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$f:\mathbb{R}\to\mathbb{R}$ is continuous and $\int_{0}^{\infty} f(x)dx$ exists could any one tell me which of the following statements are correct?

$1. \text{if } \lim_{x\to\infty} f(x) \text{ exists, then it is 0}$

$2. \lim_{x\to\infty} f(x) \text{ must exists, and it is 0}$

$3. \text{ in case if f is non negative } \lim_{x\to\infty} f(x) \text{ exists, and it is 0}$

$4. \text{ in case if f is differentiable } \lim_{x\to\infty} f'(x) \text{ exists, and it is 0}$

I solved one problem in past which says: if $f$ is uniformly continuos and $\int_{0}^{\infty} f(x)dx$ exists then $\lim_{x\to\infty} f(x)=0$, so the condition in one says $f$ is uniformly continuous? and hence $1$ is true? well I have no idea about the other statements. will be pleased for your help.

Myshkin
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    Hint: Consider a function which has "spikes" of height $1$ at every integer, but the widths of the spikes go to zero very fast. – J. J. May 30 '13 at 16:42

2 Answers2

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The point is that the limit need not exist. You can construct something according to the comment by J.J. For instance, you can consider a function such that at any integer the height is one but the width is $1/n^2$. Then the area of each rectangle is $1/n^2$ and

$$ \int_0^\infty f(x)dx=\sum\frac{1}{n^2}<\infty. $$

But the limit does not exists. You get

$$ \lim_{n\rightarrow\infty} f(n)=1, $$

$$ \lim_{n\rightarrow\infty} f(n-\delta)=0,\;\; 0<\delta<<1 $$

Thus, 1 is true. If the limit exists, it should be zero (if it isn't then the function can't be integrable, right?). 2 is not true (see the hint by J.J. and the first part of my answer above). I think that 3, 4 are not true due to the same reason.

Edit: To prove that 4 is not true you can construct a (more decaying) example and then integrate it.

guacho
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  • I edited the first sentence: when I read "the limit could not exist," it sounds to me like "there is no possibility for this limit to exist." I think you meant something more along the lines of "the limit may or may not exist," which is now how I read it. If you disagree, feel free to change it. – Stahl May 30 '13 at 17:11
  • @Stahl Yes, I completely agree with your change. – guacho May 30 '13 at 17:12
  • I do not understand width of $1\over n^2$ – Myshkin May 30 '13 at 18:14
  • At each integer $n$, I consider rectangles with height equal to 1 and width equal to $1/n^2$. – guacho May 30 '13 at 18:28
  • YOu only need a convergent series, so, any power greater than 1 ($1/n^{1+s}$) will be ok. – guacho May 30 '13 at 18:31
  • what do you mean by at each integer $n$ there is a rectangle?do you want to say: between the interval $[1,2]$ The rectangle is of area 1, then between $[2,3]$ $1/4$ is width and height $1$ but that doesnt make sense too!, anyway I am not understanding youe statement unles you draw a picture for me. – Myshkin May 30 '13 at 23:31
  • Why not? Otherwise, the functions is zero. $$ f(x)=\sum \textbf{1}_{[n,n+1/n^2]}. $$ Now $f(x)\geq0$, and $$ \int f(x)dx=\sum\frac{1}{n^2}, $$ right? Following this idea you can construct a similar continuous example, right? If you change rectangles with triangles it will be enough, I think. – guacho May 31 '13 at 07:41
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  1. True and this a proof by contradiction if $\lim_{x\to\infty}f(x)=\ell\neq 0$ and suppose $\ell>0$ (the case $\ell<0$ is similar) then there's $A>0$ s.t $$\forall x\geq A\quad f(x)\geq \frac{\ell}{2}$$ so we have $$\int_A^\infty f(x)dx\geq \frac{\ell}{2}\int_A^\infty dx=+\infty \quad \Rightarrow\Leftarrow$$

  2. and 3. aren't correct and we can construct a counterexample as suggested by J.J