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Show there is no function $f:\mathbb{R}\to\mathbb{R}$ so that $f(0) > 0$ and $f(x+y)\ge f(x)+yf(f(x))\,\forall x,y\in\mathbb{R}.$

I think one can show that $f(f(z))$ is positive for some $z$ and that $f(x)\to\infty$ as $x\to \infty$. Then one might be able to find some $a$ with $f(a) \ge a+1$.

To show $f(f(z))$ is positive for some z, suppose for all $z, f(f(z)) \leq 0.$ Then for all $y\le 0$, $$f(x+y)\ge f(x) + yf(f(x)) \ge f(x)$$ implying that $f(x)$ is decreasing.

If $f(f(x)) < 0$ for some $x$, then $$f(x+y) \ge f(x) + yf(f(x))$$ for all $y\leq 0,$ so $f(z)$ tends to infinity as $z$ tends to $-\infty$. Similarly, $f(z)$ tends to $-\infty$ as $z\to\infty.$

If $f(f(0)) = 0,$ then $f(y) \ge f(0) > 0$ for all $y.$ We know $f$ is decreasing so $f(y) \leq f(0)$ for $y\ge 0$ and hence for all $y\ge 0, f(y) = f(0).$ In particular, $f(f(0)) = f(0),$ so we get that for $y\in\mathbb{R}$ $$f(y) \ge f(0) + yf(0) = f(0)(1+y).$$ This contradicts the fact that $f(y) \leq f(0)$ for $y > 0$. Hence we must have $f(f(0)) < 0$ by assumption. But I'm not sure how to continue from here.

Bessel
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user3379
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  • Is there any continuity or differentiablity assumption on this function? because we cannot take a limit unless we know continuity. – Sam Oct 11 '22 at 17:51
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    @Sam does continuity really matter? We have $f(y) \ge g(y)$ for some function $g(y)$ that tends to infinity as $y$ tends to infinity. So for all $M, $ there exists large enough y so that $f(y) > M$, which precisely implies $\lim\limits_{y\to\infty} f(y) = \infty.$ Continuity just asserts that $\lim\limits_{y\to a} f(y) = f(a)$ for each a at which f is continuous. – user3379 Oct 11 '22 at 19:02
  • No, was ust wondering if there is more since this can be reduced to a comparison of $f'(x)$ and $f(f(x))$ for any $x$ and was wondering if differentiable was assumed. – Sam Oct 11 '22 at 19:13
  • Dunno if it'll help, but: depending on the sign of $x$, we can get this information at least (by combining the results of taking "$x = 0, y \in \mathbb{R}$" and "$y = -x$" separately): $$\begin{cases} x > 0 & \Rightarrow\quad f(f(x)) \geq f(f(0))\ x < 0 & \Rightarrow\quad f(f(x)) \leq f(f(0))\end{cases}$$ However, I feel just as stuck as you for now... – Bruno B Oct 11 '22 at 21:17
  • One thing I'll add though is in regards to your "Then one might be able to find some $a$ with $f(a) \geq a +1$": if you meant that purely from the fact that $f(x) \to \infty$, then that's wrong, an example being the $\ln$ function: $\ln(x) \to \infty$ but $\forall a \in \mathbb{R}, \ln(a) \leq a - 1 < a + 1$. If you meant something else then that could be fine though, but we'd need another condition than just going to $\infty$. – Bruno B Oct 11 '22 at 21:27
  • One last comment: "If $f(f(0))=0$, then $f(y) \geq f(0)>0$ for all $y$" is sufficient to conclude that there's a contradiction by saying it'd then be true for $y = f(0)$, no need for decreasing-ness. – Bruno B Oct 11 '22 at 21:39
  • This is an unusual functional inequality. Where did it come from? – marty cohen Oct 23 '22 at 18:25

