I have the expression

\sup_a \left\{ \inf_b f(a,b) \right\}


which renders as

If the braces \left\{ and \right\} were not part of the expression, I could use \adjustlimits to ensure that the arguments in the supremum and infimum shared the same vertical position.

Is there a way to achieve the desired effect in this case?

• just use \{ and \} not \left\{ and \right\} the over-sized braces are not helping here Dec 11 '16 at 21:01
• @DavidCarlisle: Thanks for the comment. If you're suggesting \adjustlimits \sup_a \{ \inf_b f(a,b) \}, this also does not compile. If you're suggesting removing \adjustlimits altogether, that does not produce the desired effect.
– user31162
Dec 11 '16 at 21:07
• No the {} comment was more of a side issue. I'm sure we have questions/answers on equalising the limits on operators with and without a descender, I'll see if I can find... Dec 11 '16 at 21:47

There's no hope of making \adjustlimits to work in this case.

The problem is twofold: “sup” has a descender, “inf” doesn't; “b” has an ascender, “a” hasn't.

So I give two solutions: in the first one I use a special “inf” with an artificial descender; in the second solution, I use a special “sup” with a smashed “p”.

\documentclass{article}

\usepackage{amsmath}

\makeatletter
\DeclareMathOperator*{\infd}{inf\vphantom{\operator@font p}} % inf with descender
\DeclareMathOperator*{\supx}{su\smash{\operator@font p}} % sup without descender
\makeatother

\begin{document}

$\sup_{a\vphantom{b}}\bigl\{\infd_{b} f(a,b)\bigr\}$

$\supx_{a\vphantom{b}}\bigl\{\inf_{b} f(a,b)\bigr\}$

\end{document}


• oh yes b isn't the same shape as a I forgot that:-) Dec 11 '16 at 22:44