## Convergence of continuous function

A few years back I was playing with the strong law of large numbers (essentially I wanted to understand pointwise convergence for averages over sets other than ${\{1,2,\dots,n\}}$) following Etemadi’s proof. This didn’t lead anywhere in the end (essentially because almost surely pointwise convergence is not a topological convergence, see example 40 on page 111 of this book), but somehow along the way I was faced with the following question:

Is there an infinite set ${I\subset{\mathbb N}}$ such that for all ${\alpha>1}$, the intersection ${I\cap\{\alpha^n:n\in{\mathbb N}\}}$ is finite? And what if, instead of ${\alpha>1}$ we only ask for ${\alpha>2}$?

After asking this question around I got an answer from (then a high school student) Miguel Santos and this prompted me to generalize the result. In this post I will summarize the observations I made on this subject. Later I learned that most of what follows is well known among some circles. In particular the question above is exercise I-1.24 in this book; exercises I-1.25, I-1.26 and II-1.25 of that book also follow from the results below; as well as exercise 9.12 of this other book.

— 1. Continuous functions —

Let ${f:{\mathbb R}\rightarrow{\mathbb R}}$ be a function. It is a classical fact that ${\displaystyle \lim_{t\rightarrow\infty}f(t)=L\in{\mathbb R}}$ if and only if for every sequence ${(x_n)}$ with ${\displaystyle\lim_{n\rightarrow\infty}x_n=\infty}$ we have ${\displaystyle\lim_{n\rightarrow\infty}f(x_n)=L}$.

In this post I’m are interested in relaxing the for every’ condition, reducing the set of sequences one needs to test. To make any progress I will only deal with continuous functions. As an example, we have:

Theorem 1 Let ${f:{\mathbb R}\rightarrow {\mathbb R}}$ be a continuous function and let ${L\in {\mathbb R}}$. For any non-empty open interval ${U\subset(1,\infty)}$, the following conditions are equivalent:

1. ${\displaystyle\lim_{t\rightarrow\infty}f(t)=L}$
2. ${\displaystyle\forall\alpha\in U:\ \lim_{n\rightarrow\infty}f(n\alpha)=L}$
3. ${\displaystyle\forall\alpha\in U:\ \lim_{n\rightarrow\infty}f(n^d\alpha)=L}$, where ${d>0}$ is a real constant
4. ${\displaystyle\forall\alpha\in U:\ \lim_{n\rightarrow\infty}f(\alpha^n)=L}$
5. ${\displaystyle\forall\alpha\in U:\ \lim_{n\rightarrow\infty}f(n^\alpha)=L}$

where under the limits we consider ${n\in{\mathbb N}}$ and ${t\in{\mathbb R}}$.

This will follow from Theorem 2 below.

Remark 1 In conditions (1)-(1), if we replace the quantifier ${\forall\alpha\in U}$ with the weaker condition for almost every ${\alpha}$‘ (in the sense of the Lebesgue measure) the theorem no longer holds. In fact, Lesigne showed that for every continuous integrable function ${f:{\mathbb R}\rightarrow{\mathbb R}}$ and almost every ${\alpha>0}$ one has ${\displaystyle\lim_{n\rightarrow\infty}f(n\alpha)=0}$. On the other hand, it is well known that there are continuous integrable functions which do not converge at infinity.

Recall that a set ${S\subset{\mathbb R}}$ is called nowhere dense if its closure ${\bar S}$ contains no open set, and a set is called meagre (or of the first category) if it is the countable union of nowhere dense sets.

Theorem 2 Let ${\alpha_0\in{\mathbb R}}$ and let ${g:(\alpha_0,\infty)\times(1,\infty)\rightarrow{\mathbb R}}$ be a function such that

• For each ${\alpha>\alpha_0}$, the map ${x\mapsto g(\alpha,x)}$ is increasing and unbounded.
• For each ${x\in(1,\infty)}$, the function ${\alpha\mapsto g(\alpha,x)}$ is continuous.
• For any pair ${\alpha<\beta}$, if ${n}$ is large enough depending on ${\alpha}$ and ${\beta}$, we have ${g(\beta,n)>g(\alpha,n+1)}$.