1 Answers1

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Pick $a\in \Bbb R$, Then from $P(a,x-a)$, $$\tag1f(x)\ge f(a)+(x-a)f(f(a)).$$ The line given by the expression on the right passes through $(a,f(a))$, thus showing that $f$ is convex. Then $f$ is also continuous. By swapping $x\leftrightarrow a$, also $f(a)\ge f(x)-(x-a)f(f(x))$, so that for the difference quotient either $f(f(a))\le \frac{f(x)-f(a)}{x-a}\le f(f(x))$ or $f(f(a))\ge \frac{f(x)-f(a)}{x-a}\ge f(f(x))$. By continuity of $f\circ f$ and squeeze theorem, $f$ is differentiable and $$ \tag2f'(x)=f(f(x)).$$ By induction, all higher derivatives exist ($f$ is smooth) and can be expressed as sums of products of iterates of $f$. For example, $$f''(x)=f'(x)f'(f(x))=f(f(x))f(f(f(x))) $$ and $$f''(x) = f''(x)f(f(f(x))) + f'(x)^2f''(f(x)) = f(f(x))f(f(f(x)))^2 + f(f(x))^2f(f(f(x)))f(f(f(f(x)))).$$

Suppose $f$ is constant, so $f(x)=f(0)\ne0$ for all $x$. Then $0=f'(x)=f(f(x))=f(0)\ne 0$, contradiction. We conclude that $f$ is not constant.

Non-constant convex implies not bounded from above.

Let $a,b\in\Bbb R$ with $f(a)=f(b)$. For $c\in\{a,b\}$, $t\mapsto f(c+t)$ is a solution to the ODE $y(0)=f(a)$, $y'(t)=f(y(t))$. By convexity, $f$ is Lipschitz on every bounded interval. Then by Picard-Lindelöf, this ODE has a unique global solution. This implies that $f(a+t)=f(b+t)$ for all $t$. As a periodic continuos function would be bounded, we conclude that we cannot have $a\ne b$. Hence $f$ is injective.

Suppose $f$ is not bounded from below. Then by continuity, $f$ is onto and we can find $a$ with $f(f(a))=0$. Then by convexity, $f$ has a global minimum at $a$, contradicting unboundedness. We conclude that $f$ is bounded from below.

Let $s=\inf f$. Then either $f(a)=s$ for some $a\in\Bbb R$, or $\lim_{x\to a}f(x)=s$ for some $a\in\{-\infty,\infty\}$. In the first case, $f(s)=f(f(a))=f'(a)=0$, and in the other cases, simiularly $f(s)=\lim_{x\to s}f(x)=\lim_{x\to a}f(f(x))=\lim_{x\to a}f'(x)=0$. Hence $$ \tag3f(s)=0.$$

Then $s\ne0$ as $f(0)\ne 0$. As also $s\le f(s)=0$, we have $$ \tag4s<0.$$

Now $f\circ f= f'$ is non-decreasing and bounded from below by $s$. Hence $\lim_{x\to-\infty} f'(x)=t$ for some $t$ with $s\le t\le 0$. Suppose $t<0$. Then $f(x)\to\infty$ as $x\to-\infty$ while at the same time $f(f(x))\to s$. We conclude $\lim_{x\to\infty}f(x)=s$. If on the other hand $t=0$, then necessarily $\lim_{x\to-\infty}f(x)=s$. Thus $$ \tag5\lim_{x\to\infty}f(x)=s\quad\text{or}\quad\lim_{x\to-\infty}f(x)=s.$$

Apass.Jack
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Hagen von Eitzen
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  • I feel like injective, convex, bounded from below with a point achieveing the minimum is a contradiction, which would give us the desired result? (Edit: oh wait, $s$ isn't shown to be reached, my bad!) At least intuitively, but maybe my intuition is failing me. Awesome job either way! (+1) – Bruno B Oct 21 '22 at 18:17
  • Oh it seems like this post should let us conclude, since it shows that a solution of $f' = f \circ f$ must have $f(0) \leq 0$ by looking at the position of the unique root of $f$. Nice! – Bruno B Oct 21 '22 at 19:46
  • @Bruno oh, $f(0)<0$ should mean that $f$ is decreasing (from $\infty$ to $s$) – Hagen von Eitzen Oct 21 '22 at 20:03
  • What matters for this question specifically is that we assumed $f(0) > 0$ yet it's impossible from the post I linked. But good to know still, thanks for the clarification – Bruno B Oct 21 '22 at 20:09
  • @BrunoB I found the argument for injectivity later than it appears in the answer. The way it is arranged now, there is of course immediately only a strictly Monotonien function with $s$ at one end and infinity at the other as only possibility. Much if that later part of the answer can thus be made much shorter … – Hagen von Eitzen Oct 21 '22 at 20:09