Given $\Omega \subset \mathbb{R}^{N}$ and a locally integrable function $f$, its regular distribution $T$ is defined by $\langle T,\phi \rangle = \int_{\Omega}f(x)\phi(x)dx$, where $\phi \in D(\Omega)$.
$D(\Omega)$ is the set of smooth functions with compact support $C^{\infty}_{0}(\Omega)$ with a special convergence defined by "$\phi_n \rightarrow \phi$ in D" if
There exists a set $K$ compact in $\Omega$ such that $supp(\phi_n) \subset K$ for all $n \in \mathbb{N}$.
All derivatives of $\phi_n$ converge uniformly to $\phi$ in $\Omega$.
Now, by definition, T is a distribution if it is a linear continuous functional on $D(\Omega)$. Linearity of the regular distribution is obvious from the linearity of integration. For some reason, I had much more trouble with proving that it is continuous, or in other words that $$\phi_n \rightarrow \phi \text{ in D} \implies \langle T,\phi_n \rangle \rightarrow \langle T,\phi \rangle.$$
First, I thought that I would just show that $f \cdot \phi_n$ converges uniformly to $f \cdot \phi$ and so I can put the limit inside the integral and be done with it. However, I don't think this is actually true, because $f$ need not be bounded. For example with $\Omega = (-1, 1)$ and $f = \frac{1}{\sqrt{|x|}}$, I don't think that $f \cdot \phi_n$ converges uniformly to $f \cdot \phi$ for all $\phi$.
So I decided to try to prove it by using the Dominated Convergence Theorem. I would be very grateful if you checked that the proof is valid.
Let $\phi_n \rightarrow 0$ in D. Then there exists $K$ compact in $\Omega$ s.t. $supp(\phi_n) \subset K$ for all $n \in \mathbb{N}$. $$\langle T,\phi_n \rangle = \int_{\Omega}f(x)\phi_n(x)dx = \int_{K}f(x)\phi_n(x)dx.$$
Define $h_n(x) = f(x)\phi_n(x)$.
Since $\phi_n \in D(\Omega)$, it's continuous on K and so it's bounded by some $M_n$. Also, because $\phi_n$ converges uniformly to $0$, $\lim_{n \rightarrow \infty}M_n = 0$.
By $\tilde{\phi}$ denote the supremum of $M_n$ over all $n$ (this supremum is finite thanks to the reasoning above). Then for all $n$ and for all $x$, $|\phi_n(x)| \leq \tilde{\phi}$.
Hence, $|h_n(x)| = |f(x)|\cdot|\phi_n(x)| \leq \tilde{\phi}\cdot|f(x)| = g(x).$
It remains to show that g is integrable.
$$\int_K g(x)dx = \tilde{\phi}\cdot\int_K |f(x)|dx = \tilde{\phi} \cdot \left( \int_{K_1} f(x)dx - \int_{K_2} f(x)dx \right),$$
where $K_1 \subset K$ is the closure of the set where $f(x)$ is positive, and $K_2 \subset K$ is the closure of the set where $f(x)$ is negative. Both of these sets are closed and bounded, which means that they're compact, which means that since $f$ is locally integrable, both integrals are finite.
$h_n$ converges pointwise to $f \cdot \phi$, and by the Dominated Convergence Theorem I can interchange the limit and the integral, finally giving me
$$\lim_{n \rightarrow \infty} \int_{\Omega} f(x)\phi_n(x)dx = \int_{\Omega} \lim_{n \rightarrow \infty} \left( f(x)\phi_n(x) \right)dx = 0.$$
By linearity, this shows that $T$ is continuous.
I'm pretty sure I'm missing something much easier, but do you think that the proof is correct? Thank you!
The proof seems valid but I don't see any reason to split $K$ into $K_1$ and $K_2.$ Since $f \in L^1_{\text{loc}}(\Omega)$ you already know that $\int_K |f(x)| \, dx < \infty.$
Shorter proof
Take $\phi_k \to 0$ in $D(\Omega)$ with support in $K.$ Then, $$ | \langle f, \phi_n \rangle_\Omega | = | \langle f, \phi_n \rangle_K | = \left| \int_K f(x) \, \phi_n(x) \, dx \right| \leq \int_K |f(x)| \, |\phi_n(x)| \, dx \\ \leq \int_K |f(x)| \, \|\phi_n\|_K \, dx = \int_K |f(x)| \, dx \cdot \|\phi_n\|_K = \|f\|_K \|\phi_n\|_K \to 0 $$ since $\|f\|_K = \int_K |f(x)| \, dx < \infty$ and $\|\phi_n\|_K = \sup_K |\phi_n| \to 0.$