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

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About Joel Moreira

PhD Student at OSU in Mathematics. I'm portuguese.
This entry was posted in Analysis, Combinatorics and tagged , . Bookmark the permalink.

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