Proving a positive function doesn't exist with the condition $f(x+y)\geq yg(f(x)) \ \ , \ \ x>0 \ \ , \ \ y>0$

Question

This is an old question from Nieuw Archief voor Wiskunde 23(1975) p.242. I don't have access to this journal but I would really like to see the solution. Here is the question:

Let $g$ be a positive and continuous function on $(0,\infty)$ with the property that $$\int_1^\infty\frac{ds}{g(s)}<\infty$$Prove that there exists no positive and continuous function $f$ on $(0,\infty)$ such that $$f(x+y)\geq yg(f(x)) \ \ , \ \ x>0 \ \ , \ \ y>0$$

Any help would be appreciated, even if it is digging the original journal and the solution posted there.

Answer 1

(I am an amateur Mathematician and not an expert - I'm just answering Mathematics question for fun in my spare time. Would be nice if one could provide feedback if any error or mistakes are present - that would be helpful.)

Question: Show that there exists no positive and continuous function $f$.

Since we are trying to prove that there does not exist such $f$, it is much easier to proof such statements using a logical trick, which is to proof its contrapositive instead.

For a logical statement If (Set) A $\Rightarrow$ Then (Set) B, then its contrapositive, which means the same thing, is just If Not (Set) B $\Rightarrow$ Then Not (Set) A. For example, If Exist (Set) Rains $\Rightarrow$ Then Bring (Set) Umbrella. Then its contrapositive, which means the same thing, will be If Not Bring (Set) Umbrella $\Rightarrow$ Then Not Exist (Set) Rain.

Making proofs using the contrapositive is much easier for not exists proofs because it requires you to find not not $=$ yes functions that exists, which is much easier than finding functions that do not exists.

So here, the contrapositive would be:

Prove that there exists positive and continuous function $f$ on $(0, \infty)$ such that $$f(x + y) \lt yg(f(x)), \,x \le 0, \,y \le 0$$

Proof:

1. Integrate equation on both sides from [1, $\infty$]

$$\int_{1}^{\infty} f(x+y) \,ds \le \int_{1}^{\infty} yg(f(x)) \,ds$$ where, $s = f(x)$

2. We re-express 1) wrt to $s = f(x)$.

$$\int_{1}^{\infty} f(x+y) \,ds \le \int_{1}^{\infty} yg(s) \,ds$$

3. We know by simple example inspection (??not sure if this is true.. looking for feedback?),

$$\int f(x+y) \,ds = \int [f(x)/x \times f(x+y)] \,ds$$

Hence,

$$\int_{1}^{\infty} [f(x)/x \times f(x+y)] \,ds \le \int_{1}^{\infty} yg(s) \,ds$$

4. Since $y$ is negative, we assume $g(s)$ to be negative to establish a very loose positive function bound to the most right of the range because we are just trying to prove existence and not trying to find exact bounds (??looking for feedback?). Because $y$ is negative and $g(s)$ is negative, making $ yg(y) $ positive, we can show that

$$\int_{1}^{\infty} yg(s) \,ds \gt \int_{1}^{\infty} g(s) \,ds$$

5. Hence, we can approach and bound the inequality from the right-side using above (e.g. if 3 < 5, and we know 10 > 5 like above, then we can write 3 < 10)

$$\int_{1}^{\infty} [f(x)/x \times f(x+y)] \,ds \le \int_{1}^{\infty} g(s) \,ds$$

6. Since we know that $f(x+y) \le f(x) \times f(y)$,

we can now break $f(x+y)$ apart into individual $f(x)$ to proof some statement about $f(x)$

$$\int_{1}^{\infty} [f(x)/x \times f(x+y)] \,ds \le \int_{1}^{\infty} g(s) \,ds$$

$$\int_{1}^{\infty} [f(x)/x \times (f(x) \times f(y))] \,ds \le \int_{1}^{\infty} g(s) \,ds$$

$$\int_{1}^{\infty} [(f(x)^2)/x \times f(y)] \,ds \le \int_{1}^{\infty} g(s) \,ds$$

7. And since we also know that $$\int_{1}^{\infty} g(s) \, ds \gt \int_{1}^{\infty} \frac{ds}{g(s)} $$ .
8. We can, again, approach and bound the inequality from the right-side using above

$$\int_{1}^{\infty} [(f(x)^2)/x \times f(y)] \,ds \le \int_{1}^{\infty} g(s) \,ds$$

$$\int_{1}^{\infty} [(f(x)^2)/x \times f(y)] \,ds \le \int_{1}^{\infty} \frac{ds}{g(s)} $$

9. Since as given above,

$$\int_{1}^{\infty} \frac{ds}{g(s)} < \infty$$

10. We can show that

$$\int_{1}^{\infty} [(f(x)^2)/x \times f(y)] \,ds \le \int_{1}^{\infty} \frac{ds}{g(s)} $$

$$\int_{1}^{\infty} [(f(x)^2)/x \times f(y)] \,ds \lt {\infty} $$

11. (??After which, I am not sure how to approach... but should finally show the following form??):

$$0 \lt \int_{1}^{\infty} [f(x)^2] \,ds \lt {\infty} $$

which shows and prove that there exists a positive and continuous function $f$ on $(0, \infty)$