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Let $f$ be a locally integrable function, $f\in L_{\operatorname{loc}}^{1}(\mathbb{R}^n)$. Prove that the operator $$T_f:\phi\to\int_{\mathbb{R}^n}f(x)\phi(x)dx$$ is a distribution. (See Example 3.7 here)

I know that $T_f$ is linear, and I also know that $$T_{f}(\phi)\le ||\phi||_{\infty}\int_{K}|f(x)|dx$$ But why does this inequality implies that "$\phi_{n}\to\phi$ means $T_{f}(\phi_n)\to T_f(\phi)$"? Thank you for your attention!

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Try to use Lebesgue theorem. – Tomás Mar 5 '13 at 13:05
up vote 2 down vote accepted

The convergence $\phi_n \to \phi$ in $\mathcal{D}$ implies in particular $$\|\phi_n-\phi\|_{\infty} \to 0 \qquad (n \to \infty)$$ Thus $$|T_f(\phi_n)-T_f(\phi)| =|T_f(\phi_n-\phi)| \leq \|\phi_n-\phi\|_{\infty} \cdot \int_K |f(x)| \, dx \to 0 \qquad (n \to \infty)$$ where $K$ compact is chosen such that $\text{supp} \phi_n \subseteq K$ for all $n \in \mathbb{N}$ (see Definition 3.1).

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