emmmmm请问怎么能让这个(by lemma )靠右呢
\begin{align*}
\| S+T \|^{2}_{A} &= \omega ^{2}_{A}(S+T) \\
&\leq \sup_{\substack{\|x\|_{A}=1}}(|\langle (S+iT)x,x \rangle _{A} |^{2}\\
\text{(by Lemma)} \\
&\leq \sup_{\substack{\|x\|_{A}=1}}(|\langle (S+iT)x,x \rangle _{A} |^{2}\\
\end{align*}
看看下面的代码:
\documentclass{article}
\usepackage{mathtools}
\usepackage{eqparbox}
\usepackage[showframe]{geometry}
\begin{document}
\begin{flalign}
& & a &= bbbbbb &\nonumber \\
& & cccc & = dd & \text{(using Eq. 1)} & \nonumber \\
& & e &= f & \text{(using Thm. 2)}
\end{flalign}
\begin{flalign}
& & a &= bbbbbb \nonumber \\
& & cccc & = dd & \eqparbox{C}{(using Eq. 1)} & \nonumber \\
& & e &= f & \eqparbox{C}{(using Thm. 2)}
\end{flalign}
\end{document}
除去by lemma
后,如果只有两列对齐的情况下,在\text{(by Lemma)}
前简单地添加&&
即可。
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{align*}
c^2 &= a^2 + b^2 \\
&= a^2 + b^2 \\
&&\text{(by Lemma)}\\
&= a^2 + b^2 \\
\end{align*}
\end{document}
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{subequations}
\begin{align*}
\| S+T \|^{2}_{A} & = \omega ^{2}_{A}(S+T) \\
& \leq \sup_{\substack{\|x\|_{A}=1}}(|\langle (S+iT)x,x \rangle _{A} |^{2} \\
\shortintertext{\hfill(by Lemma)}
& \leq \sup_{\substack{\|x\|_{A}=1}}(|\langle (S+iT)x,x \rangle _{A} |^{2} \\
\end{align*}
\end{subequations}
\end{document}
\documentclass{article}
\usepackage{calc}
\usepackage{amsmath}
\begin{document}
\begin{align*}
M ={} & a_1 + a_2 + a_3 + a_4 + a_5 + a_6 \\
& \makebox[\widthof{$a_1 + a_2 + a_3 + a_4 + a_5 + a_6$}][r]{(by lemma)} \\
\leq{} & b_1 + b_2 + b_3 + b_4 + b_5
\end{align*}
\end{document}
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{alignat*}{3}
&& \| S+T \|^{2}_{A}& = \omega ^{2}_{A}(S+T) && \\
&& & \leq \sup_{\|x\|_{A}=1}(|\langle (S+iT)x,x \rangle _{A} |^{2} && \\
&& & & \text{\llap{(by Lemma)}} & \\
&& & \leq \sup_{\|x\|_{A}=1}(|\langle (S+iT)x,x \rangle _{A} |^{2} && \\
\end{alignat*}
\end{document}
请问,如果想让(by lemma)和上一行的公式右对齐呢?