Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the following functional $$ \langle u , \phi \rangle = \int_0^{\infty} \phi(t) \frac{\mathrm{d}t}{t^{\alpha}} $$ I want to show that it is a functional in $\mathcal{D'}{(\mathbb{R})}$. Because of the compact support of $\phi$, the indefinite integral could be made definite by an upper bound $C$. $$ \langle u , \phi \rangle = \int_0^{C} \phi(t) \frac{\mathrm{d}t}{t^{\alpha}} $$ So, first linearity is clear because of this property from the integral. $$ \langle u , a \phi + b \psi \rangle = \int_0^{C} (a \phi(t) + b \psi) \frac{\mathrm{d}t}{t^{\alpha}} = a \int_0^{C} \phi(t)\frac{\mathrm{d}t}{t^{\alpha}} + b \int_0^{C} \psi(t)\frac{\mathrm{d}t}{t^{\alpha}} = a\langle u , \phi \rangle + b\langle u, b\psi \rangle $$ For continuity i have to consider a sequence $\phi_k \to \phi$, but thats were i stuck, i have no idea how to show that $\langle u, \phi_k \rangle \to \langle u, \phi \rangle$? Do you have any hints for me ?

share|cite|improve this question
You need an assumption on $\alpha$: if $\alpha=-1$ and $\varphi=1$ on a neighborhood of $0$ and is non-negative, $u_3(\varphi)$ is not well-defined. And why the notation $u_3$ for this functional? – Davide Giraudo Jun 22 '12 at 16:16
ah, ok just consider $\alpha \ne 1$. the notation comes because its the third functional from a textbook, the other ones i could solve i guess, but this one i find hard. but i will change it to $u$. – Stefan Jun 22 '12 at 16:20
I do not understand: if $\phi=1$ near $0$, $\int \frac{1}{t^\alpha}dt$ might not converge? – Siminore Jun 22 '12 at 16:22
What textbook do you use? – martini Jun 22 '12 at 16:45
@DavideGiraudo Figuring out valid values of $\alpha$ might be part of the exercise, don't you think? – user20266 Jun 22 '12 at 16:48
up vote 4 down vote accepted

Any locally integrable function $f\colon\mathbb{R}\to\mathbb{R}$ (that is, $\int_K|f(x)|\,dx<\infty$ for all compact $K\subset\mathbb{R}$) defines a distribution on $\mathbb{R}$ through $$ \langle u_f,\phi\rangle=\int_{-\infty}^{\infty}f(x)\,\phi(x)\,dx\quad\forall \phi\in\mathcal{D}(\mathbb{R}). $$ Since $\phi$ has compact support and $f$ is locally integrable, $u_f$ is well defined. Linearity is obvious. As for continuity, suppose $\phi_n\to\phi$ in $\mathcal{D}$. This means in particular that there exists a compact $K\subset\mathbb{R}$ such that the support of $\phi_n$ is contained in $K$ for all $n$ and that $\phi_n$ converges uniformly to $\phi$ on $K$. Since $f$ is integrable on $K$, it follows that $$ \lim_{n\to\infty}\langle u_f,\phi_n\rangle=\lim_{n\to\infty}\int_Kf(x)\,\phi_n(x)\,dx=\int_Kf(x)\,\phi(x)\,dx=\langle u_f,\phi\rangle. $$ In your question $f(x)=\chi_{(0,\infty)}(x)x^{-\alpha}$, where $\chi_A$ is the characteristic function of a set $A$. This function is locally integrable if an only if $\alpha<1$. If $\alpha\ge1$, as Davide's comment shows, is not the functional is not defined.

On the other hand, your functional defines a distribution on $(0,\infty)$, since $x^{-\alpha}$ is localy integrable on $(0,\infty)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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