This works:

String rex = "^\\d+\\.\\s\\p{Lu}+.*";

System.out.println("1. PTYU fmmflksfkslfsm".matches(rex));
// true

System.out.println(". PTYU fmmflksfkslfsm".matches(rex));
// false, missing leading digit

System.out.println("1.PTYU fmmflksfkslfsm".matches(rex));
// false, missing space after .

System.out.println("1. xPTYU fmmflksfkslfsm".matches(rex));
// false, lower case letter before the upper case letters

Breaking it down:

• ^ = Start of string
• \d+ = One or more digits (the \ is escaped because it's in a string, hence \\)
• \. = A literal . (or your original [.] is fine) (again, escaped in the string)
• \s = One whitespace char (no need for the {1} after it) (I'll stop mentioning the escapes now)
• \p{Lu}+ = One or more upper case letters (using the proper Unicode escape — thank you, tchrist, for pointing this out in your comment below. In English terms, the equivalent would be [A-Z]+)
• .* = Anything else

See the documentation here for details.

You only need the .* at the end if you're using a method like String#match (above) that will try to match the entire string.

(answer changed after OP added more details)

Your string

String inputText = "\1f\1e\1d\02002868BF03030000000000000000S023\1f\1e\1d\03\0d";

Doesn't actually contains any \ literals because according to Java Language Specification in section 3.10.6. Escape Sequences for Character and String Literals \xxx will be interpreted as character indexed in Unicode Table with octal (base/radix 8) value represented by xxx part.

Example \123 = 1*8² + 2*8¹ + 3*8⁰ = 1*64 + 2*8 + 3*1 = 64+16+3 = 83 which represents character S

If string you presented in your question is written exactly the same in your text file then you should write it as

String inputText = "\\1f\\1e\\1d\\02002868BF03030000000000000000S023\\1f\\1e\\1d\\03\\0d";

(with escaped \ which now will represent literal).

(older version of my answer)

It is hard to tell what exactly you did wrong without seeing your code. You should be able to find at least \1, \1, \1, \0 since your regex can match one \ and one hexadecimal character placed after it.

Anyway this is how you can find results you mentioned in question:

String text = "\\1f\\1e\\1d\\020028";
Pattern p = Pattern.compile("\\\\[a-fA-F0-9]{2}");
//                                          ^^^--we want to find two hexadecimal 
//                                               characters after \
Matcher m = p.matcher(text);
while (m.find())
    System.out.println(m.group());

Output:

\1f
\1e
\1d
\02