title: Prove if h(n) > f1(n) - f2(n) > 0 and f1(n) = Ω(g(n) and f2(n) = O(g(n)) then h(n) = Ω(g(n)
body: <p>I proceeded as follows :</p>
  
  <pre><code>f1(n) &gt; c1*g(n) , for all n&gt;n1;  (*because f1(n) = Ω(g(n))*)
  f2(n) &lt; c2*g(n) , for all n&gt;n2;  (*beacuse f2(n) = O(g(n))*)
  Thus, h(n) &gt; c1*g(n) - f2(n) &gt; c1*g(n) - c2*g(n) &gt; (c1 - c2)*g(n), for all n&gt;max(n1,n2)
  </code></pre>
  
  <p>Now the problem is for h(n) = Ω(g(n)) to hold in accordance with my proof, c1 has to be greater than c2, because the constants in the O and Ω notation have to be positive. I am not able to eliminate that premise. </p>
  
  <p>Can anyone please help me with that. Thanks</p>