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Let $f:\mathbb{R}\to[0,+\infty)$ be a function such that $\int_{-‎\infty‎}^{+\infty}f(x)\,dx=1$. My question is:

Is the sequence $\big(f(n)-f(n+1)\big)$ convergent?

I found that there exist some $f$ such that the sequence $f(n)$ is not convergent, but in my research I arrived at the mentioned question. I tried to make contradiction with the definition of improper integral by using the property $f$ "positive". but I could not achieve the goal. Any help is appreciated.

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  • $\begingroup$ Which hypothesis are you assuming about $f$, other than the fact that $\int_{-\infty}^\infty f(x)\,\mathrm dx=1$ and that $f(\mathbb{R})\subset[0,\infty)$? $\endgroup$ Feb 3, 2019 at 18:41
  • $\begingroup$ @José Carlos Santos, that 's it. Do you mean it happens under some extra conditions? $\endgroup$ Feb 3, 2019 at 18:50
  • $\begingroup$ I have posted an answer. $\endgroup$ Feb 3, 2019 at 18:59
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    $\begingroup$ I think that if you assume that $f$ is uniformly continuous and bounded, the answer is yes. $\endgroup$
    – N. S.
    Feb 3, 2019 at 19:16

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For each $n\in \mathbb{N}$ define $f_n(x):\mathbb{R} \to [0,1]$ as follows: $f_n(2n) = 1$, $f_n(x) = 0 $ for $|x - 2n|\geq 2^{-n}$ and $f$ is linear on the intervals $[2n - 2^{-n}, 2n]$ and $[2n , 2n + 2^{-n}]$. The graph of $f_n$ is simply a thin triangle with height $1$, having base of length $2\times 2^{-n}$, equal edges and area $2^{-n}$. Define $$ f(x) := \sum\limits_{n=1}^\infty f_n(x), \qquad x \in \mathbb{R}. $$ Then $f$ is non-negative and $$ \int\limits_{\mathbb{R}} f(x) dx = \sum\limits_{n=1}^\infty \int\limits_{\mathbb{R}} f_n(x) dx = \sum\limits_{n=1}^\infty \frac{1}{2^n} = 1. $$ However, $$ f(2n) - f(2n+1) = 1 \text{ and } f(2n + 1 ) - f(2n+2) = -1 \text{ for all } n\in \mathbb{N}, $$ hence the sequence $f(n) - f(n+1)$ does not converge.

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  • $\begingroup$ @ Hayk Would you give an example such that $f$ is continuous? $\endgroup$ Feb 12, 2019 at 15:15
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    $\begingroup$ @soodehMehboodi, the $f$ defined above is continuous (just draw the graph of each $f_n$ to see what they actually are). With a bit more effort one may even modify $f$ to make it $C^\infty$ (infinitely differentiable), by smoothing the peaks of $f_n$s (e.g. via convolution with a properly chosen smooth function). What you cannot do, is making $f$ uniformly continuous, this was already pointed out in the comments and other answers to your original question. $\endgroup$
    – Hayk
    Feb 12, 2019 at 16:10
  • $\begingroup$ @ Hayk Firstly, sorry I thought that your example is not continuous. Also, I have worked on continuous functions and I need a continuous counter example. thanks . $\endgroup$ Feb 12, 2019 at 17:19
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Just take$$f(x)=\begin{cases}1&\text{ if }x\in[0,1]\\x^2+1&\text{ if }x\in\mathbb{N}\\0&\text{ otherwise.}\end{cases}$$Then your conditions hold, but $\displaystyle\lim_{n\to\infty}\bigl(f(n)-f(n+1)\bigr)$ diverges.

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  • $\begingroup$ @ José Carlos Santos Would you give an example such that $f$ is continuous? $\endgroup$ Feb 12, 2019 at 15:15
  • $\begingroup$ Yes, I can give such an example. $\endgroup$ Feb 12, 2019 at 15:27
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Without extra conditions the answer is no. Consider for example the function $$f(x)=\begin{cases} 2^{n} &\text{if $x\in[n,n+1/2^{2n+1}]$ with $n\in\mathbb{N}$,}\\ 0 &\text{otherwise.} \end{cases}$$ Then $$\lim_{n\to+\infty}\big(f(n)-f(n+1)\big)=\lim_{n\to+\infty}\big(2^n-2^{n+1}\big)=\lim_{n\to+\infty}(-2^n)=-\infty$$ and $$\int_{-‎\infty‎}^{+\infty}f(x)\,dx=\sum_{n=0}^{\infty}\frac{2^n}{2^{2n+1}} =\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}=1.$$ P.S. As noted by N. S. in a comment above, if $f$ is uniformly continuous then the required limit exists and it is zero. See $f$ uniformly continuous and $\int_a^\infty f(x)\,dx$ converges imply $\lim_{x \to \infty} f(x) = 0$

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  • $\begingroup$ how about $f$ has continuous derivative? $\endgroup$ Feb 3, 2019 at 19:33
  • $\begingroup$ @soodehMehboodi That is not enough, but bounded derivative suffices. $\endgroup$
    – Robert Z
    Feb 3, 2019 at 19:37
  • $\begingroup$ @ Robert Z Would you give an example such that $f$ is continuous? $\endgroup$ Feb 12, 2019 at 15:16

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