I came across this question

Does there exist a function $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that $f(x+y)>f(x)(1+yf(x))$ and $x,y\in \mathbb{R}^+$

and I didn't know how to begin on it.

  • $\begingroup$ try $y=0$ in the equation... $\endgroup$ – marwalix Aug 1 '11 at 5:32
  • $\begingroup$ @marwalix, please read the question carefully. $\endgroup$ – picakhu Aug 1 '11 at 5:34
  • 4
    $\begingroup$ @picakhu, I think marwalix's point is that if you substitute y=0 into the inequality it becomes f(x) > f(x) which is a contradiction, so that there can be no such function... $\endgroup$ – Ben Blum-Smith Aug 1 '11 at 5:39
  • 1
    $\begingroup$ you can try $y=0$ because $y$ does not necessarily belong to the domain of $f$. $x+0$ belongs to the domain whenever $x$ belongs to the domain and in the right hand term you only see $f(x)$. I read the question very carefully $\endgroup$ – marwalix Aug 1 '11 at 5:40
  • 1
    $\begingroup$ @Srivatsan: it seems all you can show is that if such f exists, it cannot be differentiable, right? We can tell that, at least within each integer unit, f is increasing: f(x+1)>f(x)(1+f(x)), so that f(x+1)-f(x)>$f(x)^2$>0, and f is then increasing in [n,n+1], for all n $\endgroup$ – gary Aug 1 '11 at 5:56

Here's a slightly different approach.

Suppose for contradiction that such an $f$ exists. In fact we will reach the same contradiction even if $f$ only satisfies $f(x+y) \geq f(x)(1+yf(x))$ for $x,y>0$.

Define a sequence $x_0 < x_1 < x_2 < \ldots$ via the recurrence $$x_0 := 2011 \text{ and } x_{n+1} := x_n + 1/f(x_n) \text{ for } n>0.$$

Making repeated use of the bound $f(x + 1/f(x)) \geq f(x)(1+f(x)/f(x)) = 2f(x)$, we find that $f(x_n) \geq 2^n f(x_0)$ for all $n$.

Taking the reciprocal of the latter inequality and summing a geometric series enables us to establish $$ x_n = x_0 + \sum_{i=0}^{n-1} (x_{i+1} - x_i) = x_0 + \sum_{i=0}^{n-1} 1/f(x_i) \leq x_0 + \sum_{i=0}^{n-1} 1/(2^i f(x_0)) < x_0 + 2/f(x_0) =: M.$$

But now, the sequence $x_0,x_1,\ldots$ is bounded (by $M$) while the sequence $f(x_0),f(x_1),\ldots$ diverges to infinity. It follows that there is some integer $N$ such that $f(x_N) > f(M)$ even though $x_N < M$. However, it has been observed such an $f$ must be monotone, so we have a contradiction.

  • $\begingroup$ Much more succinct, nice! (I wish I could upvote this; apparently I exhausted my votes for today :-() $\endgroup$ – Srivatsan Aug 1 '11 at 15:03
  • 2
    $\begingroup$ @Mike, how does one come up with such an elegant solution? What is the motivation to try something like $x_{n+1}=x_n+1/f(x_n)$? $\endgroup$ – picakhu Aug 1 '11 at 15:31
  • $\begingroup$ @picakhu Indeed. While my proof divides up the interval into $\epsilon$-sized intervals (so that one can approximate a finite Riemann sum), the idea of using adaptive increments $x \leftarrow x+1/f(x)$ has vastly improved the proof. It makes sense in retrospect, but it's a beautiful idea. $\endgroup$ – Srivatsan Aug 1 '11 at 17:31
  • $\begingroup$ @Srivatsan, your proof seems to be utilizing some traditional ideas. i.e. consider small $\epsilon$, and then you bump into a recursion which you manage to play with and tweak. However to realize that one can find a series $x_0,x_1,...$ such that $f(x_i)\geq g(i) f(x_0)$ has to be the crux. How one gets to that eludes me. Also, note that its a nice telescope that Mike saw would be useful. $\endgroup$ – picakhu Aug 1 '11 at 17:36
  • $\begingroup$ @picakhu: I'm glad you're happy with it! It occurred to me late at night, a mosquito had woken me up >:[, that the inequality could be interpreted as saying "increasing $x$ by $y$ results in multiplying the value of $f$ at $x$ by at least $1+yf(x)$". Then I just worked out what $y$ would be needed if one wished to at least double the value, and the rest was easy! $\endgroup$ – Mike F Aug 2 '11 at 21:36

There exists no such $f$. My solution is formally not based on analysis, but it is inspired by my analysis "solution" in the comments. Hopefully there's no bug here :-).

Pick a sufficiently small $\epsilon > 0$. Then, for $n \in \mathbb N_{0}$, we have: $$ f(1+(n+1)\epsilon) - f(1+n\epsilon) > \epsilon f(1+ n\epsilon)^2. $$ I find it easier to work with a related sequence $u_n$ defined as $u_n = \epsilon f(1+n\epsilon)$. Plugging this in the above equation, we get the pleasant recursive inequality $u_{n+1} > u_n + u_n^2$, with the base condition $u_0 = \epsilon f(1)$.

Proposition. If $u_k \geq \eta$, then $u_{k + \lceil 1/\eta \rceil} \geq 2 \eta$.