Then, for any continuous function ${f:{\mathbb R}\rightarrow{\mathbb R}}$ and ${L\in {\mathbb R}}$, either ${\lim_{t\rightarrow\infty} f(t)=L}$ or the set

$\displaystyle A=\left\{\alpha\in(\alpha_0,\infty):\lim_{n\rightarrow\infty}f(g(\alpha,n))= L\right\}$

is meagre.

Remark 2 If we combine Theorem 2 for the function ${g(\alpha,x)=\alpha x}$ with Lesigne’s result mentioned in Remark 1, we conclude that given any continuous integrable function ${f}$ that doesn’t converge to ${0}$ at infinity, the set

$\displaystyle \left\{\alpha\in(0,\infty):\lim_{n\rightarrow0}f(n\alpha)=0\right\}$

is both meagre and has full Lebesgue measure.

Proof:

Let ${g}$ be a function under the conditions of the theorem and let ${f:{\mathbb R}\rightarrow {\mathbb R}}$ be a continuous function. Assume that, for some ${L\in {\mathbb R}}$, the set

$\displaystyle A:=\left\{\alpha\in(\alpha_0,\infty):\lim_{n\rightarrow\infty}f(g(\alpha,n))=L\right\}$

is not meager. We will prove that ${\displaystyle\lim_{t\rightarrow\infty}f(t)=L}$. For each ${n\in{\mathbb N}}$ let ${f_n(\alpha)=f(g(\alpha,n))}$. The second condition on ${g}$ implies that each ${f_n}$ is a continuous function.

Fix ${\epsilon>0}$ and for ${m\in{\mathbb N}}$, let

$\displaystyle \begin{array}{rcl} B_m&:=&\{\alpha\in(\alpha_0,\infty):(\forall n\geq m)\ |f(g(\alpha,n))-L|\leq\epsilon\}\\&=&\bigcap_{n=m}^\infty f_n^{-1}([L-\epsilon,L+\epsilon]) \end{array}$

Each ${\alpha\in A}$ is in some ${B_m}$, in other words ${\displaystyle A\subset\bigcup_{m=1}^\infty B_m}$. Since we are assuming that ${A}$ is not meager, some ${B_m}$ is somewhere dense, i.e. the closure ${\overline{B_m}}$ must contain an open interval.

The crucial observation now is that each ${B_m}$ is a closed set, because each ${f_n}$ is continuous and ${[L-\epsilon,L+\epsilon]}$ is a closed set. We conclude that some ${B_m}$ is somewhere dense, and hence it contains an open interval, say ${(\alpha,\beta)}$.

By the third condition on ${g}$, there is some ${N\in{\mathbb N}}$ such that, for ${n\geq N}$, we have ${g(\beta,n)>g(\alpha,n+1)}$. By the first condition on ${g}$, the function ${x\mapsto g(\beta,x)}$ is increasing and unbounded. This implies that if ${t>g(\beta,N)}$ we have ${t\in(g(\alpha,n),g(\beta,n))}$ for some ${n\in{\mathbb N}}$. By continuity we get that ${t=g(y,n)}$ for some ${y\in(\alpha,\beta)\subset B_m}$.

We conclude that if ${t>g(\beta,N)}$ and ${n>m}$ then we have ${f(t)=f(g(y,n))=f_n(y)}$, for some ${y\in B_m}$. Thus, ${f(t)\in [L-\epsilon,L+\epsilon]}$, and because ${\epsilon>0}$ was arbitrary, this implies that ${\displaystyle\lim_{t\rightarrow\infty}f(t)=L}$ as desired. $\Box$

In order to prove Theorem 1, it will be convenient to relax the conditions for ${g}$ in Theorem 2 to make it easier to check in concrete examples. The next corollary gives us such conditions.

Corollary 3 Let ${\alpha_0>1}$ be a real number and let ${g:(\alpha_0,\infty)\times(1,\infty)\rightarrow{\mathbb R}}$ be a continuously differentiable function such that for ${\alpha>\alpha_0}$ and ${x\geq1}$ both partial derivatives are positive, and such that for ${\alpha>\alpha_0}$ we have the two conditions:

$\displaystyle \lim_{x\rightarrow\infty}g(\alpha,x)=\infty\qquad\qquad \lim_{x\rightarrow\infty}\frac{\frac{\partial g}{\partial\alpha}(\alpha,x)}{\frac{\partial g}{\partial x}(\alpha,x)}=\infty\text{ uniformly on compact sets}$

(here’s what uniformly in compact sets means). Then for any continuous function ${f:{\mathbb R}\rightarrow{\mathbb R}}$ and ${L\in {\mathbb R}}$, either ${\displaystyle\lim_{t\rightarrow\infty} f(t)=L}$ or the set

$\displaystyle \left\{\alpha\in{\mathbb R}:\lim_{n\rightarrow\infty}f(g(\alpha,n))= L\right\}$

is meagre.

