Derivate of distribution, how does it work?

Question

Let $f\in L^1$. Then $f$ define a distribution $$L_f(\varphi )=\int_{\mathbb R}f\varphi=:\left<f,\varphi \right>,\quad \varphi \in \mathcal S(\mathbb R^n).$$ The derivative of $f$ (in distribution sense) is defined by $$L_{f'}(\varphi )=\left<f',\varphi \right>=:-\left<f,\varphi '\right>.$$

This looks a bit abstract for me. Can we write it under the form $\left<g,\varphi \right>$ ? Because at the end, to says that $g$ is the derivative in distribution sense of $f$, we need to have $$\left<f',\varphi \right>=\left<g,\varphi \right>,$$ right ? I have in my course that a distribution is derivable infinitely many time, but this doesn't really mean that there is derivative, no ? In the sense that there is $g$ s.t. $$\left<f',\varphi \right>=\left<g,\varphi \right>.$$

For example, what is the derivative of $\delta _0$ ? I know that $$\left<\delta _0',\varphi \right>=-\varphi '(0),$$ but is there a derivative ?

I'm very confuse with these distribution.

Answer 1

Strictly speaking, under the restricted meaning of the notation $\langle f,\varphi\rangle$ that it seems to be assumed in the question, i.e. $$ \langle f,\varphi\rangle=\int\limits_{\Bbb R}f(x)\varphi(x)\mathrm{d}x\quad f\in L^1(\Bbb R),\, \varphi\in C_c^\infty(\Bbb R)\label{1}\tag{1} $$ you can write $$ \langle f',\varphi \rangle=\langle g,\varphi \rangle=\int\limits_{\Bbb R}g(x)\varphi(x)\mathrm{d}x \label{2}\tag{2} $$ if and only if $f$ is weakly differentiable, i.e. if there exist a function $g\in L^1_\mathrm{loc}(\Bbb R)$ such that $$ -\int\limits_{\Bbb R}f(x)\varphi^\prime(x)\mathrm{d}x=\int\limits_{\Bbb R}g(x)\varphi(x)\mathrm{d}x\quad \text{ for all }\varphi\in C_c^\infty(\Bbb R). $$ This is true if, for example, $f$ belongs to Sobolev spaces $W^{1,1}(\Bbb R)$, $W^{1,1}_\mathrm{loc}(\Bbb R)$, $W^{1,2}(\Bbb R)$ and $W^{1,2}_\mathrm{loc}(\Bbb R)$.
Otherwise, the first sides of equations \eqref{1} and \eqref{2} should be interpreted only as the duality brakets between $C^\infty_c(\Bbb R)\equiv\mathscr{D}(\Bbb R)$ and the space of continuous linear functionals on it, i.e. the space of distributions $\mathscr{D}'(\Bbb R)$: simply stated, this means that \eqref{1} is a way to write the value assumed by $f$ on $\varphi$, i.e. $f(\varphi)$ (remember that $f\in\mathscr{D}'(\Bbb R)$ is a continuous linear functional on $\mathscr{D}(\Bbb R)$). Thus the derivative of $\delta_0$ is not a function and $\langle\delta _0',\varphi \rangle$ simply means $$ \delta_0'(\varphi)\triangleq \langle\delta _0',\varphi \rangle= -\varphi '(0) $$ and it is not the first derivative of any function $g\in L^1_\mathrm{loc}(\Bbb R)$.

Notes

• $L_{f'}(\varphi )=\langle f',\varphi \rangle=:-\langle f,\varphi '\rangle$ is truly a definition for the first derivative of a distribution and \eqref{2} is valid only for particular distributions, precisely the ones resulting from integration of a test function with respect to a Sobolev function belonging to one of the spaces listed above.
• As I stated above, $\delta_0^\prime$ is not the first derivative of any function $g\in L^1(\Bbb R)$: however, it is true that every distribution $f\in\mathscr{D}'(\Bbb R)$ is locally the finite order derivative (in a precise sense) of a continuous function $g\in C^0(\Bbb R)$. You can find a proof of this fact for example in the monograph [1], §2.4, pp. 35-36.
• Finally an advice: when in doubt about concepts from the standard theory of distributions, maybe you can solve them by looking at the monograph of Vladimirov [1], which is a simple yet rigorous and fairly complete introduction to the theory.

[1] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.