Absolutly integrable functions are injective to tempered distribution?

Question

We had a theorem that$$\mathcal{L}^1(\mathbb{R}^n)\hookrightarrow \mathscr{S}'(\mathbb{R}^n)$$ Where $\mathscr{S}'$ is the space of tempered distributions. In the proof our lecturer constructed a linear map $i:f\to T_f$ where $T_f$ is regular distribution. It means that $$T_fg=i(f)g=\int_{\mathbb{R}^n} fg\mathrm{d}\lambda$$ I don't understand, why is this map injective from $\mathcal{L}$, because function $g\simeq f$, meaning that $g$ that is almost everywhere equal to $f$ maps to the same distribution, doesn't it?

Answer 1

Functions that are equal almost everywhere are identified. The Lebesgue spaces are not spaces of functions, but spaces of equivalence classes of functions.

Sometimes one also considers the spaces of functions - which are not Hausdorff in the $\lVert\,\cdot\,\rVert_p$ seminorm - and then distinguishes the spaces by using different fonts for the $L$, e.g. $\mathscr{L}$ for the non-Hausdorff space of functions, and $L$ for its quotient, the normed space of equivalence classes of functions (modulo being equal almost everywhere).

If that is the case in your lecture, and $\mathcal{L}$ denotes the space of functions instead of the space of equivalence classes of functions, then it was a mistake, and the map $\mathcal{L}^1(\mathbb{R}^n) \to \mathcal{S}'(\mathbb{R}^n)$ is not injective, and the other font ought to have been used.

If $\mathcal{L}^1(\mathbb{R}^n)$ is the Banach space of equivalence classes of (absolutely) integrable functions, the mapping is injective.

Answer 2

$L^1(\mathbb{R^n})$ to be viewed as a normed space is quotiended over the functions that are almost everywhere zero. Knowing this your function $f\mapsto T_f$ become injective.