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


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    $\begingroup$ This is false. You want uniform continuity in (i). $\endgroup$ Feb 4, 2014 at 12:27
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    $\begingroup$ Hmm. What do you mean by "$\lim_{n\rightarrow\infty} f(x)\,dx$"? $\endgroup$ Feb 4, 2014 at 12:31
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    $\begingroup$ 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. $\endgroup$ Feb 4, 2014 at 12:44
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    $\begingroup$ See math.stackexchange.com/questions/108191/… $\endgroup$ Feb 4, 2014 at 13:08
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    $\begingroup$ The first part was disproven, for example, here. The second part was disproved here (the answer builds a $C^\infty$ counterexample). $\endgroup$ Feb 4, 2014 at 13:15

2 Answers 2


(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 $\limsup_{x\to\infty}f'(x)>0$.

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    $\begingroup$ In (ii), $g$ tends to $\infty$ when $x$ approaches $1$. $\endgroup$
    – Traklon
    Feb 4, 2014 at 13:04

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 $1/x$.


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