2
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What do the curly brackets mean in this context?

$$f(x) = 2 \cdot x \cdot 1_{x>0}$$

Is this a condition?

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  • $\begingroup$ More context would be helpful here - without a bit more information it's hard to discern what's going on. One good possibility is that the characteristic function - that is, the function that takes the value $1$ at arguments in the set $A$ and $0$ at arguments not in the set $A$, is often written as $\mathbf{1}_A$. Here, that would mean the function $g(x)$ defined as $g(x) = 1, x\gt 0; g(x) = 0, x\leq 0 $. $\endgroup$ – Steven Stadnicki Feb 14 '12 at 2:10
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    $\begingroup$ As mentioned in the popup that is supposed to appear when you tagged this notation, you are supposed to have included where you saw this notation. We're answerers, not mind-readers. $\endgroup$ – J. M. is a poor mathematician Feb 14 '12 at 2:18
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    $\begingroup$ @Soo : Since your title mentions "sub curly brackets", I wonder if you meant $2\cdot x\cdot1_{\{x>0\}}$? $\endgroup$ – Michael Hardy Feb 14 '12 at 3:26
8
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The $\Bbb 1_{x>0}$ is the indicator function $^1$ that indicates if $x$ is greater than $0$ or not. $$1_{x>0}=\begin{cases} 1, ~~x>0\\0, ~~x \le 0\end{cases}$$

This makes your function into,

$$f(x)=\begin{cases}2x,~~x>0\\0, ~~x \le0 \end{cases}$$

Footnotes:

$^1$(also called the Characteristic function, denoted by $\chi_A$ is the function that takes $1$ when $x \in A$ and $0$ otherwise.)

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  • $\begingroup$ It's a convenient function, though I personally prefer Iverson brackets myself... $\endgroup$ – J. M. is a poor mathematician Feb 14 '12 at 2:18
  • $\begingroup$ I have never heard of Iverson Brackets until you mention it to me. @J.M. Thank You. Wikipedia had to rescue me! $\endgroup$ – user21436 Feb 14 '12 at 2:21
  • $\begingroup$ What a weird notational convention; I never saw this before. I suppose one could also write $0_{x\leq0}$ instead of $1_{x>0}$ :-). Long live Iverson! $\endgroup$ – Marc van Leeuwen Feb 14 '12 at 11:10
  • $\begingroup$ Dear Professor @MarcvanLeeuwen, This is weird but, I think you'd agree not as weirder as Iverson Brackets :) On a lighter note, I happen to have heard of a Saint from Poitiers, St. Montfort. $\endgroup$ – user21436 Feb 14 '12 at 11:15
  • $\begingroup$ Macaulay brackets anybody? $\endgroup$ – Severo Raz Aug 15 '18 at 9:35

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