3

I am specifically trying to write this equation:

enter image description here

5
  • Welcome to TeX SX! And where is the problem?
    – Bernard
    May 3, 2018 at 20:29
  • I am having trouble with the RHS. for that, I am trying this \arg_{\phi\in W}{inf I(\phi)} but that's not working. Probably, there's something wrong in how I write it, but I cannot find a suitable command.
    – Schneider
    May 3, 2018 at 20:35
  • Please advise if \phi\in W should be centered below "inf" or below "arg inf".
    – Mico
    May 3, 2018 at 20:38
  • It should be below "inf". Thanks.
    – Schneider
    May 3, 2018 at 20:59
  • In my opinion, it doesn't make sense for \phi \in W to be centered below "inf". That suggests that \arg is an operator being applied to \inf ..., which of course is not and cannot be the case. Centering below \arginf seems more sensible—and is also, in my experience, far more common.
    – wchargin
    May 4, 2018 at 6:38

3 Answers 3

9

You want

  1. the “arg” operator to act on what follows;
  2. the “inf” operator with limit “phi in W”;
  3. the set which you compute the infimum of.

Just do

\arg \inf_{\phi\in W} \{ I(\phi) \}

if you are in a display. If you're in an inline formula

\arg \inf\limits_{\phi\in W} \{ I(\phi) \}

but this is not recommended as it spoils line spacing.

Here's a full example also showing the spacing issues.

\documentclass{article}

\begin{document}

This is the requested formula in a display
\[
\phi^{\star} = \arg \inf_{\phi\in W} \{ I(\phi) \}
\]
and you can look at what happens if you have the formula inline, with
the limits set below $\phi^{\star}=\arg \inf\limits_{\phi\in W} \{ I(\phi) \}$
where you see that the spacing between lines is uneven and bad. Instead, type
it in normally (some text added here just to produce more lines)
like $\phi^{\star}=\arg \inf_{\phi\in W} \{ I(\phi) \}$
and the spacing between lines will be even and smooth. Some other text
added here to fill some more lines.

\end{document}

enter image description here

6

Here are two possible solutions. In the first, \phi\in W is centered below "inf" -- similar to what's shown in your screenshot -- with a little "trick" to keep the space between "arg" and "inf" fairly tight. In the second solution, \phi\in W is centered below "arg inf".

enter image description here

\documentclass{article}
\usepackage{mathtools} %for '\smashoperator' macro
\DeclareMathOperator*{\arginf}{\arg\inf}
\begin{document}
\[
\phi^{\star} = \arg\smashoperator[l]{\inf_{\phi\in W}} \{I(\phi)\}
\]

\[
\phi^{\star} = \arginf_{\phi\in W} \{I(\phi)\}
\]
\end{document}
0
5

Here's a solution with mathtools

\documentclass{article}

\usepackage{mathtools} %loads amsmath
\DeclarePairedDelimiter{\set}\{\}

\begin{document}

\[ \phi^{\boldsymbol{\star} }= \arg\inf_{\mathclap{\phi\in W}}\,\set[\big]{I(\phi)}\]%

\end{document} 

enter image description here

2
  • 1
    @Sebastiano: I don't really understand: what should be smaller?
    – Bernard
    May 3, 2018 at 21:06
  • 1
    \phi\in W for my opinion is too big and go out the command \inf.
    – Sebastiano
    May 3, 2018 at 21:23

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