Analysis Problem of the Day 63

Today’s classic problem appeared as Problem 6 on the Texas A&M August 2023 Real Analysis Qual:

Problem 6. Let $f: (0,\infty) \to \mathbb{R}$ be a continuous function such that $\lim_{n \to \infty} f(nx)=0$ for all $x>0.$ Show that $\lim_{x \to \infty} f(x)=0.$

Solution: Playing around with the statement of the problem, one might notice that if one evaluates $f$ on some interval (such as $[1,2]$ ), then one can cover the real line by multiples of the interval (such as $[2,4],[3,6],...$ ). More formally, for any $(a,b) \subseteq (0,\infty)$ and for any $N \in \mathbb{N},$ there exists $y >0$ such that for any $x \geq y, x \in (na,nb)$ for some $n \geq N.$ This can be seen by noting that the difference between consecutive elements of $\{\frac{x}{n}: n \geq N\},$ which is $\frac{x}{n(n+1)} \to 0,$ is monotonically decreasing as $n \to \infty.$ Thus, for $n \geq N$ and $y$ large enough such that $\frac{y}{n(n+1)}<\frac{b-a}{2}$ and $\frac{y}{n} \geq b,$ the differences of the elements in the set defined above are smaller than the length of $(a,b),$ meaning at least of these elements intersects $(a,b).$ It follows that $[y,\infty) \subseteq \bigcup_{n =N}^\infty (na,nb).$

Thus, to show the claim, it suffices to find an interval $(a,b)$ such that $f(nx) \leq \epsilon$ for all $n \geq N$ for some $N$ and all $x \in (a,b).$ For this, we note that the pointwise convergence guaranteed in the problem can be rephrased in measure-theoretic terms as

$\bigcup_{N \in \mathbb{N}} \bigcap_{n \geq N} \{x: |f(nx)| \leq \epsilon\}=(0,\infty).$

Defining $A_N:=\bigcap_{n \geq N} \{x: |f(nx)| \leq \epsilon\},$ we note that $\bigcup_N A_N =(0,\infty)$ and each $A_N$ is closed as an intersection of closed sets, so by the Baire category theorem at least one $A_N$ has non-empty interior, i.e. contains an open interval. This implies that for some open interval $x \in (a,b) \subseteq A_N$ and some $N,$ $n \geq N$ implies $|f(nx)| \leq \epsilon.$ But by the lemma in the previous paragraph, this implies that $|f(z)| < \epsilon$ for $z \in [y,\infty)$ for large enough $y.$ Thus, $\lim_{x \to \infty} f(x)=0,$ as desired.

Remark: The two key ingredients in this problem are the covering property and the Baire category theorem (which relies on the continuity of $f$ ). As we saw in the proof of the covering property, a sufficient condition for the statement to hold where $\lim_{n \to \infty} f(nx)=0$ is replaced with $\lim_{n \to \infty} f(a_n x)=0$ for $a_n>0$ increasing is that $\frac{1}{a_n} \to 0$ and $\frac{1}{a_{n+1}}-\frac{1}{a_n} = o(\frac{1}{a_n}).$ Thus, for example, $a_n = \log n$ works but $a_n = c^n$ for any $c>0$ does not. In general, any exponential growth will provide a boundary between allowed and disallowed sequences.

Remark: The continuity of the function is necessary as evidenced by the counterexample $f(x) = \chi_{\mathbb{Q}}(x).$

Leave a Reply Cancel reply