Uniform convergence symbol (arrow on top of arrow, same size)

Question

I am looking for a symbol that looks like two arrows of the same size, one on top of the other with (possibly) a letter on top, this is my paint sketch of it:

I've tried something like

\xrightarrow{\xrightarrow{n}}

and similar combinations but it always makes one arrow smaller.

I've seen this symbol used to say that a sequence of functions converges uniformly to another function.

You can use the solution here too: https://tex.stackexchange.com/questions/114233/no-xrightrightarrowab — koleygr, Jul 28 '17 at 03:42
The solution there works, but it doesn't look good when inline. Thanks anyway. — Cronenberg, Jul 28 '17 at 03:56

score 6 · Answer 1 · edited Feb 07 '22 at 21:40

6

With the amssymb package you can just use \rightrightarrows:

\documentclass[a4paper,12pt]{article}
\usepackage{mathtools,amssymb}
\begin{document}
\[A\overset{n}{\rightrightarrows}B\]
\end{document}

edited Feb 07 '22 at 21:40

Sebastiano

54,118

answered Feb 07 '22 at 18:24

sara

61

1

and where is n above arrows? – Zarko Feb 07 '22 at 19:23
@Zarko :-) I have put the "n" :-)))))) a contribute for a new user. – Sebastiano Feb 07 '22 at 21:41
1

Oh, dear @Sebastiano, you are so good soul. I try to encourage her to this . Well, I will do now with up-voting the answer. – Zarko Feb 07 '22 at 21:45

Steven B. Segletes · Accepted Answer · 2017-07-28T03:48:25.097

If you don't need it in smaller math styles, then this could work...

\documentclass{article}
\usepackage{stackengine,graphicx}
\stackMath
\newcommand\frightarrow{\scalebox{1}[.4]{$\rightarrow$}}
\newcommand\darrow[1][]{\mathrel{\stackon[1pt]{\stackanchor[1pt]{\frightarrow}{\frightarrow}}{\scriptstyle#1}}}
\begin{document}
$A\darrow[n] B$
\end{document}

Here is a composite alternative

\documentclass{article}
\usepackage{stackengine,graphicx,amssymb}
\stackMath
\newcommand\frightarrow{\scalebox{1}[.3]{$\rule[.45ex]{2ex}{1.5pt}%
  \kern-.2ex{\blacktriangleright}$}}
\newcommand\darrow[1][]{\mathrel{\stackon[1pt]{\stackanchor[1pt]{\frightarrow}{\frightarrow}}{\scriptstyle#1}}}
\begin{document}
$A\darrow[n] B$
\end{document}

score 4 · Answer 3 · answered Dec 21 '20 at 11:01

4

"Cheating" would be writing it as

f_n(x)^\rightarrow_\rightarrow f(x)

which results in the following formula:

answered Dec 21 '20 at 11:01

VIVID

195

Uniform convergence symbol (arrow on top of arrow, same size)

3 Answers3