6

Let $f: \mathbb R \rightarrow \mathbb R $ be a continuous function such that $\int_{0}^{\infty} \,f(x) dx$ exists.

Then Prove that incase

(i) $f$ is a non negative function, then $\lim_{x\rightarrow \infty} f(x) $ must exist and is $0$.

(ii) $f$ is a positive differentiable function , $\lim_{x\rightarrow \infty} f'(x) $ must exist and is $0$

$Attempt$: For the first part, i don't have a rigorous proof except for the fact that the given condition can be visualised geometrically. Since, the definite integral is actually calculating the area beneath the non negative function, the only way the given limit can exist when limit of f(x) itself tends to 0 at infinity.

Please give me a direction so that i can make this proof rigorous enough.

For the second part, i took an example. We know that ( leaving out the finite integration parts from $0$ to $1$ ..) $\int_{1}^{\infty} e^{-x^2} dx \leq \int_{1}^{\infty} e^{-x} dx$ and the latter converges. But the derivative of $e^{-x^2} = (-2x)e^{-x^2}$ whose integration does not exist when x $\in~[1,\infty)$ as it's a monotonic function after a finite $x$.

Any help in providing rigor to the above proof will be very helpful

Thanks

Yiorgos S. Smyrlis
  • 83,933
MathMan
  • 8,974
  • 7
  • 70
  • 135
  • 3
    This is false. You want uniform continuity in (i). – David Mitra Feb 04 '14 at 12:27
  • However, in (i), if the limit exists, it must be zero. – David Mitra Feb 04 '14 at 12:28
  • This question in an exam actually asks me to prove that limit will exist in all cases if it's a non negative function. However, i don't have a rigorous method to prove it. – MathMan Feb 04 '14 at 12:30
  • 4
    Hmm. What do you mean by "$\lim_{n\rightarrow\infty} f(x),dx$"? – David Mitra Feb 04 '14 at 12:31
  • $\lim_{x\rightarrow\infty} f(x)$ need not exist. Think of a function with "spikes" of height $1$ centered about the positive integers whose bases are of sufficiently small length so that the sum of the areas under the spikes is finite. Smoothing things out will give a counterexample to (ii). – David Mitra Feb 04 '14 at 12:33
  • I made a mistake. I have edited it now. Thanks – MathMan Feb 04 '14 at 12:33
  • You still have a "$dx$" there. This makes little sense to me... – David Mitra Feb 04 '14 at 12:34
  • @DavidMitra

    In this case, the function won't be continuous in between the integer intervals as the question demands?

     – MathMan Feb 04 '14 at 12:38
  • 6
    Take the function to be zero everywhere except for the "spikes". So $f(x)=0$ for all $x$ except for $x$ in an interval of the form $[n,n+1/n^2]$, $n>1$. On these intervals, take $f$ to be piecewise linear with value $0$ at the endpoints of the interval and value $1$ at the midpoint of the interval. This gives you a continuous function with $\int_0^\infty f$ finite and such that $\lim_{x\rightarrow\infty}f(x)$ doesn't exist. – David Mitra Feb 04 '14 at 12:44
  • 1
    See http://math.stackexchange.com/questions/108191/prove-fx-continuous-function-and-int-a-infty-fx-dx-infty-so-l – Najib Idrissi Feb 04 '14 at 13:08
  • 1
    The first part was disproven, for example, here. The second part was disproved here (the answer builds a $C^\infty$ counterexample). – TZakrevskiy Feb 04 '14 at 13:15

2 Answers2

5

(i) Not true. Let $g(x)=\max\{1-|x|,0\}$ which is continuous, nonnegative, and nonzero in $(-1,1)$, with $\int g=1$, and set $$ f(x)=\sum_{n=1}^\infty g(2^nx-n). $$ Then $f$ is continuous, $f\ge 0$, $\int_{\mathbb R} f=1$ (since $\int g(2^n x-n)\,dx=2^{-n}$) and $\limsup_{x\to\infty} f(x)=1$.

(ii) Also not true. Take $$ g(x)=\left\{\begin{array}{lll} \exp(1/(1-x^2)) &\text{if} & |x|<1, \\ 0 & \text{otherwise}. \end{array} \right. $$ Set as before $f(x)=\sum_{n=1}^\infty g(2^nx-n)$.

Then $f$ is $C^\infty$, $f\ge 0$, $\int f<\infty$, and $\limsup_{x\to\infty}f'(x)>0$.

Yiorgos S. Smyrlis
  • 83,933
4

This is false, the limit may not exist. Consider the infinite sum of $1/n^2$, and take a characteristic function of spikes in intervals of length $1/n^2$, zero otherwise. This integral, equal to $\sum 1/n^2$, converges, but the limit does not exist, not even in an almost-everywhere sense. What I think you mean is that a function in $L^1$ vanishes at infinity, which is to say that for all $\varepsilon>0$, $\mathcal{L}^n\{|f(x)|>\varepsilon\}$ is finite. If the limit of $f(x)$ does exist, then it is zero.

The condition that $f(x)\rightarrow 0$ as $x\rightarrow \infty$ is a neither necessary nor sufficient condition for integrability. The classical example is $1/x$.

michek
  • 668