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When I use \substack together with \max, I get the limits UNDER the operator, which is what I want.

However, when I use \substack with \operatorname{argmax}, the limits are subscripts, which is not what I want.

Also, when I use \substack with \max inside of

\begin{cases}
...
\end{cases}

the limits are subscripts.

Is there a way to make the limits be UNDER \operatorname{argmax} inside of a "cases" environment?

My code (with undesirable behavior):

\[
\text{where } \pd{P[r][c]}{A[i][j]} =
\begin{cases}
1 & i,j = \max_{\substack{1 \le i' \le 3 \\ 1 \le j' \le 3} } A[(r-1)\times 2+i'][(c- 
1)\times 2+j']\\[30pt]
0 & \text{ else }
\end{cases}
\]

Code that does put limits UNDER the operator:

\[
P[r][c] = \max_{ \substack{1 \le i \le 3 \\ 1 \le j \le 3} } A[(r-1)\times 
2+i][(c-1)\times 2+j]
\]
4
  • What if you use \operatorname{argmax} \limits ...?
    – Tolaso
    Commented Mar 15, 2020 at 19:02
  • 1
    I tried that, but I didn't know about the * that I needed, as you described in your answer below. Thanks!
    – Joe
    Commented Mar 15, 2020 at 19:37
  • You’re welcome ..
    – Tolaso
    Commented Mar 15, 2020 at 19:38
  • Load mathtools and use dcases, it runs the cells is display math style, normal cases use text style
    – daleif
    Commented Mar 15, 2020 at 20:34

1 Answer 1

4

OK, a slight modification.

\operatorname*{argmax} \limits_{\substack{1 \le i' \le 3 \\ 1 \le j' \le 3}}

I think this solves your problem.

Note: \operatorname{} without the asterisk cannot accept the command \limits.

The command \limits should also fix your other problem.

Also, when I use \substack with \max inside of \begin{cases} \end{cases}, the limits are subscripts.

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