# what's the ordinary derivative of the kronecker delta function?

What's ordinary derivative of the kronecker delta function? I have used "ordinary" in order not to confuse the reader with the covariant derivative. I have tried the following: $$\delta[x-n]=\frac{1}{2\pi}\int_{0}^{2\pi}e^{i(x-n)t}dt$$ but that doesn't work since. $x,n \in \mathbb{Z}$, while I look for the case $x \in \mathbb{R}$

• As a function of a single variable, $f(x) = \delta(x,n)$, which is $1$ when $x=n$ and $0$ otherwise, the function is constant at every point other than $n$ and is discontinuous at $n$. So the derivative is $0$ at every point other than $n$, and not defined at $n$. – Arturo Magidin Apr 22 '12 at 5:30
• The problem with "ordinary" the way you use it is that (IMO) most people will think you mean to specify the derivative you learned in Calc I, as opposed to the distributional derivative. – Excluded and Offended Apr 22 '12 at 5:59

Let's approach this problem using the Heaviside step function, $$H(x) = \begin{cases} 0, & x < 0 \\ 1, & x \ge 0. \end{cases}$$ Notice we choose the convention that $H(0) = 1$.

The Kronecker delta is then $$\delta_{m n} = H(m-n) + H(n-m) - 1.$$ The derivative of the Heaviside step function is the Dirac delta function, $H'(x) = \delta(x)$. Therefore, $$\begin{eqnarray*} \frac{\partial}{\partial m} \delta_{m n} &=& \delta(m-n) - \delta(n-m) \\ &=& 0 \end{eqnarray*}$$ since $\delta(-x) = \delta(x)$.

Addendum: If by ordinary derivative it is meant the derivative in the classical sense then of course the derivative is everywhere zero, except at $m = n$ where it is undefined. In order to define it at $m=n$, which seems to be the spirit of the question, we must go to the distributional derivative in which case we find the derivative is $0$.

• If you're going to talk distributional derivative, there's a shorter proof: as distributions, the Kronecker delta is equal to zero. – Excluded and Offended Apr 22 '12 at 6:00
• @Hurkyl: I am glad that I agree with your shorter proof! – user26872 Apr 22 '12 at 6:11

I take it you are defining $\delta(x,y)$ to be a function of 2 real variables given by $1$ if $x=y$ and $0$ otherwise. I suggest then that you go to the definition of the derivative for functions of two variables and see what happens when you apply it to this function, and then report back to us on your findings.

• Gerry , thanks for your prompt reply . but $n$ needs not to be a variable . – Mohammad Al Jamal Apr 22 '12 at 4:36
• In that case, Arturo's comment would seem to settle things. – Gerry Myerson Apr 22 '12 at 5:44

As Arturo wrote in a comment, the function $f:\mathbb R\to\mathbb R$ such that for all $x\in\mathbb R$ we have $$f(x)=\begin{cases}0, & \text{if x\neq0;} \\1, & \text{if x=0;}\end{cases}$$ is not continuous at $0$. It follows that it does not have a derivative at $0$. On the other hand, a trivial calculation using the definition itself of what a derivative is will show that it has a derivative at all other points, which is zero.

N.B. Of course, the above paragraph is only true if we interpret your question in classical terms. There are several other interpretations we could give to your question, and that would change the answer. We could think that you really are talking about derivatives in the sense of theory of distributions, as in oenamen's answer. You could have something else in mind. Please be more explicit.

May be it is already too late, but I will answer. If I am wrong, please correct me.

Let's have a Kronecker delta via the Fourier transform getting a $Sinc$ function: $$\delta_{k,0} = \frac{1}{a}\int_{-\frac{a}{2}}^{\frac{a}{2}} e^{-\frac{i 2 \pi k x}{a}} \, dx = \frac{\sin (\pi k)}{\pi k}$$

This function looks like: Fourier transform of "1" (Sinc) and Kronecker delta (orange dots)

Calculating the derivative we get: $$\frac{d \delta_{k,0}}{dk} = \frac{\cos (\pi k)}{k}-\frac{\sin (\pi k)}{\pi k^2} = \frac{\cos (\pi k)}{k}$$ for $k \in \mathbb Z$

On a plot it looks like Derivative of Sinc and Kronecker delta