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The following code does not work:

\begin{align*}
 \left \| x \mapsto \sup_{(y, t) \in \Gamma_x^{(A, a)}(\gamma)} & \int_{\R^d} M_{t^2}(y, z) 1_{N_\tau^c}(y, z) |f(z)| \, \d z \right \|_{L^1(\gamma)}\\
 &= \int_{\R^d} \left | \sup_{(y, t) \in \Gamma_x^{(A, a)}(\gamma)} \int_{\R^d} M_{t^2}(y, z) 1_{N_\tau^c}(y, z) |f(z)| \, \d z \right | \, \d\gamma(x)
\end{align*}

\R is defined as \newcommand{\R}{\mathbf R}.

Probably \left does not work across &'s. How do I fix this? Do I have to give the size manually? Placing \right . & \left . does not give the desired effect because they will not necessarily both be the same size.

Manually setting the size works, but I'm not sure if \left works "stepless" in the sense that it can get any size, where the \Bigg and so on are fixed. Further, it would be nice if it would work automatically so that I don't have to get back all the time to check if my delimiters are large enough.

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Please add the definition of the command \R (or the package that creates this) to your example. –  Ian Thompson Jul 20 '11 at 11:20
    
@Ian: I have done that. –  Jonas Teuwen Jul 20 '11 at 11:22
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2 Answers

up vote 3 down vote accepted

Up to the delimiters produced by \Bigg, any size produced by \left and \right can also be produced by \big, \bigg, etc. so you can manually set the size in most cases (\left and \right can create delimiters bigger than \Bigg, for use with matrices and the like). However, the example that you have provided doesn't appear to need alignment at all; you can just delete the ampersands and use multline* instead of align*.

If you have other examples where corresponding left and right delimiters appear on different lines, then you might try the breqn package.

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Thank you. The example I gave doesn't need align but I have multiple lines after that which I have deleted for the example. –  Jonas Teuwen Jul 20 '11 at 11:24
    
Okay, now I see the modification. How do I get larger delimiters than \Bigg produces if there is an & in between? The breqn package seems overkill and probably doesn't work well with my current document. –  Jonas Teuwen Jul 20 '11 at 13:03
    
@Jonas --- I can't think of a good way to get the extra large delimiters in the situation you describe, but there are other users here who will be able to help. Ask a separate question about this to attract their attention. –  Ian Thompson Jul 20 '11 at 14:57
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(Note: This answer was edited extensively by the original author, mostly to do a better job of explaining the use of struts in math formulas.)
I think a flexible (and nearly automatic) way to equate the heights of the math delimiters across the & hurdle uses the TeX command \vphantom. The \vphantom command creates a strut -- a TeX "box" with the same height/depth as its argument but with zero width, hence the box is "invisible" -- of the same height/depth as the tallest item on either side of the & divide. You then place the strut on the side of the & sign that doesn't contain this tall object; as a result, the math delimiters at either end of the formula will have the same overall heights.

The code below uses the first equation of your align example. (The second equation doesn't have this issue, so it needn't be reproduced here.) The first line in the align environment below (more or less your original code) shows the undesirable effect of not using a strut. The second line shows how the insertion of the strut on the right-hand side of the & sign serves to equate the heights of the two math delimiters. Two comments are in order: First, to separate the two parts of the formula visually, I've replaced the & sign with &\ddag; in practice, one would obviously not include the \ddag symbol. Second, to make the location of the strut in the code easily visible, I've coded the strut as an explicit macro called \hugestrut; in practice, one would just write \vphantom{...}.

The only part that's not completely automatic is that, in order to create the strut, one has to know which element in the formula has the greatest height/depth. However, finding this out should be rather easy from a visual inspection of the two sides of the & hurdle.

\documentclass{article}
\usepackage{amsmath,amssymb}
\newcommand{\R}{\mathbb{R}}
\begin{document}
%% \hugestrut is a strut, i.e., an invisible zero-width box
\newcommand{\hugestrut}{\vphantom{\sup_{(y, t) \in \Gamma_x^{(A, a)}(\gamma)}}}

\begin{align*}
%% first, without the strut
\left \|\, x \mapsto \sup_{(y, t) \in \Gamma_x^{(A, a)}(\gamma)} \right.
&\ddag \left. 
\int_{\R^d} M_{t^2}(y,z) 1_{N_\tau^c}(y,z) |f(z)| 
\,\mathrm{d} z \,\right \|_{L^1(\gamma)}\\[2em]
%% second, with the strut placed on the right-hand side of the & term
\left \|\, x \mapsto \sup_{(y, t) \in \Gamma_x^{(A, a)}(\gamma)} \right.
&\ddag \left. \hugestrut
\int_{\R^d} M_{t^2}(y,z) 1_{N_\tau^c}(y,z) |f(z)| 
\,\mathrm{d} z \,\right \|_{L^1(\gamma)}
\end{align*}  
\end{document}
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‍@Mico: I've marked your inline code with backticks `. –  Hendrik Vogt Jul 20 '11 at 12:42
    
Thanks, Hendrik! I'm still fairly new to this group, so I'm glad to get these pieces of advice. –  Mico Jul 20 '11 at 13:43
    
@Mico: Thanks. I don't understand yet how it works, but I'll study it. –  Jonas Teuwen Jul 20 '11 at 15:00
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