I have a image $U_{m \times n}:\Omega \to \mathbb R^2$, the output $P$ can be define as $$P=\mu J_{m \times n} - U$$ where $\mu = \max \{ u_{ij} : 1 \leq i \leq m, 1 \leq j \leq n\}$, $J_{m \times n}$ be the $m \times n$ matrix whose $i, j$th component is $1$: that is, the all-ones matrix. (This notation isn't quite standard, but it's as close to standard as I know. $J$ is often the all-ones matrix)

However, it is so many sentences for expression the above equation. Do we have more short and standard way to represent it? As my found, the $J_{m \times n}$ can be expressed by indicator function such as

$$P=\max (U)\times 1_{\Omega}-U$$

where $1_{\Omega}$ is Indicator function

Does it equivalent with original meaning? If not, please give me a standard and common expression in image processing. Thanks

  • $\begingroup$ I would have defined $U_{m\times n}: [1,\ldots,m]\times [1,\ldots,n] \to \mathbb{R}$ instead, if you really want to have a professional notation. Unless your image is 2-valued, which would pose another challenge in defining a scalar $\max$. $\endgroup$ – Laurent Duval Nov 15 '15 at 11:11

I think the notation $J_{m,n}$ is pretty standard for an all-ones matrix. So the only possible clarification would be to dispense with $\mu$ and directly specify what it actually is:


where $||.||_{\max}$ denotes the max norm of a matrix.

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  • $\begingroup$ Thank Matt L. How about indicator function instead of $J$. Based on your comment, it will be $$P=||U||_{\max}1_{\Omega}-U$$ $\endgroup$ – Jame Nov 1 '15 at 12:29
  • $\begingroup$ @user8430: I'm not sure if you could use it to specify the all-ones matrix. Furthermore, you would need to specify what the subset $\Omega$ is. I believe that using $J_{m,n}$ is much clearer, and probably the indicator function can't even actually be used in that way. $\endgroup$ – Matt L. Nov 1 '15 at 12:37
  • $\begingroup$ $\Omega$ is image domain, then $1_{\Omega}$ means 1 values in whole image domain. Actually, I want to reduce notation in that form. Thank for your max norm notation. For image processing, people do not like write form as size $J_{m \times n}$ $\endgroup$ – Jame Nov 1 '15 at 12:40
  • $\begingroup$ The $J_{n\times n}$ is quite standard for the counter-identity or exchange matrix too, i.e. with ones on the second diagonal. $\endgroup$ – Laurent Duval Nov 1 '15 at 14:31
  • $\begingroup$ @LaurentDuval: Do you think my $1_{\Omega}$ can represent matrix of one?However, I found that $1_{\Omega}$ likes the function that map a domain in real value. $\endgroup$ – Jame Nov 1 '15 at 14:44

I would propose \mathbb{1} from the \usepackage{bbold} (or \mathds{1}, see details and options here), with outputs a $1$ with a double bar, like for $\mathbb{R}$ (I cannot have it displayed on SE). I would stick to a simple $\max $ or $\sup$, as it is not a norm without the absolute value. Plus, image values are taken in $\mathbb{R}^2$, as in some astronomical, seismic or processed images. So, something like:

$\max_{i,j} \{ A_{i,j}\} . \mathbb{1}-A$

double-bar indicator notation

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  • $\begingroup$ I know of the max norm of a matrix (as I've used it in my answer), but I'm not sure if $\max(A)$ is common notation. Wasn't that actually the point of your comment under my answer? $\endgroup$ – Matt L. Nov 1 '15 at 15:45
  • $\begingroup$ The max of all elements can be negative for instance. Hence it is not a norm. I beleive $\max_{i,j}$ is probably more readable. In my answer I wanted to emphasize on the double-bar one. $\endgroup$ – Laurent Duval Nov 1 '15 at 15:50
  • $\begingroup$ It is a norm, because it's the maximum of the absolute values of the matrix entries (check the link). This should be OK because we're talking about images (I know in the OP it was just the max, not max |.|, but it should make no difference for the application). $\endgroup$ – Matt L. Nov 1 '15 at 16:00
  • $\begingroup$ $\mathbb{1}$ doesn't work? It seems to for me: just use \mathbb{1} (in \$ signs). $\endgroup$ – Peter K. Nov 3 '15 at 14:21
  • $\begingroup$ It does not appear per se in my browsers (recent Firefox and Chromium). I do not understand why... yet $\endgroup$ – Laurent Duval Nov 3 '15 at 14:27

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