I was trying to prove by induction that

$ $ T(n) = \begin{cases} 1 &\quad\text{if } n\leq 1\ T\left(\lfloor\frac{n}{2}\rfloor\right) + n &\quad\text{if } n\gt1 \ \end{cases} $ $ is $ \Omega(n)$ implying that $ \exists c>0, \exists m\geq 0\,\,|\,\,T(n) \geq cn \,\,\forall n\geq m$

Base case : $ T(1) \geq c1 \implies c \leq 1$

Now we shall assume that $ T(k) = \Omega(k) \implies T(k) \geq ck \,\,\forall k < n$ and prove that $ T(n) = \Omega(n)$ .

$ $ T(n) = T(\lfloor{\frac{n}{2}}\rfloor) + n \geq c\lfloor{\frac{n}{2}}\rfloor + n \geq c \frac{n}{2} -1 + n \geq n\left(\frac{c}{2} – \frac{1}{n} + 1\right) \geq^{?} cn\ c \leq 2 – \frac{2}{n} $ $

So we have proved that $ T(n) \geq c n$ in :

1) The base case for $ c \leq 1$

2) The inductive step for $ c \leq 2 – \frac{2}{n}$

Yet we have to find a value that satisfies them both for all $ n\geq 1$ , the book suggest such value is $ c = 1$ which to me is not true since :

$ $ 1 \leq 1\1\leq2 – \frac{2}{n}\implies 1 \leq 0 \text{ for n = 1} $ $

My guess would be $ 0$ but is not an acceptable value; So we just say its $ \Omega(n)$ but for $ n \gt 1$ ?Or how can we deal with it?