Distributions as Generalized functions

Question

Just started looking at distributions and got a hold of Robert Strichartz: "A guide to distribution theory and Fourier Transforms".

On p.12, he defines two locally integrable functions as equivalent if as distributions they are equal, and goes on: "Thus by ignoring the distinction between equivalent functions, we can regard the locally integrable functions as a subset of the distributions. This makes precise the intuitive statement that the distributions are a set of objects larger than the set of functions, justifying the term generalized functions".

Could someone please elaborate on this idea of distributions being generalizations of "ordinary" functions, and tell me why I should buy into Strichartz definition of equivalent functions and the idea that the distributions are generalizations conditioned I "ignore their distinction".

Answer 1

A locally integrable function $f$ defines a distribution $T_f$ with $T_f(g)=\int f(x) g(x) dx$. Obviously $f$ uniquely determines $T_f$, but $T_f$ does not uniquely determine $f$; $T_f=T_g$ if and only if $f=g$ almost everywhere. If you identify functions by this equivalence relation, then the distributions are now a strict extension of this collection of equivalence classes, since there are distributions not of the form $T_f$, such as the Dirac delta function.

Answer 2

• The real numbers are the rationals plus all the limits of sequences of rationals converging in the $|.|$ norm.

• $L^1$ is the continuous functions plus all the limits of sequences of continuous functions converging in the $\|.\|_{L^1}$ norm.

• The distributions are the locally integrable functions plus all the limits of sequences of locally integrable functions $f_n$ for which the sequence of continuous operators $C^\infty_c \to C^\infty,\phi \mapsto \phi \ast f_n $ converges ($\ast$ for the convolution).

Try with $f_n(x) = n 1_{x \in [0,1/n]}, \lim_{n \to \infty} \phi \ast f_n = \phi$ so $f_n \to \delta$ in the sense of distributions

Answer 3

Let $X$ and $Y$ be two vector spaces.

Given a map $T:X\to Y$, the inclusion $T(X)\subset Y$ is, obviously, genuine.

Assume that $T$ is linear and injective. Then, from the algebraic point of view, $X$ and $T(X)$ are indistinguishable from one another. Therefore, in this case, we can identify $X$ with $T(X)$ and see $X$ as being a subspace of $Y$. In this context, we write $X\subset Y$ (being aware that, strictly speaking, no element of $X$ belongs to $Y$). For details, see the answers here, here and here.

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Now, let us return to the particular context:

Define $T:L^1_{\operatorname{loc}}(\Omega)\to\mathcal{D}'(\Omega)$ by $T(f)=T_f$, where $T_f$ be the distribution defined by $f$. Then:

• $T$ is linear.
• $T$ is injective provided that we ignore the distinction between equivalent functions.

Therefore, from the above discussion, by ignoring the distinction between equivalent functions (which forces $T$ to be injective), we can see $X=L^1_{\operatorname{loc}}(\Omega)$ as a subspace of $Y=\mathcal{D}'(\Omega)$.