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Is there a traditional notation or name of this function:

$$ \epsilon (x) = \begin{cases} 0 & \textrm{ if }x = 0 \\ 1 & \textrm{ if }x \neq 0 \end{cases} $$

I know one can use Indicator function $1_{A}(x)$ for $A = \mathbb{R} - \{0\}$, but I still want to know if there is a traditional, simpler, notation for it.

Alessio K
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user565739
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    Not that I know of. The simplest way of writing is either $1_{\mathbb R\setminus {0}}$ or $1 - 1_{{0}}$. Maybe even $1-\delta_0$, but that is already slightly ambiguous. – 5xum May 05 '15 at 07:28
  • Well... $1_{\Re \backslash { 0 } }$ is not a function. Moreover, $1 - \delta_0$ cannot be correct because the value of $\delta_0$ is infinite at $x=0$. This is a bizarre function, and I suspect the question poser isn't quite clear on what he or she needs. One can write the function as a limit, but again, that seems rather awkward. – David G. Stork May 05 '15 at 07:31
  • I don't think there's a traditional name for this function, although I'm sure you could find it in the literature somewhere. I think the indicator function notation is fine. Or if you're writing something where you will use this a lot and that will become cumbersome, it's fine to just call it $\epsilon$ as long as you define it beforehand. – Jair Taylor May 05 '15 at 07:34
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    @David Yes, it is a function. And no, it does not take the value infinity. – Tobias Kildetoft May 05 '15 at 07:42
  • @DavidG.Stork Ohh, and by the first "it" I meant $1_{\mathbb{R}\setminus{0}}$ and for the second "it" I meant the function $\delta_0$. – Tobias Kildetoft May 05 '15 at 07:50
  • @TobiasKildetoft If $1_{\Re \backslash { 0 }}$ is a function (as you say), what is its argument? And I referred explicitly to $\delta_0$ (which I interpret to mean $\delta (x)$, as we must for the real line), which is in fact infinite at $x=0$, since its integral is 1.0: $\int \delta (x) dx = 1$. (We cannot use an indicator function for integers here.) – David G. Stork May 05 '15 at 07:53
  • @DavidG.Stork Its argument is not written when we consider it as a function (rather than considering one of its values). And it was quite clear from context that $\delta_0$ was meant as the indicator function for $0$ (no reason why we cannot use the same notation for arbitrary numbers). – Tobias Kildetoft May 05 '15 at 07:56
  • @TobiasKildetoft Could you please then write out explicitly your first function and your second with an argument (e.g., $x$), just as we must be able to do with every function (and I do and all other respondents below do)? Thanks. – David G. Stork May 05 '15 at 07:59
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    @DavidG.Stork The first one is precisely defined like the function the OP is asking about. In general it is a common notation to use $1_A$ as the indicator function for the set $A$, meaning that $1_A(x)$ is $1$ if $x\in A$ and $0$ otherwise. $\delta_0$ is a common shorthand for the indicator function for the set ${0}$. – Tobias Kildetoft May 05 '15 at 08:01
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    @DavidG.Stork why would you write it as a limit when there is a simple explicit definition as an indicator function? And why do you doubt indicator functions are functions? – Sasho Nikolov May 06 '15 at 04:39
  • @SashoNikolov "In many applications, the Dirac delta is regarded as a kind of limit (a weak limit) of a sequence of functions having a tall spike at the origin." (http://en.wikipedia.org/wiki/Dirac_delta_function) – David G. Stork May 06 '15 at 14:47
  • @DavidG.Stork Yes, because the Dirac delta is not a function, it's a distribution (in the Schwartz sense), The simple indicator function this question is about is a perfectly ordinary function. I think you may be confusing the Dirac delta and the Kronecker delta function. – Sasho Nikolov May 06 '15 at 17:03

7 Answers7

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The Kronecker delta function $\delta_{ij}$ is $1$ when $i=j$ and $0$ otherwise, so $\epsilon(x)=1-\delta_{0,x}$.

Kyle Miller
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In the Iverson bracket notation you could write $$\epsilon(x)=[x\ne0].$$

bof
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You can write $\epsilon(x)=\mathrm{sgn}^2(x)$ where $\mathrm{sgn}$ is the signum function.

Chris Culter
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1

The Heaviside step function is defined as

$$ H (x) = \begin{cases} 0 & \textrm{ if }x < 0 \\ 1 & \textrm{ if }x => 0 \end{cases} $$

Clearly

$$ H (-x) = \begin{cases} 1 & \textrm{ if }x <= 0 \\ 0 & \textrm{ if }x > 0 \end{cases} $$

Summing gives:

$$ H(x)+ H (-x) = \begin{cases} 1 & \textrm{ if }x < 0 \\ 2 & \textrm{ if }x = 0 \\ 1 & \textrm{ if }x >0 \end{cases} $$

Subtract 1

$$ H(x)+ H (-x)-1 = \begin{cases} 0 & \textrm{ if }x < 0 \\ 1 & \textrm{ if }x = 0 \\ 0 & \textrm{ if }x >0 \end{cases} $$

tomi
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This funtion is also known as the "discrete norm", which is to say an absolute value (in the sense of ring theory) whose metric induces the discrete topology. The discrete metric is $d(x,y)=1$ if $x\ne y$ and $d(x,x)=0$, while the discrete norm on any domain $X$ is $f(x)=1$ if $x\in X\setminus\{0\}$ and $f(0)=0$. For the case of $X=\Bbb R$, this reduces to your function.

I hope this answers the question posed by some of why this would be a natural or useful function, rather than a simple modification of some other function like $\delta_{x,0}$ which is useful in other contexts.

Mario Carneiro
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You could use the Heaviside step function $H(x)$. Your function is

$\epsilon(x)=|H(x)-H(-x)|$

Martigan
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  • True, but see my post which avoids using absolute function. – tomi May 05 '15 at 07:43
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    How is this better than writing out the definition? – Tobias Kildetoft May 05 '15 at 08:13
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    This isn't "traditional notation" (nor simpler) like the question asked, it seems more like obfuscation than anything.... unless this definition from the Heaviside function allows one to derive nontrivial information about the function, it seems pointless. – Thomas May 05 '15 at 08:23
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Here's one possible definition, though extremely dependent on how one defines the value $0^0$.

If you happen to subscribe to the belief that $0^0=0$ (disclaimer, I personally define $0^0=1$), then $$\epsilon(x)=x^0$$

Hayden
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