 and we will have there exists a sequence x(n) such that

/f(x(n))/ > a(n)/x(n)/+b(n)

Next for other suitable a=a1(n),b= b1(n)>0, we will have another sequence y(n) satisfying the above inequality.....

No! I don't think it works, having nothing to evaluate the difference xn-yn !

Along the same lines, I tried to prove the following.

There exist a,b>=0 and c>0 such that for every |x|>c, we have |f(x)|=0 and c>0 there exists |x|>c such that |f(x)|>a|x|+b.

Choosing any a1,b1>=0 and any c1>0, there exists |x1|>c1 such that |f(x1)|>a1|x1|+b1. Note that |x1| is strictly positive.

Now, for every positive integer n>1, we choose an=1/|xn_1|, bn=|f(xn-1+1/n)|, and cn=|xn-1|. Then there exists |xn|>cn=|xn-1| such that |f(xn)|>|xn|/|xn-1|+|f(xn-1+1/n)|, whence |f(xn)|-|f(xn-1+1/n)|>|xn|/|xn-1|. Further, we have |f(xn)-f(xn-1+1/n)|>=|f(xn)|-|f(xn-1+1/n)|, and |xn|/|xn-1|>1, since |xn|>cn=|xn-1|. Hence, we have |f(xn)-f(xn-1+1/n)|>1. However, the difference of the two sequences (xn) and (xn-1+1/n) does not necessarily converge to 0, since if (x_n) diverges, the difference xn-xn-1 may not converge to 0.

The problem is that when we choose an and bn, we do not know what xn is. First we choose an and bn (and cn), and then we get an xn that, obviously, depends on an and bn (and cn).