A [distribution](/page/Distribution) is an abstract object — a continuous linear functional on a space of test functions. It cannot be evaluated at a point, integrated against Lebesgue measure, or substituted into classical formulas. Yet the most common distributions in practice *do* come from functions: the distribution $T_f$ associated to a locally integrable function $f$ acts on a test function $\varphi$ by integration, $T_f(\varphi) = \int f\varphi \, d\mathcal{L}^n$. These are the **regular distributions**, and the map $f \mapsto T_f$ is the bridge between classical function theory and the distributional framework.

Understanding this bridge is essential because most of analysis — $L^p$ estimates, Fourier transform formulas, pointwise bounds, Sobolev norms — operates on functions, while the distributional setting provides the generality needed for PDE theory. Every time one writes an expression like $|\hat{u}(\xi)|^2$ for a [tempered distribution](/page/Tempered%20Distributions) $u \in \mathcal{S}'(\mathbb{R}^n)$, one is implicitly asserting that $\hat{u}$ is a regular distribution — represented by an actual measurable function whose absolute value can be computed pointwise. Making this assertion rigorous requires the theory developed on this page.

## Motivation

[motivation] ### Why Functions Are Not Distributions A measurable function $f: \Omega \to \mathbb{R}$ and a continuous linear functional $T: \mathcal{D}(\Omega) \to \mathbb{R}$ are fundamentally different objects. The function $f$ assigns a real number to each point $x \in \Omega$ (or at least to almost every point); the functional $T$ assigns a real number to each test function $\varphi \in \mathcal{D}(\Omega)$. The function lives on $\Omega$; the functional lives on the infinite-dimensional space $\mathcal{D}(\Omega)$. Because of this type mismatch, the statement "$f$ is a distribution" is strictly meaningless. What *is* meaningful is the statement "$f$ generates a distribution" — meaning there is a specific distribution $T_f$ that encodes all the information in $f$, constructed by integration against test functions. The question is then: does $T_f$ retain all the information about $f$? If two functions generate the same distribution, must they be equal (almost everywhere)? If so, no information is lost, and we can work with $T_f$ as a faithful proxy for $f$. ### Why Integration Against Test Functions Is the Right Construction The map $f \mapsto T_f$ defined by $T_f(\varphi) := \int_\Omega f(x)\varphi(x) \, d\mathcal{L}^n(x)$ is not the only way to associate a linear functional to a function. One could try pointwise evaluation: $E_f(\varphi) := f(x_0)$ for some fixed $x_0$. But this depends on the pointwise value of $f$, which is undefined for $L^p$ functions (defined only up to null sets). Or one could try $A_f(\varphi) := \int f \, d\mathcal{L}^n$ (ignoring $\varphi$ entirely), but this loses all local information about $f$. Integration against test functions is the right choice because it is simultaneously: - **Well-defined on $L^1_{\mathrm{loc}}(\Omega)$**: The integral $\int_\Omega f\varphi \, d\mathcal{L}^n$ is finite for every $\varphi \in \mathcal{D}(\Omega)$ because $\varphi$ has compact support and $f$ is locally integrable. - **Null-set insensitive**: If $f = g$ $\mathcal{L}^n$-a.e., then $T_f = T_g$ (integration does not see null sets), so the construction respects the equivalence relation on $L^1_{\mathrm{loc}}$. - **Faithful**: The map $f \mapsto T_f$ is injective — distinct $L^1_{\mathrm{loc}}$ functions (distinct as equivalence classes) generate distinct distributions, as proved in the [injectivity theorem](/theorems/450) below. - **Compatible with calculus**: The [distributional derivative](/page/Distributional%20Derivative) of $T_f$ recovers the classical derivative when $f$ is smooth, and extends it to non-smooth functions by integration by parts. ### The Injectivity Problem Injectivity is not obvious. The question is: if $\int_\Omega f\varphi \, d\mathcal{L}^n = 0$ for every $\varphi \in \mathcal{D}(\Omega)$, must $f = 0$ a.e.? This would fail if $\mathcal{D}(\Omega)$ were too small — for instance, if $\mathcal{D}(\Omega)$ contained only the zero function, every distribution would be zero and injectivity would be vacuous. It would also fail if we used a space of test functions that could not "detect" the values of $f$ at enough points. The proof that it works relies on mollification: if $\rho_\varepsilon$ is a standard [mollifier](/page/Mollifier), the function $y \mapsto \rho_\varepsilon(x - y)$ is a test function (for $x$ away from $\partial\Omega$), so $(f * \rho_\varepsilon)(x) = T_f(\rho_\varepsilon(x - \cdot)) = 0$. Since $f * \rho_\varepsilon \to f$ a.e. by the Lebesgue differentiation theorem, we conclude $f = 0$ a.e. [/motivation]

## Definition

[definition: Regular Distribution] Let $\Omega \subseteq \mathbb{R}^n$ be a non-empty open set and let $f \in L^1_{\mathrm{loc}}(\Omega)$. The **regular distribution** generated by $f$ is the distribution $T_f \in \mathcal{D}'(\Omega)$ defined by the map \begin{align*} T_f: \mathcal{D}(\Omega) &\to \mathbb{R} \\ \varphi &\mapsto \int_\Omega f(x)\,\varphi(x) \, d\mathcal{L}^n(x). \end{align*} [/definition]

