I've tried \operatornamewithlimits{\forall}, which looks ugly.
(Basically it is similar look as using \operatornamewithlimits{\cap}
instead of \bigcap.
Thanks in advance!
Martin
After a little bit more googling I found this
http://code.google.com/p/uwisdom/source/browse/Tex.wiki?repo=wiki
suggesting
In TeX/LaTeX, to create a large "forall" symbol (which ordinarily is
no
larger in display mode than in any other math mode), do something like
{{{
\newcommand{\bigforall}[2]{{{\raisebox{-6pt}{\mbox{\Large$\forall$}
$#1$}}\atop{\scriptstyle #2}}}
}}}
With the above definition
$$\bigforall{}{\varepsilon>0} f(x)\le 1+\varepsilon$$
does what I wanted.
Martin
You'll get a better behaviour redefining \forall as a math operator.
amsopn make it easy:
\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage{amsmath,amsopn}
\usepackage{relsize}
\makeatletter
\let\forall@mathchar \forall
\let\forall \relax \DeclareMathOperator*\forall{\forall@mathchar}
\DeclareMathOperator*\varforall{\mathchoice{\mathlarger\forall@mathchar}{\forall@mathchar}{\forall@mathchar}{\forall@mathchar}}
\begin{document}
$\forall x \in \frac A{\mathcal R}$
$$\forall x \in \frac A{\mathcal R}$$
$\forall_{\substack{x>0 \\ \varepsilon>0}} x \in \frac A{\mathcal R}$
$$\forall_{\substack{x>0 \\ \varepsilon>0}} x \in \frac A{\mathcal R}$$
$\varforall_{\substack{x>0 \\ \varepsilon>0}} x \in \frac A{\mathcal R}$
$$\varforall_{\substack{x>0 \\ \varepsilon>0}} x \in \frac A{\mathcal R}$$
\end{document}\endinput