Show that a functional is a distribution

Question

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 ?

Answer 1

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 $(0,\infty)$.