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Let $f:[0,+\infty)\longrightarrow R^{+}\bigcup\{0\}$ be a continous and for any $x\in[0,+\infty)$ the sequence $\{f(x+n)\}$ converges to zero,prove that $$\lim_{x\to+\infty}f(x)=0$$

I think this problem is wrong, so someone can take some example? Thank you, meaning that find a $f$ such:

let $f:[0,1]\longrightarrow R^{+}\bigcup\{0\}$ be a continous and for any $x\in[0,1]$ the sequence $\{f(x+n)\}$ converges to zero,prove that $$\lim_{x\to+\infty}f(x)\neq 0$$

someone tell me $$f(x)=\dfrac{x}{1+x^2\sin{x}}$$ But I think this is example is not such my meaning.Thank you and I have seen this problem :

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math110
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  • Just a remark: I don't know if this notation is common, but I know the much shorter notation $\mathbb R_0^+$ for the set $\mathbb R^+\cup {0}$. – celtschk Aug 02 '13 at 07:08
  • oh, I mean that $$f(x)\ge 0$$ – math110 Aug 02 '13 at 07:10
  • Of course. There's nothing wrong with your notation. I just wanted to make you aware of a shorter way to write it. The "I don't know if this notation is common" refers to the shorter notation I gave. – celtschk Aug 02 '13 at 07:11
  • About your question: Why do you think the problem is wrong? – celtschk Aug 02 '13 at 07:14
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    Domain of $f$ is $[0,1]$. Then what do you mean by $f(x+n)$? – Kunnysan Aug 02 '13 at 07:15
  • because I have see $f$ must uniformly continous ,then we have $$\lim_{x\to +\infty}f(x)=0$$ – math110 Aug 02 '13 at 07:16
  • @Kunnysan,I have edit,Thank you – math110 Aug 02 '13 at 07:19
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    Ah, the uniform continuity is definitely necessary - otherwise let the values of $f$ on $[n-1,n)$ be a bump of height 1 and support $(n-\frac{1}{2n},n-\frac{1}{3n})$. – Anthony Carapetis Aug 02 '13 at 07:25
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    $f(x+n)$ goes to zero as $n\to\infty$. So for $\epsilon>0$ there exist $N$ such that $|f(x+n)|<\epsilon$, $\forall n>N$. So for $|f(y)|=|f({y}+[y])|<\epsilon$ for $[y]>N$ or $\forall y>N+1$. So $\lim_{x\to\infty}f(x)=0$. What is wrong about it? – Kunnysan Aug 02 '13 at 07:29
  • Hello,@AnthonyCarapetis what's your mean? – math110 Aug 02 '13 at 07:29
  • @AnthonyCarapetis: $f$ is continuous, then bump?? – Kunnysan Aug 02 '13 at 07:31
  • @Kunnysan: the sequences $f(x+n)$ can converge at different rates for each $x$, so there is a different $N$ for each pair $(\epsilon,x)$. You need to show that for each $\epsilon$ there is a single $N$ that works for every $x$ for the full convergence to hold. – Anthony Carapetis Aug 02 '13 at 07:41
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    By bump I mean a continuous function that is zero outside the support and reaches 1 somewhere inside - the easiest example to construct is probably a piecewise-linear "upwards-pointing triangle". – Anthony Carapetis Aug 02 '13 at 07:43

2 Answers2

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I'm not sure whether your intended question includes the uniform continuity condition or not; if so then here's a proof. If not, then I described a counterexample in the comments.

Any sequence in $[0,\infty)$ can be written uniquely as $\{n_j + x_j\}$ where $n_j \in \mathbb{N}, x_j \in [0,1)$. Take any such sequence converging to $+\infty$; we must show that $f(n_j + x_j) \to 0$.

Since $[0,1]$ is compact we can pass to a subsequence (which I will refer to also as $(n_j,x_j)$ for convenience) such that $x_j \to x_0 \in [0,1]$. Then by the uniform continuity we have $$ f(n_j + x_j) \le f(n_j + x_0) + \omega(|x_j - x_0|) $$ where $\omega$ is the modulus of continuity for $f$. The first term on the RHS converges to zero by our assumptions and the second by the definition of the modulus of continuity and $x_j \to x_0$; so we have $f(n_j + x_j) \to 0$.

This argument in fact works for any subsequence; so every subsequence has a subsequence on which $f$ converges to zero, and thus $f$ converges to zero on the original sequence; so since it was arbitrary we have shown $f(x) \to 0$ as $x\to \infty$.

Anthony Carapetis
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As a counterexample take any function $f$ which is $0$ in each interval $[n+2/n,n+1]$ and is $1$ in the points $n+1/n$ for $n\in \mathbb N$. For each $x$ the sequence $f(x+n)$ will be definitely $0$.

Emanuele Paolini
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