Given a smooth projective variety $X \subset \Bbb{CP}^k$ why is it true that global sections of $O(l)|\_X, l >> 0 $ are just global sections of $O(l)$ on $\Bbb{CP}^k$ restricted to $X$? 


Here $O(l)$ is just an appropriate power of the anti-tautological line bundle on $\Bbb{CP}^k$. 



global sections of canonical line bundle of a projective variety
*This is an answer to the edited question*


Consider the short exact sequence
$$
0 \to \mathscr I\_X(l) \to \mathscr O\_{\mathbb P^k}(l) \to \mathscr O\_X(l) \to 0, 
$$
and the long exact sequence of cohomology it induces:
$$
0 \to H^0(\mathbb P^k, \mathscr I\_X(l)) \to H^0(\mathbb P^k,\mathscr O\_{\mathbb P^k}(l)) \to H^0(X,\mathscr O\_X(l)) \to H^1(\mathbb P^k, \mathscr I\_X(l)) \to\dots . 
$$


By Serre's theorem $H^1(\mathbb P^k, \mathscr I\_X(l))=0$ for $l\gg 0$ and hence the previous map $H^0(\mathbb P^k,\mathscr O\_{\mathbb P^k}(l)) \to H^0(X,\mathscr O\_X(l))$ is surjective.


You will not be able to get injectivity. In fact, the larger the $l$, the "less" injective that map is. $\dim H^0(\mathbb P^k,\mathscr O\_{\mathbb P^k}(l))$ grows at the order of $l^k$ while $\dim H^0(X,\mathscr O\_X(l))$ grows at the order of $l^{\dim X}$, so there is nothing you can do to guarantee injectivity. Quite the opposite, the only time you might have injectivity is for small $l$'s.