Proof. We have $u_{k+1} - u_k > \eta^2$, $u_{k+2}-u_{k+1} > \eta^2$, and so on. (I am implicitly using the fact that $u_n$ is monotonic.) Adding all these $\lceil 1/\eta \rceil$ inequalities, we get the claim. $\Box$

Number of iterations before reaching $1$. For $\eta \leq 1$, we have the simplification $\lceil 1/\eta \rceil \leq 2/\eta$. Also I find it convenient to use the informal language "$u$-value at iteration $i$" to denote $u_i$. The above observation says that if the $u$-value at the current iteration is at least $\eta \leq 1$, then after at most $2/\eta$ iterations, the $u$-value becomes at least $2\eta$. Now, if $2\eta \leq 1$, then we stop; else, after an additional $2/(2\eta)$ iterations, the $u$-value is at least $2^2\eta$. Using this argument repeatedly, the number of iterations before the $u$-value becomes at least $1$ is at most $$ \frac{2}{\eta} + \frac{2}{2\eta} + \frac{2}{2^2 \eta} + \ldots $$ (this is actually a finite sum with $\approx \log(1/\eta)$ terms), which is smaller than the sum of the corresponding infinite geometric series, namely $4/\eta$.

Final contradiction. Now, since $u_0 = \epsilon f(1) \stackrel{def}{=} \eta$, we have $$ 1 \leq u_{\frac{4}{\epsilon f(1)}} = \epsilon \cdot f\left(1+\epsilon \cdot \frac{4}{\epsilon f(1)} \right) = \epsilon \cdot f\left(1+\frac{4}{f(1)} \right). $$ This is a contradiction since the $\epsilon > 0$ in the above argument was arbitrary, whereas the second factor is a fixed positive quantity depending only on $f$.

Note. While no $f$ exists satisfying the requirements of the problem, we can do better if we relax the domain to some interval of the form $(0,A)$. In this case, just the function $f(x) = \frac{1}{A-x}$ works.

  • $\begingroup$ Looks good, I was working on something similar. I think you're missing an $n$ in your first inequality? $\endgroup$ – Mike F Aug 1 '11 at 7:40
  • $\begingroup$ @Mike corrected, thanks! $\endgroup$ – Srivatsan Aug 1 '11 at 7:45

Another approach: Rewrite the inequality as $(*): f(x+y)-f(x)>yf(x)^2$. So we see that $f$ is a strictly increasing function. Fix $x$ and let $y\rightarrow+\infty$, we also see that $f$ is unbounded. Hence we can pick a number $a$ such that $f(a)>1$.

Now, for any $x\ge 0$, define $g(x)=f(x+a)-f(a)$. Clearly $g(0)=0$ and $g$ is strictly increasing. Also, it is easy to show that $(**): g(x+y) - g(x)>yg(x)^2$ whenever $x\ge 0$ and $y>0$. So, the inequalities for $f$ and $g$ are similar, except that we allow $x=0$ in the inequality for $g$. However, the function $g$ has a nicer property -- by putting $x=a$ in $(*)$, we have $g(y)>y$ for all $y>0$.

Let $b>0$ and $h=b/n$. By $(**)$, we get $g(ih)-g((i-1)h)>hg((i-1)h)^2$. Summing up for $i=1,2,...,n$ and let $h\rightarrow 0$, we see that $g(b) >\int_0^b g(u)^2 du$. As $g$ is an increasing function, the Riemann integral exists. Now, apply the inequality $g(u)>u$ to the integrand, we get $$g(b)>\int_0^b g(u)^2 du > \int_0^b u^2 du = \frac{b^3}{3}.$$ Apply this to the integrand, we get $$g(b)>\int_0^b g(u)^2 du > \int_0^b \left(\frac{u^3}{3}\right)^2 du = \frac{b^7}{3^2 \times 7}.$$ Continue in this manner, we get $$ g(b) > \frac{b^3}{3},\ \frac{b^7}{3^2 \times 7},\ \frac{b^{15}}{3^4 \times 7^2 \times 15}, ... $$ The $n$-th term of the above sequence is of order $O(b^{2^{n+1}}/2^{n^3})$. Therefore, when $b>1$, the value of $g(b)$ will eventually blow up, i.e. $f(a+b)$ does not exist.

The final step above suggests that when the domain of $f$ is restricted to a small neighbourhood of zero, $g$ may not blow up [edit: or even the neighbourhood is so small that $a$ and hence $g$ are not defined] and hence a solution for $f$ can exist. For example, if $f$ is only defined on some $(0, b)$, we may take $f(x)=e^{\alpha x}$ as a solution provided that $b$ is small and $\alpha > e^{\alpha b}$.

  • 1
    $\begingroup$ A nice proof! Besides you seem to have uncovered another $f(x)$ that works for $(0, b)$ for small $b$. Wonder where this function is hiding in the other proofs. $\endgroup$ – Srivatsan Aug 1 '11 at 17:07
  • $\begingroup$ I'd noticed something somewhat relevant. If $f$ satisfies the given condition, then for $x,y >0$ and $n$ a positive integer we have $f(x+y)/f(x)=\prod_{k=0}^{n-1}f(x+\frac{(k+1)y}{n})/f(x+\frac{ky}{n})>\prod_{i=0}^{n-1} (1+\frac{y}{n}f(x+\frac{ky}{n}))\geq(1+\frac{yf(x)}{n})^n$. Taking $n \to \infty$ gives $f(x+y)\geq f(x)e^{f(x)y}$. $\endgroup$ – Mike F Aug 2 '11 at 21:25
  • $\begingroup$ Supposing $f$ is continuous at $y$ and taking $x\to 0$ from above shows $f(y) \geq A e^{Ay}$ where $A := \lim_{x \to 0^+} f(x)$. We can drop the assumption that $f$ is continuous at $y$ by considering an increasing sequence of points $y_n$ at which $f$ is continuous with $y_n \to y$. This is of course possible since $f$ is monotone and so has at most countably many discontinuities. The conclusion is that $f(x) \geq Ae^{Ax}$ holds for all $x>0$ so the "partial examples" in your answer are minimal in some sense. $\endgroup$ – Mike F Aug 2 '11 at 21:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.