Let $\Omega\subset\Bbb R^n$ be a domain (i.e. some open subset with smooth boundary). Recall that we can define the space of distributions $\mathcal{D}'(\Omega)$ as the topological dual of $C^\infty_{\mathrm c}(\Omega)$. Furthermore, for any distribution $u\in\mathcal{D}'(\Omega)$, and for $1\le p\le\infty$, we can define the $L^p(\Omega)$ norm $$\lVert u\rVert_{L^p(\Omega)}=\sup_{\lVert\varphi\rVert_{p'}=1}\lvert\langle u,\varphi\rangle\rvert$$ which corresponds with the usual $L^p$ norm whenever $u=f\in L^p$.

Now, if $\lVert u\rVert_{L^p(\Omega)}<\infty$, then $u$ can be extended (using Hahn-Banach) to some bounded linear functional $\ell\in L^{p'}(\Omega)'$. If $p>1$, then we know that $\ell=\langle f,\cdot\rangle$ for some $f\in L^p(\Omega)$, and since $C_{\mathrm c}^\infty(\Omega)\subset L^{p'}(\Omega)$, we have that $u=f$ as distributions.

But what if $\lVert u\rVert_{L^1(\Omega)}<\infty$? Then must we have $u\in L^1$?

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    $\begingroup$ If $u=\delta_x\notin L^1$ with $x\in\Omega$, then $\|u\|_{L^1(\Omega)}=1$. $\endgroup$ – user254433 Jul 8 '17 at 23:02
  • $\begingroup$ Is the issue just that my definition of $\lVert u\rVert_{L^1(\Omega)}$ was bad, or is there really no (reasonable) way of assigning a sensible $L^1$ norm to distributions? $\endgroup$ – Monstrous Moonshine Jul 8 '17 at 23:07
  • $\begingroup$ @MonstrousMoonshine Suppose $\text{supp}\,\varphi\subset K$ and $|\langle u,\varphi\rangle|\le C(K)\|\varphi\|_{L^\infty(\Omega)}$. If $C(K)\to 0$ as $dx(K)\to 0$, then $u\in L^1$. $\endgroup$ – user254433 Jul 9 '17 at 0:04
  • $\begingroup$ What do you mean by $dx(K)$? Is $K\subset\Bbb R^n$ a compact set here? $\endgroup$ – Monstrous Moonshine Jul 9 '17 at 8:32

As user254433 pointed out, having finite $L^1$ norm in the distributional sense does not imply we have an $L^1$ function -- it could be a singular finite measure instead, such as the Dirac delta.

It is difficult to characterize $L^1$ functions through their actions on some other things, because $L^1$ is not the dual space of any normed space.

One can describe $L^1$ functions as measures that are absolutely continuous with respect to the Lebesgue measure $dx$. This leads to the characterization that user254433 suggested: given an open set $\Omega$, define $$ \|u\|_{1,\Omega} = \sup \{|\langle u,\varphi\rangle|\colon |\varphi|\le 1, \ \operatorname{supp} \varphi\subset \Omega \} $$ be the "restricted $L^1$ norm of $u$". Then $u\in L^1$ if and only if for every $\epsilon>0$ there exists $\delta>0$ such that $\|u\|_{1,\Omega}<\epsilon$ whenever the Lebesgue measure of $\Omega$ is less than $\delta$.

This isn't really saying much; it's just the absolute continuity of a measure expressed in terms of a distribution.


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