$f>0$ 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? 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 $f(x) &lt; M , \forall x \in [0,1]$ then we 

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

Please help . Thanks in advance .
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 $0\le b_n-K a_m&lt;a_m$. 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 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 $x_0&lt;x_1&lt;x_3\dots$ such that the sequence of $g(x_n)/x_n$ converges. (Bolanzo-Weierstrass)

Now we have shown that $|g(x_n)/x_n - L| &lt; \epsilon$ for infinitely many x_n. We must show that $|g(x)/x - L|&lt;\epsilon$ 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 $a&lt;x_{n+1}-x_n$

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

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

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