Weak versus locally integrable

derivatives and delta functions

derivatives and delta functions

University of Michigan

Physics Department

Ann Arbor, MI 48109–1120

Physics Department

Ann Arbor, MI 48109–1120

Revised: February 25, 2016

Contents

In one dimension, the delta function $\del(x)$ is the classic
example of a generalized function which is not locally Lebesgue
integrable, i.e., not in $\Loneloc$, but is the weak derivative
of a generalized function $\theta(x)$, the Heaviside function,
which *is* locally integrable and has an $\Loneloc$
derivative. To put it another way, the Heaviside function is an
example of a locally integrable function with weak and locally
integrable derivatives that do not coincide.

This note summarizes how that works in $d$ dimensions for
integral and fractional inverse powers $\norm{x}^{-n}$, $x \in
\Rd$, $n \ge 0$, and their role for integer $n$ in the
traditional application of the $d$-dimensional
Gauss-Ostrogradsky divergence
theorem to arrive at the $d$-dimensional delta function and
its derivatives. A good reference for all of that is Laurent
Schwartz, *ThÃ©orie des Distributions*, Hermann, revised
edition, 1966.

The last section, on nonintegrable inverse powers of distance, raises the question of how to write integer inverse powers as invariant derivatives of locally integrable powers, or powers multiplied by a logarithm.

Let $f \in \SRd$, the Schwartz space of rapidly decreasing smoooth functions, and $h \in \SpRd$, the Schwartz space of tempered distributions. This discussion keeps a distinction between the two common notations for the linear functional defined by $h$ on $\SRd$, $\bk{h}{f}$ and $\int\!hg\,\dx$. Namely, we use the second notation only when $hg$ is actually an absolutely integrable function, and the integral is a Lebesgue integral. The two coincide when $h \in \Loneloc \cap \Sp$.

In particular we want to distinguish between the locally integrable derivative of a locally integrable function, and the weak derivative. The notation $\partial$ is used for the $d$-dimensional gradient or derivative operator.

*Definition 1*: The *weak derivative*
$\partialw$ of $h \in \Loneloc(\Rd)\cap\SpRd$ is defined
to be its distributional derivative:
\begin{equation}
\bk{\partialw h}{f} = - \bk{h}{\partial f}\,.
\end{equation}

Here is the basic fact about weak and $\Loneloc$ derivatives:

*Lemma 1*: Let $h$ and $\partial h$ belong to
$\Loneloc(\Rd)\cap\SpRd$. Then $\partial h = \partialw h$
iff the following is a Lebesgue integral identity for $f \in \SRd$:
\begin{equation}
\int(\partial h)\,f \dx = - \int\! h\,\partial f\,\dx\,,
\qquad
\dx \equiv \dx_1...\dx_d \qed
\end{equation}

An important example of the relationship between integrable and weak derivatives is provided by the function $h = \norm{x}^{-n}$ (Euclidean norm), which is locally integrable only when $n < d$. We use the notation $\norm{x} = r$. The gradient \begin{equation} \partial\frac{1}{r^n} = -n\frac{\hat{x}}{r^{n+1}}\,, \qquad \hat{x} = \frac{x}{r}, \qquad \norm{\hat{x}} = 1 \end{equation} exists everywhere except at $x = 0$, but is locally integrable iff $\npone < d$. In the following, $n$ can be either fractional or integral.

First, we need a basic fact about tempered distributions.

*Lemma 2 (dominated convergence property)*: Let $h$ be
any locally integrable, polynomial bounded function. Then
at any point $a \in \Rd$, the tempered distribution defined by
$h$ obeys the limit
\begin{equation}
\bk{h}{f} = \limepsz\int\limits_{\norm{x-a} > \vareps}\! h\,f\,dx\,.
\end{equation}

*Proof*: For $\eps > 0$, the integral
\begin{equation}
\int\limits_{\norm{x-a} > \vareps}\! h\,f\,dx
= \int\! \theta\left(\norm{x-a}-\vareps\right)\,h\,f\,\dx
\end{equation}
defines a tempered distribution, because $\theta\,h f$ is
absolutely integrable. The limit $\vareps \rightarrow 0$ exists
by Lebesgue dominated convergence, and defines a distribution by
the weak convergence property of tempered distributions. $\tqed$

The main result of this section is the following.

