Nice problem. Is it yours?

This is how I see it.

Use L'Hospital for f = (f e^(-x))/e^(-x) to conclude that f and thus f' have finite limit at infinity.

This implies that e^(-x) f has limit 0 at infinity and so there is c > 0 such that

| e^(-x) f | < 1 on (c, infinity)

Since 1 < (1/c) x on (c, infinity, | e^(-x) f | < (1/c) x on (c, infinity) so

| f | < (1/c) x e^x on (c, infinity)

Let b = 1 + the maximum of | e^(-x) f | on [0, c]

Thus | e^(-x) f | < b on [0, c] or |f|< b e^(-x) on [0, c]

Now if you take a = 1/c you get the conclusion.

I may be wrong but I don't think I used the fact that the derivative is continuous.

Yes, it's my, GM 12/2001, I found it today, searching for GM problems with inequalities.

Yes, It seems your solution don't use the continuity of the derivative.

Unfortunately, I don't remember my initial solution!

I think me too I used function h=e^(-x)f. Oh, yes! h' has limit 0 at infinity. Now I use h' continuous and it results h' is bounded on [0,infinity), then h is Lipschitz. It results /h/ is Lipschitz, and from here

/h(x)/=0, m = sup/h'/.

Now we take a=m and b=/h(0)/=/f(0)/.

Hi Dinu, thank you for pointing to this interesting problem.

I would like to share consequences of the conclusion of the fact obtained by @Gabriel T. Prajitura in his first step, which reads:

f is continuous in I , with finite limit at oo.

In particular,

. . . . f is bounded in I. . . . . (*)

____________________

Corollary. For every bounded positive function g:I->(0,oo) and every function f satisfying assumption of the problem there exists a constant A such that for every x ε I we have

. . . . . |f(x)| < A / g(x) . . . . (**)

In particular for g(x) = e^{-x} one gets that for some constant A

. . . . |f(x)| < A e^x. . . . . (***)

etc.

Gabriel T. Prajitura Thanks for solution! Clearly the continuity of f' is not used, so it's not necessary.

Joachim Domsta Dear Joachim, thanks for joining the discussion.

Dear All, I am sorry that I didn't notice an omission of the assumptions in this inference:

>

as a counter example one may use f(x) = e^x :-(

I think we can show Dinu’s inequality with fewer assumptions regarding the function f and for a more general version of the function g, by noting that if a real function f is bounded on a subset A of R and a real function g is bounded below by a positive real number on A, then there exists k>0 such that |f(x)|=m for some m>0. Choosing k=M/m>0, we have kg(x)>=km=M>=|f(x)|; i.e., |f(x)|0. Further, f is continuous on the closed and bounded interval [0,x0], and thus bounded there. As a result, f is bounded on [0,infinity). Now, considering the function g such that g(x)=exp(x), we have that g is bounded below by 1>0 on [0,infinity). Consequently, there exists k>0 such that |f(x)|0 and a any nonnegative real number, we have kexp(x)

Spiros Kontantogiannis

Indeed your post is just a more detailed prove of my corollary (**) from (*).

The issue is now that condition (*) follows from incorrectly obtained in the first post existence of limit of f from the Dinu assumptions. The incorrectness of (*) based on the Dinu assumption simply is proven by the counterexample

f(x) = e^x, ..for x>=0.

Indeed, it satisfies the Dinu assumptions since f-f'=0 in I=[0,oo). However this f is not bounded.

Joachim Domsta , thanks. I confused Dinu's assumption about f and its derivative. I thought he assumed that both f and its derivative have finite limit at infinity, instead of their difference.

Joachim Domsta Spiros Konstantogiannis Thanks for your valuable comments!

Dear All,

The proof by Dinu can be slightly enhanced allowing for strengthening the claim as follows:

. . . h(x) := f(x) e^{-x}

. . . . . . . .= f(0) + int_[0,x] h'(u) du

Since by assumptions f'-f is bounded hence

. . . |h'(u)| == |f'(u)-f(u)| e^{-u}

. . . . . . . . =< m e^{-u}

where m denotes the supremum of |f'-f| . Consequently for every x we have

|f(x)| e^{-x} =< |f(0)| + int_[0,x] |h'(u)| du

. . . =< |f(0)| + m int_[0,x] e^{-u} du

. . . =< |f(0)| + m

After multiplication by the exp function we obtain the following stronger claim

. . . |f(x)| =< ( |f(0)| + m ) e^x . . . (*)

This estimate cannot be substantially stronger since for f=exp we get h=0 and f(0)=1 and m=0 which implies equality in the inequality (*).

Note that L'Hôpital's rule cannot be used here, as suggested by Professor Pratijura, since f(x)e^{-x} need not have limit 0, in general. In fact, the function f(x)=e^x satisfies that f-f' is zero (and has finite limit at infinity), but f has no finite limit (nor f', either).