$f>0$ on real line ; $f(x+y)\le f(x)f(y) , \forall x,y \in \mathbb R$ ; $f([0,1])$ is bounded set ; does $\lim_{x \to \infty}(f(x))^{1/x}$ exist?

Question

Let $f: \mathbb R \to (0,\infty)$ be a function such that $f(x+y)\le f(x)f(y) , \forall x,y \in \mathbb R$ and $f$ is bounded on $[0,1]$ ; then does the limit $\lim_{x \to \infty}(f(x))^{1/x}$ exists ?

What I have found is $f(x)\le f(x/n)^n , \forall x \in \mathbb R , \forall n \in \mathbb N$ ; so say if $f(x) < M , \forall x \in [0,1]$ then we

get $f(x)\le f\Big(\dfrac x{[x]+1}\Big)^{[x]+1}\le M^{[x]+1} \le M^{2x} , \forall x>1$ ; so that $(f(x))^{1/x}$ remains bounded for large $x$ .

Please help . Thanks in advance .

This strongly reminds me of this: Fekete's lemma for real functions. Notice that after taking logarithm, you get a subadditive function. The Kuczma's book (see the question I linked for the full reference) has a whole chapter on subadditive functions, so you might find something useful there. — Martin Sleziak, May 20 '16 at 14:01
See also this related question: http://math.stackexchange.com/questions/100945/limit-of-a-function-satisfying-an-inequality — Martin Sleziak, May 20 '16 at 14:15

G.Kós · Answer 1 · 2016-04-30T06:49:23.897

The answer is yes.

Let $g(x)=\log f(x)$. We want to prove that $\lim\limits_{x\to\infty}\dfrac{g(x)}{x}$ exists.

Let $M:=\max\left(0,\log\sup_{[0,1]}f\right)$. We have $g(x+y)\le g(x)+g(y)$ for $x,y\ge0$ and $g(x)\le M$ for $0\le x\le 1$. By repeating the sub-additive property we can see that $$ g(kx) \le k \cdot g(x) \quad \text{for $k=1,2,\ldots$}; $$ for $k=\lfloor x\rfloor+1$ this provides $$ g(x) \le \Big(\lfloor x\rfloor+1\Big) \cdot g\left(\frac{x}{\lfloor x\rfloor+1}\right) \le (x+1)M \quad\text{for all $x\ge 0$.} $$

Take two sequences, $a_1,a_2,\ldots$ and $b_1,b_2,\ldots$ of positive reals such that $a_n\to\infty$, $b_n\to\infty$, $\dfrac{g(a_n)}{a_n}\to\liminf\limits_{x\to\infty}\dfrac{g(x)}{x}$ and $\dfrac{g(b_n)}{b_n}\to\limsup\limits_{x\to\infty}\dfrac{g(x)}{x}$.

Consider an arbitrary pair $n,m$ of indices. Let $K=K_{n,m}=\left\lfloor\frac{b_n}{a_m}\right\rfloor$, so $0\le b_n-K a_m<a_m$. Then $$ g(b_n) \le g(K a_m) + g(b_n-K a_m) \le K g(a_m) + (b_n-Ka_m+1)M \le K g(a_m) + (a_m+1) \cdot M. $$ (If $K=0$ then $g(b_n) \le (b_n+1)M \le (a_m+1) \cdot M$.) Dividing by $b_n$, $$ \frac{g(b_n)}{b_n} \le \frac{K g(a_m) + (a_m+1) \cdot M}{b_n} = \frac{K_{n,m} a_m}{b_n} \cdot \frac{g(a_m)}{a_m} + \frac{(a_m+1) \cdot M}{b_n}. $$ Now fix $m$ and take limits with $n\to\infty$. (Update: some explanation is added:) On the LHS, by the definition of $b_n$, $\frac{g(b_n)}{b_n}\to\limsup_{x\to\infty}\frac{g(x)}{x}$. Since $a_m$ is fixed, we have $\frac{b_n}{a_m}\to\infty$, so $\frac{K_{n,m} a_m}{b_n}=\frac{\lfloor b_n/a_m\rfloor}{b_n/a_m}\to1$. In the last fraction the numerator $(a_m+1) \cdot M$ is fixed, the denominator $b_n$ tends to $\infty$. Therefore, $$ \limsup_{x\to\infty}\frac{g(x)}{x} \le \frac{g(a_m)}{a_m}. $$ Now take $m\to\infty$ to get $$ \limsup_{x\to\infty}\frac{g(x)}{x} \le \liminf_{x\to\infty}\frac{g(x)}{x}. $$ Done.

