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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".

undefined
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3 Answers3

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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.

Ian
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  • This is what I'm getting at. I have two distinct objects $f$ and $g$ to begin with ($\exists x \in \Omega: f(x) \neq g(x)$), introduce the new notion of a distribution for $T_f$ and $T_g$, respectively, and find that $T_f=T_g$. I then say that given $f$ and $g$ are treated as equivalent, the distributions can be considered a strict extension of these former objects. Just doesn't sound very genuine to me. I would have guessed that the set of functions considered in some sense was not sufficient for whatever use we had in mind and so needed to extend it while maintaining their difference. – undefined Feb 19 '19 at 10:35
  • @undefined You're getting stuck on a technicality. The fact that $T_f$ doesn't uniquely determine $f$ is not really all that important. The important thing about distribution theory is being able to work with distributions that aren't given by integration against functions, and yet still manipulate them in a useful way (e.g. differentiate them). – Ian Feb 19 '19 at 14:18
  • I think that we could add that functions that work the same in integration are often identified. One reason for this is that otherwise for example $|f| = \int |f(x)| , dx$ wouldn't be a norm, since $|f| = 0$ would not require $f(x) \equiv 0.$ Also for example $f(0) = 1,\ f(x) = 0 \text{ otherwise}$ fulfills $|f| = 0.$ – md2perpe Feb 19 '19 at 20:25
  • @md2perpe It is true that this identification is useful in certain other contexts, though it is also not universally useful, for example measure zero issues are quite annoying in the theories of stochastic processes and PDEs. – Ian Feb 19 '19 at 20:28
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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.

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 $Y=\mathcal{D}'(\Omega)$.

Pedro
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  • A late thanks Pedro. Could you please add some comments on the continuity of this map and the notion of a continuous embedding into $D'(\Omega)$, not being a normed space. In particular, I'd like to continuously embed a Banach space into $D'(\Omega)$, where $\Omega=\mathbb{T}$, the unit circle. It is related to my question here: https://math.stackexchange.com/questions/3230961/definition-a-banach-space-continuously-embedded-in-the-space-of-distributions – undefined May 30 '19 at 22:09
  • God bless you... – davise Dec 14 '23 at 09:53
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  • 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

reuns
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