*Theorem 1*: When the gradient of $r^{-n}$ is $\Loneloc$
at the origin, it coincides with the weak gradient.

*Proof*: Since $r^{-n}$ has no $\Loneloc$ derivative when
$\npone \ge d$, we need only consider $\npone < d$. The
following sequence of equations, explained below, shows that the
weak and locally integrable derivatives coincide:
\begin{align}\label{wdef}
\bk{\partialw r^{-n}} {f}
& = -\bk{r^{-n}}{\partial f}
\\[1ex]\label{limint:gradf}
& = -\limepsz \int\limits_{r>\vareps}
\!r^{-n}\,\partial f\,dx
\\[1ex]\label{lim:1}
& = \limepsz\intgteps\partial r^{-n}\,f\,dx
- \limepsz\intgteps\partial\left(r^{-n}\,f\right)\,dx
\\[1ex]\label{lim:2}
& = \int\! \partial r^{-n}\,f\,dx
+ \limepsz\inteqeps \hat{x}\,f \vareps^{d-n-1}\,\dOme
\\[1ex]\label{int:f}
& = \int\! \partial r^{-n}\,f\,dx
\\[1ex]\label{int:gradf}
& = -\!\int\!r^{-n}\,\partial f\,\dx
\end{align}
Here are the steps:

(i) Equation \eqref{limint:gradf} follows from the r.h.s. of the definition \eqref{wdef} by the dominated convergence property.

(ii) The integral in \eqref{limint:gradf} converges to the integral over $\Rd$ as $\vareps$ approaches zero, by dominated convergence. The result is recorded in the last step, Eq. \eqref{int:gradf}.

(iii) The first integral in \eqref{lim:1} also converges by dominated convergence. The second integral in that equation gives the second integral in \eqref{lim:2} by the Gauss-Green theorem, where $\hat{x}$ is minus the unit normal to the finite part of the surface surrounding the volume outside the $\pdmone$-sphere at radius $\vareps$, with the surface integral at infinity contributing zero. Here $\dOme$ is the solid angle element subtended by the $\pdmone$-sphere.

(iv) The second integral in \eqref{lim:2} goes away, because $d \mt n \mone > 0$, and every component has absolute value bounded by \begin{equation}\label{solidangle} \int\abs{f \vareps^{d-n-1}}\,\dOme\, \leq\, \vareps^{d-n-1} \sup\, \abs{f} \int\!\dOme\,, \qquad \int\!\dOme = \SA\,. \end{equation} The result is Eq. \eqref{int:f}, which together with Eq. \eqref{int:gradf} shows that the $\Loneloc$ derivative of $r^{-n}$ coincides with its weak derivative, thus proving the lemma. $\tqed$

The total solid angle in \eqref{solidangle} is the finite area of the unit hypersphere in $d$ dimensions.

To see in more detail what goes wrong when $n \ge \dmone$, consider first the case $n = \dmone$. The argument above changes in steps (iii) and (iv). The limit in \eqref{lim:2} is not the problem. It still exists and still vanishes: \begin{equation} \limepsz\inteqeps \hat{x}\,f \vareps^0\,\dOme = f(0) \int\!\hat{x}\,\dOme = 0\,. \end{equation} The unit vector $\hat{x}$ depends only on the angles in $\dOme$, and not on $\vareps$, and averages to zero. The problem is that the integrand in the first integral in \eqref{lim:1} is no longer locally integrable at the origin.

