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on 2^O(lg n) being a polynomial

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Here's an argument to prove that 2^O(lg n) is a polynomial.

Fact.  f(n) in 2^O(lg n) is a polynomial.

Proof.  By definition, f(n) in 2^O(lg n) means that there's a real c > 0
and natural n(0) such that for all n >= n(0)

  0 <= f < 2^{c lg n} = 2^{lg(n^c)} = n^c

QED.

It's hard for me to say what's precisely wrong here, but I disagree with
the result.  I think there are functions in 2^O(lg n) which are not
polynomials.  For example,

  2^{lg(lg n)} n^c is in 2^O(lg n)

because 2^{lg(lg n)} n^c < 2^{c lg n} for c = 1 and n(0) = 2.  But
clearly 2^{lg(lg n)} n^c is not a polynomial because lg(lg n) is not a
constant.

What's wrong with that argument?  Or where am I wrong?  Thank you.

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on 2^O(lg n) being a polynomial Daniel Bastos <dbastos@toledo.com> - 2017-03-24 22:25 -0300
  Re: on 2^O(lg n) being a polynomial Kaz Kylheku <336-986-7787@kylheku.com> - 2017-03-25 02:39 +0000
  Re: on 2^O(lg n) being a polynomial Ben Bacarisse <ben.usenet@bsb.me.uk> - 2017-03-28 14:25 +0100
    Re: on 2^O(lg n) being a polynomial Kaz Kylheku <336-986-7787@kylheku.com> - 2017-03-28 17:45 +0000
    Re: on 2^O(lg n) being a polynomial Daniel Bastos <dbastos@toledo.com> - 2017-03-28 20:28 -0300
      Re: on 2^O(lg n) being a polynomial Ben Bacarisse <ben.usenet@bsb.me.uk> - 2017-03-29 14:12 +0100

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