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I have to construct a continuous function such that

$$\int_{-\infty}^\infty |f(x)| \, dx<\infty$$

but

$$\lim_{x\to \infty}|f(x)|$$

does not exists.

I have already known one messy example that deals with lines and minimum distance. I just want to see different examples. I know we can construct a triangle with fixed height and decreasing base such that the area get's smaller and the integral is like the geometric series. However, the fixed height makes the limit inexistent. I just dont know how to describe that properly. Thanks.

José Carlos Santos
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Dr Richard Clare
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  • Let $g(x) = (1-|x|) 1_{|x| < 1}$ which is continuous and compactly supported on $[-1,1]$. Then try $f(x) = \sum_{n=1}^\infty c_n g(n^2 (x-n))$ for some sequence $c_n$ – reuns Aug 22 '17 at 19:00
  • @reuns I seem to remember having already seen this suggestion, formulated exactly like that. Am I mistaken? – Did Aug 22 '17 at 19:39

3 Answers3

2

Hint:

$$ f(x) = \sum_{n\in\mathbb{Z}} n \exp\left[-n^6(x-n)^2\right] $$ does the job. Can you prove it?

Jack D'Aurizio
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  • For the integration, is similar to integrate the normal distribution. and the limit clearly does not exist for infinitely many values of n, it oscillates. Great example. – Dr Richard Clare Aug 25 '17 at 16:21
1

Define:

  • $f(x)=0$ when $x<0$.
  • In $[0,1]$, the graph of $f$ is a line segment from $(0,0)$ to $\left(\frac12,1\right)$ and a line segment from $\left(\frac12,1\right)$ to $(1,0)$.
  • In $[1,2]$, the graph of $f$ is a line segment from $(1,0)$ to $\left(1+\frac14,1\right)$, a line segment from $\left(1+\frac14,1\right)$ to $\left(1+\frac12,0\right)$ and a line segment from $\left(1+\frac12,0\right)$ to $(2,0)$.
  • In $[2,3]$, the graph of $f$ is a line segment from $(2,0)$ to $\left(2+\frac18,1\right)$, a line segment from $\left(2+\frac18,1\right)$ to $\left(2+\frac14,0\right)$ and a line segment from $\left(2+\frac14,0\right)$ to $(3,0)$.
  • And so on...

I shorter way of defining $f$ is: $f(x)=0$ if $x<0$ or $x>\lfloor x\rfloor+2^{-\lfloor x\rfloor}$ and$$f(x)=1-2\left|2^{\lfloor x\rfloor}\bigl(x-\lfloor x\rfloor\bigr)-\frac12\right|=1-\left|2^{\lfloor x\rfloor+1}\bigl(x-\lfloor x\rfloor\bigr)-1\right|$$otherwise.

José Carlos Santos
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0

$$ \text{For } x>0 \text{ let } f(x) = \begin{cases} 1 & \text{if } n < x < n + \dfrac 1 {2^n} \text{ for } n = 1,2,3,\ldots, \\ 0 & \text{otherwise}. \end{cases} $$ For $x<0,$ just make this an even function.

Then $\lim\limits_{x\,\to\,\infty} |f(x)| $ does not exist, but $\displaystyle\int_{-\infty}^\infty |f(x)|\,dx < \infty.$

PS: I notice that I neglected continuity. That is easily dealt with by making these pulses triangular rather than rectangular. Thus $$ f(x) = \begin{cases} 2^n(x-n) & \text{if } n<x<n+\dfrac 1 {2^n} \\ 2^n(n+1-x) & \text{if } n+\dfrac 1 {2^n}, < x < n + \dfrac 2 {x^n}, \\ 0 & \text{otherwise}. \end{cases} $$

Michael Hardy
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  • @RichardClare : I'm just putting tall but narrow pulses on the graph of $f$ at each integer. So $f$ alternates between $0$ and $1$, so it cannot approach a limit as $x\to\infty.$ But the sum of the areas of those pulses is the integral of $f$ over the whole line, and that sum is finite. – Michael Hardy Aug 22 '17 at 19:39
  • @RichardClare Sorry but what is "confusing" here? – Did Aug 22 '17 at 19:40
  • ok, I see that I neglected continuity. So I've changed the pulses from rectangles to triangles, so that $f$ will be continuous. – Michael Hardy Aug 22 '17 at 19:47
  • @MichaelHardy is not $2^n$ in the second definition instead of $x^n$? – Dr Richard Clare Aug 22 '17 at 20:15
  • @RichardClare : Because it's the slope of the sides of the triangle: one side has slope $2^n$ and the other $-2^n.$ Take a minute and draw the picture. That should clarify this. – Michael Hardy Aug 22 '17 at 21:40