Here's an example of a test function $f$ for which integrability explicitly fails. Let $f$ be the product of two smooth functions $f_1$ and $f_2$. Let $f_1$ be rapidly decreasing at infinity, nonzero at the origin, and a function only of $r$. Let $f_2$ be a smooth, nonrotation-invariant function on the unit $\pdmone$-sphere. It depends only on the angles in $\dOme$, and not on $r$. Because \begin{equation} \dx = r^{d-1}\dr\,\dOme\,, \end{equation} the first limit in \eqref{lim:1} becomes: \begin{align} \limepsz\intgteps\partial r^{-(d-1)}\,f_1\,f_2\,dx & = -\pdmone\limepsz\intgteps f_1 \frac{\dr}{r}\, \int\!\hat{x}\,f_2\, \dOme \\[1ex]\label{lim:3} & = -\pdmone\,b\,\limepsz\intgteps f_1 \frac{\dr}{r}\,. \end{align} Here $b$ is the generally nonzero $d$-vector resulting from the $r$-independent, nonsymmetric $\dOme$ integration, and the limit in \eqref{lim:3} is logarithmically divergent.

In the remaining case, $n > \dmone$, there continues to be no $\Loneloc$ derivative for $r^{-n}$, and moreover the second limit in \eqref{lim:2} diverges.

Without loss of generality, we consider only distributions with support at $x = 0$. It is well-known that any such distribution is a finite linear combination of the delta function and its derivatives.

Denote the area of the unit hypersphere by \begin{equation} S_d = \SA\,. \end{equation} Below we reproduce the classic proof of the classic formula for the delta function in $d>0$ dimensions: \begin{alignat}{2}\label{laplace:d} -\sgn(\dmtwo)\laplac \frac{1}{r^{d-2}} & = \abs{\dmtwo}\,S_d\,\delta(x) & & (d \ne 2) \\[1ex]\label{laplace:2} -\laplac\, \ln\frac{1}{r}\, & = S_2\,\delta(x) & \qquad & (d = 2) \end{alignat} See the end of this section for a list with the coefficients spelled out, through four dimensions.

We actually do the proof for arbitrary derivatives of the delta function, for which we use the Schwartz derivative monomial notation: \begin{align} m & = (m_1,...m_d) \qquad |m| = m_1 + ... + m_d \\[1ex] D^m & = \pnderbot{m_1}{x_1}...\pnderbot{m_d}{x_d} \end{align} In the statement and proof of the theorem we keep the pedantic weak derivative notation, to make explicit what quantities are and are not $\Loneloc$. Often one does not care, and any special notation is left out, as in Eqs. \eqref{laplace:d} and \eqref{laplace:2}, with the understanding that the weak derivative is intended. That's part of the charm of generalized functions—they can be differentiated without worry.

*Theorem 2*: For dimension $d > 0$:
\begin{alignat}{2}
-\sgn(\dmtwo)\,\Dw{m}\,\dotp{\partialw}{\partial} \frac{1}{r^{d-2}}
& = \abs{\dmtwo}\,S_d\,\Dw{m}\,\delta(x)
&
(d\, & \ne 2)
\\[1ex]
-\Dw{m}\,\dotp{\partialw}{\partial}\, \ln \frac{1}{r}\,
& = S_d\,\Dw{m}\,\delta(x)
&
\qquad (d\, & = 2)
\end{alignat}

