The question is the following:

$f(x)$ is uniformly continuous on $[0, \infty)$ and for any $x > 0$, $\lim\limits_{n\to \infty}f(x+n) = 0$, where $n \in \mathbb{Z}_{>0}$. Prove that $\lim\limits_{x\to \infty} f(x) = 0$.

Hint: Divide $[0, 1]$ into small equal-length intervals

I do not understand what this question means. What exactly it is asking to be proved? Wouldn't it be obvious that $\lim\limits_{x\to \infty} f(x) = 0$ since $\lim\limits_{n\to \infty}f(x+n) = 0$. Also, what is the meaning or help from the hint?

Thanks for your help! I am so confused...

  • 2
    $\begingroup$ It would be instructive to see a counterexample when $f$ is merely continuous. $\endgroup$ – Alex Provost Mar 24 '16 at 19:00
  • 1
    $\begingroup$ @AlexProvost: I agree. Here is a counterexample. In the interval $[n, n+1)$, let the graph of $f$ be an isosceles triangle starting at $(x=n,y=0)$ with width $1/(n+1)$ and height $1$. Set $f = 0$ elsewhere in $[n,n+1]$. So, within $[n,n+1)$, the support of $f$ is $[n, n+1/(n+1)]$. It's easy to see that for any real $x$, we have $f(x+n) = 0$ for all sufficiently large $n$, so $\lim_{n \to \infty}f(x+n) = 0$. But within any interval $[n,n+1)$ the function takes on both $0$ and $1$ as values, so $\lim_{x \to \infty} f(x)$ does not exist. Obviously $f$ is not uniformly continuous. $\endgroup$ – Bungo Mar 24 '16 at 20:06

Fix $\varepsilon>0$. Since $f$ is uniformly continuous, there exists $\delta>0$ such that $x,y\geq 0$ and $|x-y|<\delta$ implies that $|f(x)-f(y)|<\varepsilon$.

Now choose $m\in\mathbb{N}$ such that $\frac{1}{m}<\delta$, and let $x_1=\frac{1}{m},x_2=\frac{2}{m},\dots,x_m=1$. By hypothesis, for each $1\leq i\leq m$ we can choose an $N_i$ such that $|f(x_i+n)|<\varepsilon$ for all $n\geq N_i$. Taking $N=\max\{N_1,\dots,N_m\}$, it follows that $$ |f(x_i+n)|<\varepsilon $$ for all $n\geq N$, and all $1\leq i\leq m$.

For the last step, suppose that $x>N$. There is an integer $n\geq N$ such that $y=x-n\in[0,1]$, hence there is some $i$ such that $|y-x_i|\leq \frac{1}{m}<\delta$. Therefore also $$|(x_i+n)-x|=|(x_i+n)-(y+n)|<\delta$$ hence $$ |f(x)|\leq |f(x)-f(x_i+n)|+|f(x_i+n)|<\varepsilon+\varepsilon=2\varepsilon$$

Since $\varepsilon$ was any positive real number, this shows that $\lim_{x\to\infty}f(x)=0$.


Wouldn't it be obvious that $\lim\limits_{x \to \infty} f(x) = 0$ since $\lim\limits_{n \to \infty} f(x + n) = 0$.

Yes, that's exactly the point. However, you will have to incorporate the definition of uniform continuity in your answer. (If $f$ were not uniformly continuous, that limit of $f(x+n)$ might not exist.)

  • $\begingroup$ But what is the meaning of that hint? $\endgroup$ – David To PokTo Mar 24 '16 at 18:35

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