Proof: Let ${g}$ be a function that satisfies the conditions of this corollary. We need to show that ${g}$ satisfies the conditions of Theorem 2. The first condition follows from the fact that for all ${\alpha>\alpha_0}$ we have ${\displaystyle\lim_{x\rightarrow\infty}g(\alpha,x)=\infty}$ and the partial derivative ${\frac{\partial g}{\partial x}}$ is positive. Since ${g}$ is continuous, the second condition holds as well.

To check the third condition let ${\beta>\alpha>\alpha_0}$ and fix some positive integer ${n}$. We need to show that ${g(\beta,n)>g(\alpha,n+1)}$. Let ${h(t)=g\big(\alpha(1-t)+\beta,n+1-t\big)}$. Note that ${h(1)=g(\beta,n)}$ and ${h(0)=g(\alpha,n+1)}$. The Mean Value Theorem now implies that ${h(1)-h(0)=h'(c)}$ for some ${c\in(0,1)}$. We have

$\displaystyle h'(c)=(\beta-\alpha)\frac{\partial g}{\partial\alpha}\big(h(c)\big)-\frac{\partial g}{\partial x}\big(h(c)\big)$

From the hypothesis on ${g}$ we can choose ${N\in {\mathbb N}}$ such that for ${x\geq N}$ and all ${a\in[\alpha,\beta]}$ we have

$\displaystyle (\beta-\alpha)\frac{\partial g}{\partial \alpha}(a,x)>\frac{\partial g}{\partial x}(a,x)$

Since both partial derivatives of ${g}$ are positive we have that ${h(c)\in[\alpha,\beta]\times[n,n+1]}$. Therefore, if ${n>N}$, then ${g(\beta,n)-g(\alpha,n+1)=h'(c)>0}$. $\Box$

We can now easily deduce Theorem 1 from this corollary.

Proof: (of Theorem 1) Condition (1) trivially implies the others. Now consider the functions ${g_1(\alpha,t)=\alpha t}$, ${g_2(\alpha,t)=t^d\alpha}$, ${g_3(\alpha,t)=\alpha^t}$ and ${g_4(\alpha,t)=t^\alpha}$. Each of those functions satisfy the conditions of Corollary 3.

By Baire’s category theorem, the open interval ${U}$ is not meagre. Therefore, it follows from Corollary 3 that any of the conditions (1) to (1) imply that ${\displaystyle\lim_{t\rightarrow\infty}f(t)=L}$. $\Box$

— 2. Combinatorial consequences —

The techniques used to prove Theorem 2 can be adapted to study convergence of sequences. This in turn can be used to show that certain sparse sets of ${{\mathbb N}}$ have infinite intersection. We use the classical notation ${\lfloor x\rfloor}$ to denote the largest integer no greater than the real number ${x}$.

Theorem 4 Let ${I\subset{\mathbb N}}$ be an infinite set and let ${U\subset(1,\infty)}$ be a non-empty open interval. Then

1. The intersection ${I\cap\{\lfloor \alpha n\rfloor:n\in{\mathbb N}\}}$ is infinite for some ${\alpha\in U}$.
2. For each real ${d>0}$, the intersection ${I\cap\{\lfloor \alpha n^d\rfloor:n\in{\mathbb N}\}}$ is infinite for some ${\alpha\in U}$.
3. The intersection ${I\cap\{\lfloor n^\alpha\rfloor:n\in{\mathbb N}\}}$ is infinite for some ${\alpha\in U}$.
4. The intersection ${I\cap\{\lfloor \alpha^n\rfloor:n\in{\mathbb N}\}}$ is infinite for some ${\alpha\in U}$.

Taking $U=(1,\infty)$ or $U=(2,\infty)$, part 3 of this theorem answers the question raised in the introduction of this post.