That $T_f$ is a distribution requires verification. Linearity follows from linearity of the Lebesgue integral. For continuity, one checks the semi-norm condition from the [characterisation of distributions](/theorems/449): if $\mathrm{supp}(\varphi) \subseteq K$ for a compact $K \subset \Omega$, then

\begin{align*} |T_f(\varphi)| &\le \int_K |f(x)||\varphi(x)| \, d\mathcal{L}^n(x) \le \left(\int_K |f| \, d\mathcal{L}^n\right) \sup_K |\varphi|. \end{align*}

The constant $C_K := \int_K |f| \, d\mathcal{L}^n$ is finite because $f \in L^1_{\mathrm{loc}}(\Omega)$ and $K$ is compact. The bound $|T_f(\varphi)| \le C_K \sup_K |\varphi|$ has order $N_K = 0$: the distribution $T_f$ depends only on the values of $\varphi$, not on any of its derivatives. This means every regular distribution has **order zero** — a characterisation that fails for singular distributions (for example, $\delta_0'(\varphi) = -\varphi'(0)$ has order $1$).

[definition: Singular Distribution] A distribution $T \in \mathcal{D}'(\Omega)$ is **singular** if there is no $f \in L^1_{\mathrm{loc}}(\Omega)$ such that $T = T_f$. [/definition]

[example: The Dirac Delta Is Singular] For $x_0 \in \Omega$, the Dirac delta $\delta_{x_0}(\varphi) := \varphi(x_0)$ is a distribution. It is singular: if $\delta_{x_0} = T_f$ for some $f \in L^1_{\mathrm{loc}}(\Omega)$, then $\int_\Omega f\varphi \, d\mathcal{L}^n = \varphi(x_0)$ for all $\varphi \in \mathcal{D}(\Omega)$. Choose a sequence $\varphi_k \in \mathcal{D}(\Omega)$ with $0 \le \varphi_k \le 1$, $\varphi_k(x_0) = 1$, and $\mathrm{supp}(\varphi_k) \subseteq B(x_0, 1/k)$. The left side satisfies $|\int f\varphi_k \, d\mathcal{L}^n| \le \int_{B(x_0,1/k)} |f| \, d\mathcal{L}^n \to 0$ as $k \to \infty$ (since $f \in L^1_{\mathrm{loc}}$ and the measure of $B(x_0, 1/k)$ tends to zero). But the right side is $\varphi_k(x_0) = 1$ for every $k$. Contradiction. [/example]

## Injectivity of the Canonical Embedding

The most important structural property of the map $f \mapsto T_f$ is that it is injective: distinct $L^1_{\mathrm{loc}}$ functions generate distinct distributions.

[quotetheorem:450]

The injectivity has several important consequences. First, it justifies the *identification* of $f$ with $T_f$: since the map is injective, we can (and do) write $f \in \mathcal{D}'(\Omega)$ as shorthand for $T_f \in \mathcal{D}'(\Omega)$, and no ambiguity arises. Second, it gives a precise meaning to the question "is the distribution $T$ a function?": this means "does there exist $f \in L^1_{\mathrm{loc}}(\Omega)$ with $T = T_f$?", and if so, $f$ is unique up to null sets.

Third, it ensures that operations defined distributionally are consistent with classical operations. If $f$ is smooth, the distributional derivative $\partial^\alpha T_f$ agrees with the classical derivative: $\partial^\alpha T_f = T_{\partial^\alpha f}$, because integration by parts gives $(\partial^\alpha T_f)(\varphi) = (-1)^{|\alpha|}T_f(\partial^\alpha \varphi) = (-1)^{|\alpha|}\int f \, \partial^\alpha\varphi \, d\mathcal{L}^n = \int (\partial^\alpha f)\varphi \, d\mathcal{L}^n = T_{\partial^\alpha f}(\varphi)$ (the boundary terms vanish because $\varphi$ has compact support). Without injectivity, the equality $\partial^\alpha T_f = T_{\partial^\alpha f}$ would not pin down $\partial^\alpha f$ uniquely.

[example: When Two Functions Generate the Same Distribution] Let $f(x) = x$ and $g(x) = x$ for $x \ne 0$, $g(0) = 7$. Then $f = g$ $\mathcal{L}^1$-a.e. on $\mathbb{R}$ (they differ only at the single point $0$, which has measure zero), so $T_f = T_g$ in $\mathcal{D}'(\mathbb{R})$. This is consistent with injectivity: $f$ and $g$ represent the same element of $L^1_{\mathrm{loc}}(\mathbb{R})$, so they are not "distinct" in the relevant sense. The distribution $T_f$ cannot see pointwise values on null sets — only the equivalence class of $f$ in $L^1_{\mathrm{loc}}$. [/example]

## Regular Tempered Distributions

The same construction works in the [tempered distribution](/page/Tempered%20Distributions) setting, with a stronger integrability condition. For $f \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$, the functional $T_f(\varphi) = \int f\varphi \, d\mathcal{L}^n$ is a distribution on $\mathcal{D}(\mathbb{R}^n)$, but it may not extend continuously to $\mathcal{S}(\mathbb{R}^n)$ — because Schwartz functions have rapid decay but not compact support, and $f$ might grow too fast at infinity for the integral to converge.

The condition for $T_f$ to define a tempered distribution is that $f$ must not grow faster than a polynomial at infinity. Specifically, $T_f \in \mathcal{S}'(\mathbb{R}^n)$ whenever $f$ satisfies