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J - 30

'`z'(]-32*1=I.)&.(a.&i.)1!:1]1

J is read right-to-left, so to break this down:

1. Prompt user for input: 1!:1]1
2. Perform algorithm in code-point-space: &.(a.&i.)
3. Identify character range for each letter; the characters between codepoints "`" and "z" are considered lowercase: 1=I..
4. For each lower-case codepoint, subtract 32: `']-32* ...'
5. Note that step (2) creates an implicit step (5): we started out by projecting from character to integer domain, so now that we're finished, we map those integers back onto characters.

Obviously this particular implementation only considers ASCII; but the approach could be extended to at least the basic multilingual plane in Unicode.