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Let $f: \Bbb N^{\star}\to \Bbb N^{\star}$ a bijective functions such that exists $$\lim _{n \to \infty} {\frac {f(n)} {n}}.$$ Find the value of this limit.

I noticed that, if $f(n)=n$, then the limit is $1$. I couldn't make more progress. Can you help me?

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    I think you can argue by contradiction. Something like, if the limit is $L>1+2\epsilon$, then eventually $f(n)>(L+\epsilon)n$, so in particular, $f$ must have covered $1$ through $n$ already, but then $f$ is going to grow too fast. Hand wavy, but I think that's the idea. – 2017-01-03

3 Answers 3

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Let $L = \lim_{n \to \infty} \frac{f(n)}{n}$. Note $L \geq 0$. Consider the cases:

$L > 1$:

Let $\epsilon > 0$ be such that $L > 1 + \epsilon$. Then $\exists N \in \Bbb N$ such that $\frac{f(n)}{n} > 1 + \epsilon \quad \forall n \geq N$ $\iff f(n) > (1+\epsilon)n$.

Take $M > N$ sufficiently large such that $\epsilon M > 1 \iff (1 + \epsilon)M > M + 1$. Then $\forall n > M, f(n) > M + 1$, so all of $f^{-1}(1), f^{-1}(2), \dots f^{-1}(M + 1)$ must be given by $n_1, n_2, \dots, n_{M+1} \leq M$, but then by the pigeon-hole principle, at least two of the $n_i$ are equal which is a contradiction, as $f$ is a bijection.

$0 \leq L < 1$:

Take $\epsilon > 0$ such that $L < 1 - \epsilon$. Proceed similarly to the $L > 1$ case. $\exists N \in \Bbb N$ such that $\frac{f(n)}{n} < 1 - \epsilon \quad \forall n \geq N \iff f(n) < (1-\epsilon)n$. Choose $M > N$ sufficiently large such that $\epsilon M > 1 \iff (1-\epsilon)M < M - 1$. Then $\forall n > M, f(n) < M - 1$, but this is clearly a contradiction since $f$ is a bijection.

Hence $L = 1$.

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    I'm not OK with your second part $0\le L<1$. First very slight remark, one has to assume $L+\epsilon<1-\epsilon$ (which we can always do, for sure) to have $f(n)<(1-\epsilon)n$, $\forall n\ge N$. But which I don't agree with is how you pass from $"(1−\epsilon)MM$, $f(n)$\forall n\le M$, $f(n) – 2018-02-06
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    But one remedies with the following. Still assume $L+2\epsilon<1$ and take $N$ such that $f(n)<(L+\epsilon)n<(1-\epsilon)n$, for all $n\ge N$. Take now $M>N$ such that $\frac{N}{M}<\epsilon$. This gives $f(M)<(1-\epsilon)M$n\in\{N,...,M\}$ one has $f(n)<(1-\epsilon)n\le (1-\epsilon)M$f$ induces an injection from $\{N, ..., M\}$ to $\{1,...,M-N-1\}$, which gives a contradiction by the pigeonhole principle. – 2018-02-06
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    Yes you're quite right, the latter case is wrong as it stands - the $(1-\epsilon)n$ turned into a $(1-\epsilon)N$ in my head at some point I assume. Regarding your first remark, yes - we require $1 - \epsilon$ bounded away from $L$ by some positive distance, I agree, but at no point am I taking $1-\epsilon \to L$, so it is (albeit perhaps implicitly) incorporated into $L < 1 - \epsilon$. In particular, your inequality says nothing more than mine - it's equivalent to $L < 1 - 2 \epsilon = 1 - \epsilon '$. Although, it would be clearer to explicitly state the use of $L+ \delta < 1 - \epsilon$. – 2018-02-07
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Let $f$ be permutation of $\mathbb{N^*}$. Suppose the inequality $f(n) \leq n$ holds for only a finite set of integers $A$. Let $m = \max A$, then $f \left( [m + 1,\infty) \cap \mathbb{N} \right ) \subset [m + 2,\infty) \cap \mathbb{N}$, thus $f^{-1} (\left \{ 1,m+1 \right \}) \subset \left \{ 1,m \right \}$, which is not possible.

Thus $f(n) \over {n}$ $ \leq 1$ holds infinitely many times.

Applying the same reasoning to $f^{-1}$ we get : $\frac{f^{-1}(n)}{n} \leq 1 $ i.e $f(n) \over {n}$ $ \geq 1$ holds infinitely many times.

Since the limit exists, $\lim \frac{f(n)}{n} = 1$ .

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Because this limit is finite, exist $M$ and $n_0$ such that $f(n)/nn_0$ ie is bounded ae.

Now, because $f(n)n_0$ then $f=O(n)$ and by definition the limit request is 1

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    But what if $f=O(n^k)$ is not true? – 2017-01-03
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    There are many bijections that are not of the form $O(n^k)$ for any $k$. – 2017-01-03
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    @s.harp I Just edit – 2017-01-03
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    Not following. Which step of your argument fails for $f(n)=2n$? – 2017-01-03
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    I'm hesitant to comment, because the answer keeps changing fundamentally, but you don't use that $f$ is a bijection anywhere, and of course the statement is not true if $f$ is not a bijection ( such as $f(n)=2n$ ) – 2017-01-03