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

I was thinking about the following problem:

Let $f:\mathbb R\rightarrow \mathbb R$ be a continuous function such that $\int_{0}^{\infty}f(x)dx$ exists. Then which of the following statements are correct?

(a) If $\lim_{x\to\infty}f(x)$ exists, then $\lim_{x\to\infty}f(x)=0,$

(b) The limit $\lim_{x\to\infty}f(x)$ must exist and is zero,

(c) In case $f$ is a nonnegative function, the limit $\lim_{x\to\infty}f(x)$ must exist and is zero,

(d) In case $f$ is a differentiable function, the limit $\lim_{x\to\infty}f'(x)$ must exist and is zero.

If I take $f(x)=e^{-x}$, so that the given condition is satisfied then we see that options (a) and (c) are correct. But I am not sure about the choice given in (b) and (d). But If I have to prove it in general, then how can I prove it? I mean an alternative better approach. Please help. Thanks in advance for your time.

Statement (a):

Let's say $\lim_{x\to \infty} = a$. if $a\neq 0$, then by definition of limit, there is a number $X$ such that for any $x \geq X$, we have $|f(x)| \geq a/2$. This means that from that point on, the integral $\int_0^xf(x)\,dx$ will increase (or decrease, depending on whether $a$ is positive or negative) by at least $a/2$ for each unit we increase $x$, and thus the improper integral doesn't exist. So if $a$ exists and the integral is finite, then $a = 0$ by this contradiction.

Statement (b) and (c):

I do these together because they can be disproven with the same counter-example. Start out with the function that is constantly $0$. Now, take the interval $[0.5, 1.5]$, and raise the graph as a triangle with a top in the point $(1, 2)$. Now the total integral is $1$. Then for the interval $[1.75, 2.25]$, raise a triangle so the top is at $(2, 2)$. The total integral is now $1.5$. Continue doing this around every integer point along the $x$ axis, always halving the interval length. You will end up with a function that integrates to $2$, but has bumps all over with height $2$. So no limit. It is also non-negative, so it fulfils (c).

Statement (d):

Taking the function above, you can "smooth out" the corners so that the function is differentiable, and the change to the total integral is finite. Now you have a differentiable function which has derivatives of whatever size you want, if you just go far enough out, so the derivative has no limit.

• If you smooth out the corner then how do you know that the function will be differentiable? I doubt because the bumps will keep shrinking..@Arthur – cmi Nov 18 '18 at 7:40
• @cmi If the bumps are copies of, say, a single period of $1+\cos(x/2^{n+1}\pi)$ instead of being triangles, then they are definitely differentiable. – Arthur Nov 18 '18 at 8:26
• Can you please tell me what is " single period of $1+cos(x/2^{n+1}\pi)$"?@Arthur – cmi Nov 18 '18 at 11:54
• @cmi Pick a natural number $n$ (use a relatively small one, like $0$ or $1$ or $2$). Draw the graph of $1+\cos(x/2^{n+1}\pi)$. See that it consists entirely of bumps. Take one of those bumps and use it in my example. Then increase $n$, and take one of those bumps. And so on. These will be differentiable, although for each bump the maximum value of the derivative will get larger and larger, which was the entire point of using something like that as an example. – Arthur Nov 18 '18 at 11:56

Hint: Consider only non-negative functions. Does a finite integral of $f$ imply that $f$ takes small values? Certainly not. Does a finite integral of $f$ imply that $f$ is small on a "large" set? Yes. But: If the set where $f$ can be large has to be small, does that imply that $f$ has to converge to zero at infinity?

Try to construct an example of a non-negative $f$ which is non-zero for every $x\in\mathbb{Z}$ but still has finite integral...

Regarding b) and c) (and d) )

Consider a function that contains of triangles with height 1 on an interval $[n,n+2^{-n}]$ for every natural number (including zero) $n$ and is zero everywhere else.

This function is

• continuous (+)
• integrable (++)
• has no limit (+++)

(+): This function looks as follows \begin{align} f(x)=\begin{cases} 0 &\mbox{if } x\notin [n,n+2^{-n}] \text{ for any $n$} \\ u_n(x) &\mbox{if } x\in[n,n+2^{-(n+1)}] \text{ for an $n$} \\ d_n(x) &\mbox{if } x\in[n+2^{-(n+1)},n+2^{-n}] \text{ for an $n$} \\ \end{cases} \end{align} Where \begin{align} u_n(x) = (x-n)\cdot 2^{(n+1)} \end{align} is the left part of the triangle and the right part is given by

\begin{align} d_n(x) = -(x-(n+2^{-(n+1)})\cdot 2^{(n+1)} \end{align}

This function is continuous, which can be seen by inserting the values $n$, $n+2^{-(n+1)}$ and $n+2^{-n}$.

(++): This function is integrable. \begin{align} 0<\int_0^{\infty} f(x)\, dx = \sum_0^{\infty} 2^{-(n+1)}<\sum_0^{\infty} 2^{-n}=2 \end{align}

(+++): The limit does not exist. Consider the following sequences which both reach infinity. \begin{align} (x_n)_n=n \rightarrow \infty\\ (y_n)_n=n+2^{-(n+1)} \rightarrow \infty \end{align} But we have \begin{align} f((x_n)_n) = 0 \, \forall n \end{align} and \begin{align} f((y_n)_n) = 1 \, \forall n \end{align}

As $f(x)\geq0$ this also proves c) wrong.

For d) instead of triangles you could use a function similar to the Gaussian function which lies below the triangle. You could therefore build the convolution $\tilde f = f*G$ with $G$ a Gaussian function. By that, you would have properties (+) and(++) and additionally a differentiable function whose derivative does not converge.

Here are some hints:

(a) If the limit exists, but it is not zero, then what happens?

(b) Try to find a function that has infinitely many "bumps", whose total area is finite, so that the integral is finite but the limit of the function is not defined. This will also address (c) and (d).