*Proof*: First we need some convenience definitions:
\begin{alignat}{3}\label{csubd:ne2}
h\, & = \sgn(\dmtwo)\frac{1}{r^{d-2}}
& \qquad
\partial h\, & = -\abs{\dmtwo} \frac{x}{r^d}
& \qquad
(d\, & \ne 2)
\\[1ex]
h\, & = \ln\frac{1}{r}
&
\partial h\,& = -\frac{x}{r^2}
&
(d\, & = 2)
\\[1ex]\label{csubd:gt2}
\partial h\, & = - c_d\,\frac{x}{r^d}
&
&
&
(d\, & > 0)
\end{alignat}
Note that in every dimension both $h$ and $\partial h$ are
$\Loneloc$ and polynomial bounded. Furthermore, the divergence
$\dotp{\partial}{\partial h}$ is defined, and vanishes at $x \ne
0$ because
\begin{equation}
\dotp{\partial\,}{\frac{x}{r^n}} = (d - n)\frac{1}{r^n}
\qquad
x \ne 0\,.
\end{equation}
Then we calculate:
\begin{align}\label{defweakderivs}
\bk{\Dw{m}\,\dotp{\partialw}{\partial}\,h}{f}
& = (-1)^{|m|+1}\int\!\dotp{(\partial\,h)}{\partial\,D^m f}\,\dx
\\[1ex]\label{divh}
& = (-1)^{|m|+1} \limepsz
\intgteps\!\dotp{(\partial\,h)}{\partial\,D^m f}\,\dx
\\[1ex]\label{divall}
& = (-1)^{|m|+1} \limepsz\intgteps
\dotp{\partial}{\left[(\partial\,h)\,D^m f\right]}\,dx
\\[1ex]\label{surfint}
& = (-1)^{|m|} \limepsz\inteqeps
\dotp{\hat{x}}{(\partial\,h) D^m f}\,\vareps^{d-1}\dOme
\\[1ex]\label{surfresult}
& = -(-1)^{|m|} D^m f(0)\, c_d\!\int\!\dOme
\\[1ex]\label{finresult}
& = -c_d\,S_d\,\bk{\Dw{m}\,\del}{f}
\end{align}
Here are the steps:

Eq. \eqref{defweakderivs}: The definition of $\Dw{m}\partialw$, and local integrability.

Eq. \eqref{divh}: The dominated convergence property of tempered distributions.

Eq. \eqref{divall}: Integration by parts in \eqref{divh}, and vanishing of the divergence of $\partial h$.

Eq. \eqref{surfint}: The $d$-dimensional divergence theorem.

Eq. \eqref{surfresult}: Convergence in \eqref{surfint} is dominated by $\sup \abs{D^m f}$, the definition of $\partial h$ in \eqref{csubd:gt2}, and the fact that, at $r = \vareps$, \begin{equation} \frac{\dotp{\hat x}{x}}{r^d} = \frac{1}{r^{d-1}} = \frac{1}{\vareps^{d-1}}\,. \end{equation}

Eq. \eqref{finresult}: The definition of $\Dw{m}\del$.

Eq. \eqref{finresult} proves the theorem, by inspection of the definition of $c_d$ in Eqs. (\ref{csubd:ne2}–\ref{csubd:gt2}). $\tqed$

Here is the promised list of spelled-out formulas, with the weak derivative understood: \begin{alignat}{2} -\onderbot{2}{x}\onderbot{m}{x}\left(-\abs{x}\right)\, & = 2\onderbot{m}{x}\del(x) & \qquad (d\, & = 1) \\[1ex] -\laplac D^m\ln{\frac{1}{r}}\, & = 2\pi\, D^m\del(x) & \qquad (d\, & = 2) \\[1ex] -\laplac D^m\!\frac{1}{r}\, & = 4\pi\, D^m\del(x) & \qquad (d\, & = 3) \\[1ex] -\laplac D^m\!\frac{1}{r^2}\, & = 2\pi^2 D^m\del(x) & \qquad (d\, & = 4) \end{alignat}

In one dimension, fractional and integral inverse powers are definable as tempered distributions, which agree with the ordinary functions at $x \ne 0$, by taking derivatives of locally integrable functions: \begin{align} \label{xfracpow} \abs{x}^\alpha & = \frac{1}{(\alp\pt 1)...(\alp\pt n)} \onderbot{n}{x} \left[ (\sgn x)\,\abs{x}^{\alp+n} \right] \qquad \alp + n > -1 \\[1ex]\label{xpow} \frac{1}{\abs{x}^n} & = \frac{1}{(n\mone)!}\onderbot{n}{x} \left[ (-\sgn x)^n \ln \frac{1}{\abs x} \right] \end{align} In Eq. \eqref{xfracpow}, $\alp$ is not allowed to be a negative integer between $-1$ and $-n$.

