I'm not a math student. I just want to touch distribution theory briefly. My question is, is a function with good properties defined in a closed interval a generalized function?
For example, $f(x)=x$ only defined on $[0,1]$, is $f\in\mathscr{D}'(\mathbb{R})$
and:$$\langle f,\varphi\rangle=\int_0^1 f\varphi dx$$
($\varphi\in\mathscr{D}(\mathbb{R}^n)$ is a test funtion)
Please note that the $f\in\mathscr{D}'(\mathbb{R})$ is not the $f\in\mathscr{D}'(\Omega)$($\Omega$ is an open set of $\mathbb{R}^n$)
And whether any ideal function $f(x)$ only defined on $[a,b]$, such as a function that is absolutely integrable within a defined closed interval, satisfies this, and satisfies: $$\langle f,\varphi\rangle=\int_a^b f\varphi dx$$
Because I only found $f\in L_{loc}^1(\mathbb{R}^n)$ that satisfy this，which defined $\langle f,\varphi\rangle=\int_{\mathbb{R}^n} f\varphi dx$, so $f\in \mathscr{D}'(\mathbb{R}^n)$, But the integral region is $\mathbb{R}^n$, which makes me confused.
A not too ill-behaved function on a subregion of $\mathbb R^n$ can be extended to a distribution on all of $\mathbb R^n.$
For example, if $f \in L^1_\text{loc}([a,b])$ then we can define a distribution $T_f$ on $\mathbb R$ by $$ \langle T_f, \phi \rangle = \int_a^b f(x) \, \phi(x) \, dx. $$
We can even extend $x^\alpha,$ for any $\alpha<0,$ defined on $(0,\infty)$ to a distribution on $\mathbb R.$