Remark 3 The strength of Theorem 4 lies in the fact that ${I}$ can be any infinite set. Thus we can bootstrap the theorem and get, for instance, that there exist ${\alpha}$ and ${\beta}$ in any given small non-empty open interval such that the intersection

$\displaystyle \{\lfloor n^\alpha\rfloor:n\in{\mathbb N}\}\cap\{\lfloor \beta^n\rfloor:n\in{\mathbb N}\} \ \ \ \ \ (1)$

contains infinitely many prime numbers or infinitely many square numbers (say). In fact, it follows from Corollary 6 below that we can choose ${\alpha}$ and ${\beta}$ such that the set (1) contains infinitely many primes and infinitely many squares.

In order to prove Theorem 4 we need to establish an analogue to Theorem 2 for functions which are not necessarily continuous but which are constant on each interval ${[n,n+1)}$. To do that we need to assume that ${g(\alpha,t)}$ increases with ${\alpha}$ for each fixed ${t}$. This is not a very strong condition, in particular it is also present in Corollary 3.

Theorem 5 Let ${\alpha_0\in{\mathbb R}}$ and let ${g:(\alpha_0,\infty)\times(1,\infty)\rightarrow{\mathbb R}}$ be a function satisfying the conditions of Theorem 2 with the additional property that, for each ${x>1}$, the function ${\alpha\mapsto g(\alpha,x)}$ is increasing. Then for any sequence ${(x_k)}$ taking values on ${{\mathbb R}}$ and any ${L\in {\mathbb R}}$, either ${\lim_{k\rightarrow\infty} x_k=L}$ or the set

$\displaystyle A=\left\{\alpha\in(\alpha_0,\infty):\lim_{n\rightarrow\infty}x_{\lfloor g(\alpha,n)\rfloor}=L\right\}$

is meagre.

To make the analogy with the Theorem 2 take ${f(t)=x_{\lfloor t\rfloor}}$. This function is usually not continuous so we can not apply the Theorem 2 directly, but it is constant on each interval ${[n,n+1)}$, so we can modify the argument a little to cover this case.

Proof: Let ${g}$ be a function that satisfies the conditions and let ${(x_n)}$ be a real valued sequence. Assume that, for some ${L\in{\mathbb R}}$, the set

$\displaystyle A=\left\{\alpha\in(\alpha_0,\infty):\lim_{n\rightarrow\infty}x_{\lfloor g(\alpha,n)\rfloor}=L\right\}$

is not meagre. We will prove that ${\displaystyle\lim_{k\rightarrow\infty}x_k=L}$.

Fix ${\epsilon>0}$. For ${n\in{\mathbb N}}$ let

$\displaystyle D_n:=\{\alpha\in(\alpha_0,\infty):|x_{\lfloor g(\alpha,n)\rfloor}-L|<\epsilon\}$

and for ${m\in{\mathbb N}}$, let

$\displaystyle B_m:=\{\alpha\in(\alpha_0,\infty):(\forall n\geq m)|x_{\lfloor g(\alpha,n)\rfloor}-L|<\epsilon\ \}=\bigcap_{n=m}^\infty D_n$

By construction, each ${\alpha\in A}$ is in some ${B_m}$, in other words ${\{B_m\}_{m=1}^\infty}$ is a countable covering of ${A}$. Since ${A}$ is assumed to be not meagre, some ${B_m}$ must be somewhere dense and thus ${\overline{B_m}}$ must contain an open interval, say ${(\alpha,\beta)\subset\overline{B_m}}$.

Unlike in the proof of Theorem 2, the set ${B_m}$ is not necessarily closed, but we claim that if an open interval is contained in ${\overline{B_m}}$, then it is actually contained in ${B_m}$. Since

$\displaystyle \overline{B_m}\subset\bigcap_{n=m}^\infty \overline{D_n}$

we have ${(\alpha,\beta)\subset \overline{D_n}}$ for all ${n\geq m}$. Define ${E:=\{t>0: |x_{\lfloor t\rfloor}-L|\leq\epsilon\}}$. Clearly ${E}$ is the union of intervals of the form ${[a,b)}$ with integer end points. We can write ${D_n:=\{\alpha>\alpha_0:g(\alpha,n)\in E\}}$ and, moreover, the function ${\alpha\mapsto g(\alpha,n)}$ is continuous and increasing, so ${D_n}$ is also the union of some intervals of the form ${[a,b)}$. This implies that, since ${(\alpha,\beta)\subset\overline{D_n}}$, we actually have that ${(\alpha,\beta)\subset D_n}$. Since this is true for each ${n\geq m}$ we showed that ${(\alpha,\beta)\subset B_m}$, proving the claim.

