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 $\Bbb{CP}^k$.
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.
This is not true. The restriction map $H^0(\Bbb{CP}^k, \mathcal{O}(l))\rightarrow H^0(X,\mathcal{O}_X(l))$ is neither injective nor surjective in general. The kernel gives the space of hypersurfaces of degree $l$ containing $X$. Surjectivity for all $l$ means that $X$ is projectively normal, a property which does not hold in general.
