0
$\begingroup$

I am a bit confused with the notion of local integrability. As far as I know:

a function $f(x)$ is locally integrable in (say) $\mathbb{R}$ if $\int_A |f(x)|dx$ exists for every bounded region $A$ in $\mathbb{R}$.

My first question is the existence part? Does this mean it shouldn't be infinite?

Secondly, let me take an example (taken from "Generalized Functions Theory and Technique" by Ram P. Kanwal)

$f(x) = \left\{\begin{array}{lr} \dfrac{1}{\sqrt{x}}, & \text{for } x>0\\ 0, & \text{for } x<0\\ \end{array}\right.$.

The function above is claimed to be locally integrable. How can one go about establishing this fact? Even more interestingly its derivative, except at $x=0$ is claimed not to be locally integrable

$f^\prime(x) = \left\{\begin{array}{lr} -\dfrac{1}{2}x^{-3/2}, & \text{for } x>0\\ 0, & \text{for } x<0\\ \end{array}\right.$.

Any suggestions on how to quickly test for local integrability would be great.

$\endgroup$
1
  • 2
    $\begingroup$ Yes, the integral needs to be finite. Try to integrate the derivative over $(0,1)$. $\endgroup$
    – dannum
    Mar 30, 2017 at 14:07

1 Answer 1

3
$\begingroup$

Yes, in order to be locally integrable the integrals need to be finite. One way to see that $f$ is locally integrable is that $f(x) \geq 0$, and we can use the $p$-test for convergence from calculus for example to show that $\int_{A} f(x)\,dx < \infty$ for any set bounded set $A\subset \mathbb{R}$ which contains $0$. After that you can use continuity to show that $\int_{A} f(x)\,dx < \infty$ for all bounded sets $A\subset\mathbb{R}$ such that $\lbrace 0 \rbrace \not\subset A$.

As for its derivative not being locally integrable, you just need to find a bounded set $A\subset\mathbb{R}$ such that $\int_{A} f'(x)\,dx$ does not converge. Again you can use the $p$-test for convergence to show that $\int_{0}^{1} f'(x)\,dx$ does not converge, and so $f'$ is not locally integrable.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .