Let $(x_n)_{n \ge 1}$ be a sequence of real numbers satisfying
$$x_{m+n} \le x_n+x_m$$
$m,n \ge 1$. Show that $\lim \limits_{n \to \infty} \dfrac{x_n}{n}$ exists and is equal to $\inf \left \{\dfrac{x_n}{n}; n\ge 1 \right \}$.
Asked
Active
Viewed 1,525 times
3
-
2Are you sure this is the question? What about $x_n = 2^n$? Is it actually $x_{m+n} \le x_n + x_m$? – A.S Apr 04 '13 at 02:01
-
May I suggest that you change the title for a more explicit one? – Julien Apr 04 '13 at 02:54
-
1A proof of Fekete's lemma (together with some references and related facts) is also given in an answer to this question: http://math.stackexchange.com/questions/57455/existence-of-a-limit-associated-to-an-almost-subadditive-sequence/ – Martin Sleziak Apr 19 '14 at 21:28
1 Answers
11
This was false with the reverse inequality. Take $x_n=n^2$, for instance. It would still have been true with $\sup$ instead of $\inf$, as it follows from the following.
Now it is true: if $x_n$ is subadditive, i.e. $$ x_{n+m}\leq x_n+x_m\qquad\forall n,m\geq 1 $$ then $\lim \frac{x_n}{n}=\inf \frac{x_n}{n}$. Note that moreover this limit/inf can take any value in $[-\infty,+\infty)$. To see why the case $-\infty$ can occur, take $x_n=-n^2$. For real values, take $x_n=an$.
This is called Fekete's lemma. See here for a proof by one of our colleagues. The strategy is to prove that $$ \inf \frac{x_n}{n}\leq \liminf \frac{x_n}{n}\leq\limsup\frac{x_n}{n}\leq\frac{x_k}{k}\qquad\forall k\geq 1. $$ The first two inequalities are trivial. The last one follows from a neat application of the Euclidean division. Then take the inf on the rhs to see that all these quantities are equal. So $\frac{x_n}{n}$ tends to $\inf \frac{x_n}{n}$. This is in $[-\infty,\infty)$. As observed earlier, all these cases can occur. Only the case $+\infty$ must be excluded, as the inf of a nonempty set is $<+\infty$.
Note: taking the exponential, you get a similar statement for submultiplicative positive sequences. A famous application of this is the fact that $\|a^n\|^\frac{1}{n}$ converges for every element $a$ in a Banach algebra. With more work, one can show that the limit is the spectral radius of $a$. That's called the spectral radius formula.
