If my @capitals array contains only 'London' ...

An array contains only 'London' if it has only one element and that element is the string 'London':

c:\@Work\Perl\monks>perl -wMstrict -le
"my $string = 'London';
 my @capitals = split /,/, $string;
 if (@capitals == 1 && $capitals[0] eq 'London') {
   print q{contains only 'London'}
   }
"
contains only 'London'
[download]

... OR it contains all elements of my @capitals.

I don't understand what this requirement means. Can you please elaborate?

I'm not sure I understand your problem very well, in your first sentence you talk about a string but show something that looks like an array... See How do I post a question effectively?

Anyway, to check if a string is included in another, the easiest way is to use the index function. Or to check if a string is present in an array, use grep.

#!/usr/bin/perl
use strict;
use warnings;

my @strings = ('London, Paris','','London');

foreach my $string (@strings)
{
   my @capitals= split(/,/,$string); 

   print "list of capitals: @capitals\n";
   print "number of capitals:".@capitals."\n";
   print "number of capitals:",scalar @capitals,"\n";
   print "lots of capitals\n\n" if (@capitals>1);
   print "one or no capitals\n\n" if (@capitals <= 1);
}

__END__
list of capitals: London  Paris
number of capitals:2
number of capitals:2
lots of capitals

list of capitals: 
number of capitals:0
number of capitals:0
one or no capitals

list of capitals: London
number of capitals:1
number of capitals:1
one or no capitals
[download]

Just test $#capitals

say $#capitals ? 'Lots of capitals' : "Just one: @capitals";
[download]

Watch out, though. $#capitals would evaluate to true (-1) if the list has no capitals:

perl -lE "@capitals=(); $,=', '; print scalar(@capitals), $#capitals, 
+$#capitals ? 'Lots' : 'One';
0, -1, Lots

perl -lE "@capitals=qw(London); $,=', '; print scalar(@capitals), $#ca
+pitals, $#capitals ? 'Lots' : 'One';
1, 0, One

perl -lE "@capitals=qw(London Paris); $,=', '; print scalar(@capitals)
+, $#capitals, $#capitals ? 'Lots' : 'One';
2, 1, Lots
[download]

There's yet another pitfall associated with the use of  $#array to determine the number of elements in an array. It's possible to assign to  $[ (see perlvar) to change the base index of Perl arrays. Doing so will result in your source entry appendage(s) receiving a sharp whack, so Don't Do That!^SM

Of course, it's possible to use  $#array for this purpose in a base-agnostic way:

c:\@Work\Perl\monks>perl -wMstrict -le
"print qq{perl array base is $[};
 ;;
 my @ra = ();
 print qq{A: array (@ra) is empty} if $#ra + 1 - $[ == 0;
 ;;
 @ra = (1);
 print qq{B: array (@ra) is empty} if $#ra + 1 - $[ == 0;
"
perl array base is 0
A: array () is empty

c:\@Work\Perl\monks>perl -wMstrict -le
"$[ = 1;
 print qq{perl array base is $[};
 ;;
 my @ra = ();
 print qq{A: array (@ra) is empty} if $#ra + 1 - $[ == 0;
 ;;
 @ra = (1);
 print qq{B: array (@ra) is empty} if $#ra + 1 - $[ == 0;
"
Use of assignment to $[ is deprecated at -e line 1.
perl array base is 1
A: array () is empty
[download]

But an array  @array evaluated in scalar context always yields the number of array elements, so I don't understand why one would ever do this in any other way.

BTW: It's possible to assign a number other than 0 or 1 to $[, but then you're really on thin ice, so Double-Plus-Don't Do That!

---

Give a man a fish:  <%-{-{-{-<

Indeed, but the possibility of "no capitals" was not in the spec. :-)