2
$\begingroup$

Let $h:[1, \infty)\rightarrow \mathbb R$ a continuous non-negative function, such that $\int_{1}^{\infty} h(x)\ dx$ converges. does $h$ must be bounded in $[1, \infty)$?

I tried to prove it by showing that if $\int_{1}^{\infty} h(x)\ dx$ converges, then by the definition the $\lim_{b \to \infty}\int_{1}^{b} h(x)\ dx$ exists. Can I conclude that from the existence of this limit, the function $h$ is bounded?

$\endgroup$
  • $\begingroup$ No. It's the same phenomenon that happens for the question: Why do we need "$\lim_{x\to\infty} f'(x)$ exists" in the theorem "if $f$ is a $C^1$ strictly increasing function such that $\lim_{x\to \infty} f(x)=L\in\Bbb R$ and $\lim_{x\to \infty} f'(x)=M\in\Bbb R$, then $M=0$"? $\endgroup$ – user228113 May 3 '17 at 9:47
5
$\begingroup$

This statement is false. Consider a function $f$ which is usually $0$, except for a triangular "bump" that happens at each natural number. The bump has width $1/n^3$ and height $n$, and so it has area $\frac{1}{2n^2}$. Then we have $$\int_1^\infty f(x) dx = \sum_1^\infty \frac{1}{2n^2} = \pi^2/12 < \infty$$ But $f$ is not bounded, since the bumps get arbitrarily large.

More formally, if $x$ is a real number greater than or equal to $1$ then we may uniquely express $x$ in the form of $x=n+\varepsilon$, where $n\in \mathbb{N}$ and $\varepsilon \in [0,1)$. Then let $f(x) = 0$ if $\varepsilon > 1/n^3$, $f(x) = (2/n^2)(x-n)$ if $\varepsilon < 1/(2n^3)$, and $f(x) = -(2/n^2)(x-n-1/(2n^3))$ otherwise.

$\endgroup$
2
$\begingroup$

Simple answer: no. $$ f(x)=\sum_{n\geq 1}n\,e^{-n^6(x-n)^2} $$ is a continuous, positive and integrable function on $\mathbb{R}^+$, but it is unbounded since $f(n)\geq n$.

$\endgroup$
1
$\begingroup$

enter image description here

Answer with this beautifully drawn picture using paint :D. The domain is off though.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.