Spacing around longer function argument's parentheses

Question

The line

i(x,y,z,t) = \delta(x-\ell_x(z)) \delta(y-\ell_y(z)) \hat{\imath}(\xi(z),t)

produces (for my eyes, at least) too little space between the successive function calls. It is difficult to visually distinguish the three calls. I guess the underlying reason is that the expressions that serve as function arguments are relatively long. For comparison,

i(x,y,z,t) = \delta(x) \delta(y) \hat{\imath}(\xi(z),t)

already looks a lot better in my opinion.

I know that I could

1. Manually adjust spacing using \, etc., including the option to define a macro that does this automatically.
2. Insert \cdot as a means of adding seperating space.

However, my question is: Is there a canonical way (e.g. a package) of dealing with this problem? I imagine I'm not the first one to encounter this problem... still, I wasn't able to find useful, general solutions to the problem.

A possible solution might be something along the lines of

i(x,y,t) = \delta \funcargs{x-\ell_x(z)} \delta \funcargs{y-\ell_y(z)} \hat{\imath} \funcargs{\xi(z)}{t}

Note that I find the use of \cdot to be unsatisfying since it introduces inconsistency in the document (sometimes having the dot and sometimes not, with the meaning being the same).

Answer 1

First, let me give a paraphrasing from The TeXbook showing the amounts of spaces between different types of math atoms:

where

• Ord includes things like a, \theta, and things inside \mathord or just plain {}'s;
• Op includes things like \sin, \gcd, and things inside \mathop;
• Bin includes things like \pm, \otimes, and things inside \mathbin;
• Rel includes things like \leq, \mid, and things inside \mathrel;
• Open includes things like \langle, (, and things inside \mathopen;
• Close includes things like \rangle, ), and things inside \mathclose;
• Punct includes things like ,, ., and things inside \mathpunct;
• Inner includes things like \cdots and things inside \mathinner.

Here is an example plain-tex document illustrating a couple of different ways:

\let\funarg\mathinner
\def\funop#1{\mathop{{}#1}}
$$ i(x,y,z,t) = \delta(x-\ell_x(z)) \delta(y-\ell_y(z)) \hat\imath(\xi(z),t), $$
$$ i\funarg{x,y,z,t} = \delta\funarg{(x-\ell_x(z))} \delta\funarg{(y-\ell_y(z))} \hat\imath\funarg{(\xi(z),t)}, $$
$$ \funarg{i(x,y,z,t)} = \funarg{\delta(x-\ell_x(z))} \funarg{\delta(y-\ell_y(z))} \funarg{\hat\imath(\xi(z),t)}, $$
$$ \funop i(x,y,z,t) = \funop\delta(x-\ell_x(z)) \funop\delta(y-\ell_y(z)) \funop{\hat\imath}(\xi(z),t), $$
\bye

So as you can see, you probably want the last option of using a \mathop atom for the function "name".

With LaTeX, this could be defined as follows (the second row below shows the output of your original code):

\documentclass{article}
\newcommand\funop[1]{\mathop{{}#1}}
\begin{document}
$i(x,y,z,t) = \funop\delta(x-\ell_x(z)) \funop\delta(y-\ell_y(z)) \funop{\hat\imath}(\xi(z),t)$

vs.

$i(x,y,z,t) = \delta(x-\ell_x(z)) \delta(y-\ell_y(z)) \hat{\imath}(\xi(z),t)$
\end{document}

(The reason I didn't include the parenthesis in the command is that I felt it would be too much effort to measure the arguments to decide on the size of the fences. I think it's simpler to just manually define the height with \bigl( etc., and also it's nicer to see the parens in the manuscript instead of {}'s)

Answer 2

Inserting a thinspace (\,) between the multiplicative factors should work. Inserting \cdot in these locations should work too. You could also combine the thinspaces with making the outer parentheses a tad bigger than the inner ones (by using \bigl( and \bigr)). The three possibilities are illustrated below:

\documentclass{article}
\begin{document}
$i(x,y,z,t) = \delta(x-\ell_x(z)) \, \delta(y-\ell_y(z)) \, \hat{\imath}(\xi(z),t)$

$i(x,y,z,t) = \delta(x-\ell_x(z)) \cdot \delta(y-\ell_y(z)) \cdot \hat{\imath}(\xi(z),t)$

$i(x,y,z,t) = \delta\bigl(x-\ell_x(z)\bigr) \, \delta\bigl(y-\ell_y(z)\bigr) \, 
    \hat{\imath}\bigl(\xi(z),t)\bigr)$
\end{document}

One can automate the third method by creating a macro such as

\def\myfunc#1{\bigl(#1\bigr)\,}

and then typing

$i(x,y,z,t) = \delta\myfunc{x-\ell_x(z)} \delta\myfunc{y-\ell_y(z)} 
    \hat{\imath}\myfunc{\xi(z),t)}$