答案1
你要
- “arg” 运算符对后续操作进行操作;
- 极限为“phi in W”的“inf”算子;
- 计算其最小值的集合。
做就是了
\arg \inf_{\phi\in W} \{ I(\phi) \}
如果您在显示中。如果您在内联公式中
\arg \inf\limits_{\phi\in W} \{ I(\phi) \}
但不建议这样做,因为它会破坏行距。
这是一个完整的例子,也展示了间距问题。
\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}
答案2
这里有两种可能的解决方案。第一种,\phi\in W
位于“inf”下方居中 - 类似于屏幕截图中显示的 - 并采用一点“技巧”来保持“arg”和“inf”之间的空间相当紧密。第二种解决方案,\phi\in W
位于“arg inf”下方居中。
\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}