Statements on function with finite integral over $[0, \inf[$

Question

Let $f: [0, \infty[\to[0, \infty[$ be a continuous function such that: $$\int_0^\infty f(x) dx < \infty$$ Which of the following statements are true?

• The sequence $\{f(n)\}_{n\in\mathbb{N}} $ is bounded.
• $f(n) \to 0$ as $n\to \infty $
• The series $\sum_{n=1}^\infty is convergent.

Intuitively i feel each option is true.

For (a),if $f(n)$ is unbounded,since $f$ is non negative,the integral cannot be finite. For (b),if $f(n)$ does not tend to $0$,then again integral cannot be finite. For (c),since the series should take a value lesser than or equal to the integral of $f$,the series should be convergent(since partial sums must be bounded).

However the answer says none of the options are correct! Where am i going wrong?

Answer 1

Make up the function $f$ the following way (for $n\in\mathbb N, n\ge 2$):

• $f(n)=n$
• Linear in $[n-\frac{1}{n^3}, n]$ so that $f(n-\frac{1}{n^3})=0$
• Linear in $[n, n+\frac{1}{n^3}]$ so that $f(n+\frac{1}{n^3})=0$
• $f(x)=0$ otherwise.

In other words, at each $n\ge 2$, the function has a "spike" of height $n$, width $\frac{2}{n^3}$, and area $\frac{1}{n^2}$.

Such a function is an obvious counterexample for all three statements, yet $\int_0^{\infty}f(t)dt=\sum_{n=2}^{\infty}\frac{1}{n^2}$ - (absolutely) convergent.

Answer 2

Roughly speaking, the counterexample is a function that has a lot of tall thin spikes, but is zero elsewhere.

Explicitly, consider $$f(x) = \begin{cases}n & x \in [n, n+\frac{1}{n^3}) \\ 0 & \text{otherwise}\end{cases}$$

You complain that this is not continuous, but intuitively you can "smooth" it out to get a similar continuous example.

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Explicitly, let $g(x)$ be a "tent" function or a "bump" function supported on $[-1,1]$ such that $\int_{-1}^1 g(x) \mathop{dx} = c$ and $g(0) = 1$. Then consider $$f(x) = \sum_{n \ge 1} n \cdot g(n^3 (x - n)).$$ I believe then $\int_0^\infty f(x) \mathop{dx} = c \sum_{n \ge 1} \frac{1}{n^2} < \infty$, and $f(n)=n$. I may be wrong on the specific calculations, but hopefully the intuition is clear. Other answerers have provided more explicit examples for you...

Answer 3

Hint: Start considering the function $$ f(x)=\sum_{n=1}^\infty n\chi_{[\,n-1/(2^nn),n+1/(2^nn)\,]}(x), $$ How does the graph of $f$ looks like? The function $f$ is not continuous, but it can be slightly modified so that it is continuous (just “change the rectangles by triangles”). Why is it a counterexample?

[Here $\chi_A(x)$ denotes the characteristic function of the set $A$; i.e. it is the function whose value at $A$ is $1$ if $x\in A$, and $0$ otherwise.]

Answer 4

$a)$ In the words of angryavian (who deleted his/her answer), consider as a counterexample a function which has infinitely many, thin spikes, progressively taller spikes and is zero elsewhere. The height of the spikes is unbounded, but the width at the bottom of each spike could be made small enough to make the area under the $k^{th}$ spike equal to $\dfrac {1}{2^k}$, so that the integral is $\dfrac 12 + \dfrac 14 + \dfrac 18 + \cdots$ which is finite.

$b)$ The same function as in $a$. We see that for that function, the limit doesn't exist.

$c)$ Same function again, just makes the peaks of infinitely many of the spikes on the integers.