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I'm using \DeclarePairedDelimiter from the mathtools package to create a paired delimiter \DeclarePairedDelimiter{\norm}{\lVert}{\rVert} but I want the same baseline for the subscript with and without using the starred version of the paired delimiter.

The first term in the picture shows the undesired baseline without using the starred version. The second term shows the desired baseline, and the third term shows a workaround to obtain the desired baseline with the unstarred version.

enter image description here

However, I would like to prevent adding the \mathopen{} command around every call. Is it possible to add the \mathopen{} command to the \DeclarePairedDelimiter command?

MWE

\documentclass{article}
\usepackage{mathtools}
\DeclarePairedDelimiter{\norm}{\lVert}{\rVert}
\begin{document}
$\norm{a}_{\infty}$, $\norm*{b}_{\infty}$, $\mathopen{\norm{c}}_{\infty}$
\end{document}
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    Can you explain the use case of \norm*{a}_{\infty} over \norm{a}_{\infty}? (To me, employing \norm*{a}_{\infty} over \norm{a}_{\infty} flies in the face of good math typography.) Incidentally, \mathopen{\norm{a}}_{\infty} is needlessly complicated, as {\norm{a}}_{\infty achieves the exact same result.
    – Mico
    Apr 5, 2021 at 10:14
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    Related, I think...tex.stackexchange.com/questions/364308/… Apr 5, 2021 at 10:23
  • @Mico \norm{a} does not automatically scale the delimiter. However, using \norm*{} for inline math, the automatic scaling sometimes produces vertical bars that are too large. Thank you for the information that \mathopen{\norm{a}}_{\infty} achieves the same as {\norm{a}}_{\infty} @StevenB.Segletes You are right. I didn't find this question before. Thank you!
    – brnk
    Apr 5, 2021 at 10:30
  • I know that \norm{a} does not scale the delimiter. My question was why one would use \norm*{a} when there's no conceivable need to perform scaling.
    – Mico
    Apr 5, 2021 at 10:34
  • @Mico I just used \norm*{a} as an example to show the desired subcript baseline. If \norm{a} would yield the same baseline, there wouldn't be a reason to use \norm*{a} instead.
    – brnk
    Apr 5, 2021 at 10:36

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