# Show that a functional is a distribution

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 ?

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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
ah, this accept rate thing means i need to accept answers, sorry guys i didn't know that i had to do this, i am new to your community, will do it now! –  Stefan Jun 22 '12 at 16:25
What textbook do you use? –  martini Jun 22 '12 at 16:45
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 $(0,\infty)$.