I follow the proof up to the penultimate display line. Could you explain why, there, the inequality $$\lim\frac{g(b_n)}{b_n}\le \frac{g(a_m)}{a_m}$$holds, please? — John Bentin, Apr 29 '16 at 13:55
Thanks! That's all clear now (+1). This answer should be accepted, and you should get the bounty. — John Bentin, Apr 30 '16 at 18:53
@user141614 Limit does not exist if $\frac{g(x)}x$ is not monotonic. — Yuri Negometyanov, May 01 '16 at 07:36
@user141614 : $(f(x))^{1/x}$ is bounded from " above" from our calculations , hence so is $g(x)/x={\log f(x)}/x$ , so we can meaningfully talk about $\lim \sup g(x)/x$; but we don't know how close to $0$ does $(f(x))^{1/x}$ might get , so when we take " logarithm" of this , it might get blowed towards $-\infty$ , so I think there is no guarantee that $g(x)/x$ is bounded from below ; in that case are you considering $\lim \inf g(x)/x \in \mathbb R \cup {-\infty}$ ? — , May 01 '16 at 08:18
Yuri Negometyanov: ???? Limit exists if and only if limsup $\le$ liminf. — G.Kós, May 01 '16 at 08:59
user228169: liminf and limsup can be either finite, $+\infty$ or $-\infty$. If $\lim g(x)/x=-\infty$ then $\lim f(x)^{1/x}=0$. — G.Kós, May 01 '16 at 09:00

Yuri Negometyanov · Answer 2 · 2016-05-02T14:23:29.620

0

At first, $$\ln\lim_{x\to+\infty}{f(x)^{1/x}} = \lim_{x\to+\infty}\frac{\ln{f(x)}}{x},\quad x>0, f(x)>0.$$ Let $$g(x)=\frac{\ln{f(x)}}{x},$$then $$(x+\theta)g(x+\theta)<xg(x)+\theta g(\theta),\quad \theta\in[0,1], x\in(0,\infty),$$ $$\begin{cases} (x+\theta)(g(x+\theta)-g(x))\leq\theta(g(\theta)-g(x)),\quad \theta\in[0,1], x\in(0,\infty),\\ (x+\theta)(g(x+\theta)-g(\theta))\leq x(g(x)-g(\theta)),\quad \theta\in(0,1], x\in(0,\infty), \end{cases} $$ $$ \begin{cases} g(x+\theta)\leq g(x),\text{ when }g(\theta)\leq g(x),\\ g(x+\theta)\leq g(\theta), \text{ when }g(x)\leq g(\theta). \end{cases} $$ So $$g(x+\theta)\leq\max(g(x),G),\quad x\in(0,\infty),\quad G=\max_{\theta\in[0,1]} g(\theta).$$ Thus, $g(x)$ is bounded above for $x\in(0,\infty).$

When $g(x)\to -\infty,$ then $f(x)^{1/x}\to0,$ and we can account that $g(x)$ is lower bounded for $x\in(0,\infty).$

So, required limit exists if g(x) is monotonic.