Equation \eqref{xpow} chooses a particular way of resolving the delta function derivative ambiguity in the definition of inverse integer powers. Namely, it obeys the following distributional identity.

*Lemma 3*: Let $\abs{x}^{-n}$ be defined as a tempered
distribution for integer $n \ge 0$ by Eq. \eqref{xpow}. Then
\begin{equation}
x^n \frac{1}{{\abs x}^n} = (\sgn x)^n\,.
\end{equation}

*Proof*: This is easily proved by using the dominated
convergence property and integration by parts on
\begin{equation}
\BK{\varphi}{x^n \frac{1}{{\abs x}^n}}
= \frac{(-1)^n}{(n\mone)!}\limepsz\int\limits_{\abs x > \vareps}\!
\onderbot{n}{x}\left(x^n\varphi\right)
\sgn x \ln\frac{1}{\abs{x}^n}\,\dx\,. \qed
\end{equation}

It is known that inverse integer powers of the $n$-dimensional
distance $\norm x = r$ are definable as tempered distributions
which coincide with the ordinary functions at $x \ne 0$, by the
*partie finie* construction, which is laid out in Schwartz.
It is also known that every tempered distribution can be
written as a finite sum of derivatives of locally integrable,
even continuous functions. Since the distance is Euclidean
invariant, it is not surprising that such a sum can be written
in terms of invariant derivatives.

Some useful formulas for arbitrary real $\alpha$ at $x \ne 0$: \begin{alignat}{2} \partial\,r\, & = \frac{x}{r^2} & \partial \ln r\, & = \frac{x}{r^2} \\[1ex] \partial \frac{1}{r^\alp}\, & = -\frac{\alp\,x}{r^{\alp+2}} & \partial\left(\frac{1}{r^\alp} \ln \frac{1}{r}\right)\, & = -\frac{x}{r^{\alp+2}}\left(1 + \alp\ln\frac{1}{r}\right) \\[1ex] \dotp{x}{\partial}\frac{1}{r^\alp}\, & = -\frac{\alp}{r^\alp} & \dotp{x}{\partial}\left(\frac{1}{r^\alp}\ln\frac{1}{r}\right)\, & = -\frac{1}{r^\alp}\left(1 + \alp\ln\frac{1}{r}\right) \\[1ex]\label{laplacpow} \laplac\frac{1}{r^\alp}\, & = \frac{\alp\,(\alp \pt 2 \mt d)}{r^{\alp+2}} &\quad\, \laplac\left(\frac{1}{r^\alp}\ln\frac{1}{r}\right)\, & = \frac{1}{r^{\alp+2}}\left[2\alp \pt 2 \mt d + \alp(\alp \pt 2 \mt d)\ln\frac{1}{r}\right] \end{alignat}

The first equation in \eqref{laplacpow} defines a tempered distribution for any fractional power that is not locally integrable by iteration of the laplacian on an integrable power. It also serves for integer inverse powers with $\alp \pt 2 < d$.

The second equation in \eqref{laplacpow} works for $\alp \pt 2 = d$: \begin{equation} \laplac\left(\frac{1}{r^{d-2}}\ln\frac{1}{r}\right) = \frac{d\mt2}{r^d} \end{equation} Now that $r^{-d}$ is defined as a distribution, the laplacian can be iterated on it to get the remaining inverse integer powers, via the first equation in \eqref{laplacpow}.

Again, the definition for inverse integer powers $n \ge d$ picks out a particular resolution of the delta function derivative ambiguity at $x = 0$.

*Lemma 4*: As defined above by derivatives, $\norm{x}^{-n}$
for integer $n \ge d$ obeys the distributional identity:
\begin{alignat}{2}
x\,r^{n-1}\!\frac{1}{r^n}\, & = \frac{x}{r}
& \qquad
& n\; \text{odd}
\\[1ex]
r^n \frac{1}{r^n}\, & = 1
&
& n\; \text{even}
\end{alignat}

*Proof*: Apply the same argument as in the proof of
Lemma 3. $\tqed$