By the hypothesis, there is some ${N}$ such that for ${n\geq N}$ we have ${g(\beta,n)>g(\alpha,n+1)}$. Thus for ${k>g(\beta,N)}$ we have ${k\in(g(\alpha,n),g(\beta,n))}$ for some ${n}$. This implies that ${k=g(x,n)}$ for some ${x\in(\alpha,\beta)}$ and so if ${n>N}$ we have ${x_k=x_{g(x,n))}\in (L-\epsilon,L+\epsilon)}$. Since ${\epsilon>0}$ was arbitrary, we conclude that ${\displaystyle\lim_{k\rightarrow\infty}x_k=L}$ as desired. $\Box$

We are now ready to give a quick proof of Theorem 4. In fact, we can prove a stronger statement.

Corollary 6 Let ${I_1,I_2,\dots}$ be a countable family of infinite subsets of ${{\mathbb N}}$. Let ${U\subset(1,\infty)}$ be a non-empty open interval. Then

1. There exists ${\alpha\in U}$ such that for each ${k\in{\mathbb N}}$, the intersection ${I_k\cap\{\lfloor \alpha n\rfloor:n\in{\mathbb N}\}}$ is infinite.
2. For each real ${d>0}$ there exists ${\alpha\in U}$ such that for each ${k\in{\mathbb N}}$, the intersection ${I_k\cap\{\lfloor \alpha n^d\rfloor:n\in{\mathbb N}\}}$ is infinite.
3. There exists ${\alpha\in U}$ such that for each ${k\in{\mathbb N}}$, the intersection ${I_k\cap\{\lfloor n^\alpha\rfloor:n\in{\mathbb N}\}}$ is infinite.
4. There exists ${\alpha\in U}$ such that for each ${k\in{\mathbb N}}$, the intersection ${I_k\cap\{\lfloor \alpha^n\rfloor:n\in{\mathbb N}\}}$ is infinite.

Proof: (of Theorem 4) We prove only the part (4) since the same proof works for the other cases with a different (but obvious) choice of ${g}$.

Let ${g(\alpha,t)=t^\alpha}$, let ${(I_k)_{k=1}^\infty}$ be infinite subsets of ${{\mathbb N}}$ and fix an open interval ${U\subset(1,\infty)}$. Observe that ${g}$ satisfies the conditions of Theorem 5. Define the indicator sequence ${(x_n^{(k)})}$ of ${I_k}$ by letting ${x_n^{(k)}=1}$ if ${n\in I_k}$ and ${x_n=0}$ if ${n\notin I_k}$. For each ${k\in{\mathbb N}}$ define

$\displaystyle U_k:=\big\{\alpha\in U:I_k\cap\{\lfloor n^\alpha\rfloor:n\in{\mathbb N}\}\text{ is finite }\big\}$

For each ${k\in{\mathbb N}}$ and each ${\alpha\in U_k}$ we have

$\displaystyle \lim_{n\rightarrow\infty}x^{(k)}_{\lfloor g(\alpha,n)\rfloor}=\lim_{n\rightarrow\infty}x^{(k)}_{\lfloor n^\alpha\rfloor}=0$

Since ${I_k}$ is infinite, the limit ${\displaystyle\lim_{n\rightarrow\infty}x^{(k)}_n}$ is not ${0}$. It follows from Theorem 5 that ${U_k}$ is a meager set. Thus the union ${U_0=\bigcup_{k\geq1}U_k\subset U}$ is also meager.

By the Baire Category theorem, ${U}$ is not meagre, thus ${U\neq U_0}$. Taking any ${\alpha\in U\setminus U_0}$ we have that, for every ${k\in{\mathbb N}}$, the intersection ${I_k\cap\{\lfloor n^\alpha\rfloor:n\in{\mathbb N}\}}$ is infinite. $\Box$

Advertisements
This entry was posted in Analysis, Combinatorics and tagged , . Bookmark the permalink.