edited May 02 '16 at 14:23

answered Apr 30 '16 at 08:21

Yuri Negometyanov

28,026

Can you explain the 3rd line from the end? The function $h(x)=\begin{cases}1&\text{$x$ is an integer}\2&\text{$x$ is not an integer}\end{cases}$ satisfies $h(x+y)\le\max(h(x),h(y))$, but it is not non-decreasing... – G.Kós May 01 '16 at 05:45
@user141614 Thank you for comment! Corrected. – Yuri Negometyanov May 01 '16 at 07:12
When you define $G$, how do we know, that $g(\theta)$ is bounded in $\theta\in(0,1)$? – G.Kós May 01 '16 at 09:29
@user141614 from initial condotions – Yuri Negometyanov May 01 '16 at 19:56
Sorry, I do not understand it. If $\ln f(\theta)\in(-\infty,\ln M]$ and $\theta\in(0,1]$ then $g(\theta)=\frac{\ln f(\theta)}{\theta}\in?$ The definition of $G$ involves $g(0)=\frac{\ln f(0)}{0}$... – G.Kós May 02 '16 at 06:13
@user141614 $\frac {f(x)}x\to-\infty$ and $x\in(0,\infty)$, so $f(x)\to 0$. But it becames clearly after little corrections( – Yuri Negometyanov May 02 '16 at 07:01
My last attempt. Suppose that $f(x)\equiv2$. What is the numerical value of $G$? – G.Kós May 02 '16 at 09:31
@user141614 ok, example isn't right. Corrected – Yuri Negometyanov May 02 '16 at 14:23

Doug M · Answer 3 · 2016-04-22T16:24:53.640

I had a lengthy answer and it just crashed, so I will sketch the proof.

$g(x) = \ln f(x), g(x+y) \leq g(x)+g(y)$ show that $\lim_\limits{x\to\infty} g(x)/x$ exists. $g(x)/x$ is easier to work with than $f(x)^{1/x}$

$g(x)/x$ is bounded. There exists a sequence of $x_0<x_1<x_3\dots$ such that the sequence of $g(x_n)/x_n$ converges. (Bolanzo-Weierstrass)

Now we have shown that $|g(x_n)/x_n - L| < \epsilon$ for infinitely many x_n. We must show that $|g(x)/x - L|<\epsilon$ when $x\ne x_n$

Knowing that there is an upper bound for $g(x)/x$, you should be able to show that there is a maximal distance $d$ that $g(x)$ can possibly get way from the line $y = Lx$ and use that to show that when $x$ gets large $d/x$ goes to $0.$

continued....

How do we bound $d$? If $M$ is the upper bound of $g(x)$ in $[0,1]$ then $g(x)-g(x_n)\leq M(x-x_n)+M$ or $g(x_n+a) \leq M(a+1)$

And $a<x_{n+1}-x_n$

$\frac{g(x_n+a)}{x_n+a} - L \leq \frac{(M-L)a+M}{x_n+a}$

$\frac{g(x)}{x} - L \leq \frac{(M-L)a+M}{x}$

$(M-L)a+M$ is finite, and when $x$ get to be large enough, $|\frac{g(x)}{x} - L|<\epsilon$ for any epsilon.

Is the meaning of your last paragraph is to show that $|g(x)-Lx|$ is bounded for karge $x$ ? But I have tried and tried and unable to show it ; could you please elaborate some more ? You can delete your beginning arguments as I have understood those , to prevent your answer from crashing ; Thanks in advance — , Apr 22 '16 at 14:56
How did you get $g(x_n+a) \le M(a+1) $ ? Where did the $g(x_n)$ vanish to ? — , Apr 23 '16 at 07:31
In line 3: $g(x)/x$ not necessarily bounded from below; try $f(x)=e^{-x^2}$. — G.Kós, May 01 '16 at 09:32

$f>0$ on real line ; $f(x+y)\le f(x)f(y) , \forall x,y \in \mathbb R$ ; $f([0,1])$ is bounded set ; does $\lim_{x \to \infty}(f(x))^{1/x}$ exist?

3 Answers3