3-4 Asymptotic notation properties

Let \(f(n)\) and \(g(n)\) by asymptotically positive functions. Prove or disprove each of the following conjectures.

a. \(f(n) = O(g(n))\) implies \(g(n) = O(f(n))\).

b. \(f(n) + g(n) = \Theta(\min(f(n), g(n)))\).

c. \(f(n) = O(g(n))\) implies \(\lg(f(n)) = O(\lg(g(n)))\), where \(\lg(g(n)) \ge 1\) and \(f(n) \ge 1\) for all sufficiently large \(n\).

d. \(f(n) = O(g(n))\) implies \(2^{f(n)} = O(2^{g(n)})\).

e. \(f(n) = O((f(n))^2)\).

f. \(f(n) = O(g(n))\) implies \(g(n) = \Omega(f(n))\).

g. \(f(n) = \Theta(f(n / 2))\).

h. \(f(n) + o(f(n)) = \Theta(f(n))\).

a. Disprove, \(n = O(n^2)\), but \(n^2 \ne O(n)\).

b. Disprove, \(n^2 + n \ne \Theta(\min(n^2, n)) = \Theta(n)\).

c. Prove, because \(f(n) \ge 1\) after a certain \(n \ge n_0\).

\[ \begin{aligned} \exists c, n_0: \forall n \ge n_0, 0 \le f(n) \le cg(n) \\\\ \Rightarrow 0 \le \lg f(n) \le \lg (cg(n)) = \lg c + \lg g(n). \end{aligned} \]

We need to prove that

\[\lg f(n) \le d\lg g(n).\]

We can find \(d\),

\[d = \frac{\lg c + \lg g(n)}{\lg g(n)} = \frac{\lg c}{\lg g(n)} + 1 \le \lg c + 1,\]

where the last step is valid, because \(\lg g(n) \ge 1\).

d. Disprove, because \(2n = O(n)\), but \(2^{2n} = 4^n \ne O(2^n)\).

e. Prove, \(0 \le f(n) \le cf^2(n)\) is trivial when \(f(n) \ge 1\), but if \(f(n) < 1\) for all \(n\), it's not correct. However, we don't care this case.

f. Prove, from the first, we know that \(0 \le f(n) \le cg(n)\) and we need to prove that \(0 \le df(n) \le g(n)\), which is straightforward with \(d = 1 / c\).

g. Disprove, let's pick \(f(n) = 2^n\). We will need to prove that

\[\exists c_1, c_2, n_0: \forall n \ge n_0, 0 \le c_1 \cdot 2^{n / 2} \le 2^n \le c_2 \cdot 2^{n / 2},\]

which is obviously untrue.

h. Prove, let \(g(n) = o(f(n))\). Then

\[\exists c, n_0: \forall n \ge n_0, 0 \le g(n) < cf(n).\]

We need to prove that

\[\exists c_1, c_2, n_0: \forall n \ge n_0, 0 \le c_1f(n) \le f(n) + g(n) \le c_2f(n).\]

Thus, if we pick \(c_1 = 1\) and \(c_2 = c + 1\), it holds.