Q: If $||f-g_n|| \to \inf_{g\in S} ||f-g||$ and $||g_n-g_0||\to 0 ,$then $ \inf_{g\in S}||f-g||=||f-g_0||$ If $||f-g_n|| \to \inf_{g\in S} ||f-g||$ and $||g_n-g_0||\to 0 ,$then $ \inf_{g\in S}||f-g||=||f-g_0||$
Note: S is a closed subspace of Hilbert Space S.
I am just wondering how to show $||f-g_0||$ is actually the infimum? Does the following work? By the triangle inequality, $||f-g_0||\leq ||f-g_n||+||g_n-g_0||$
and taking limit as $n\to \infty$ we get $||f-g_o||\leq \inf_{g\in S}||f-g||$. But since $g_0$ is in S (as S closed), then the only way for that inequality to hold is if $||f-g_o||=\inf_{g\in s}||f-g||$ ? Is my argument correct? How to make it more precise with epsilons? Thanks.
edit: I guess an even more general result follows, using continuity of the norm..

A: Your argument is correct. 
The only step that needs more rigour is showing that $||f-g_0|| \leq ||f - g_n|| + ||g_n-g_0||$ implies $||f-g_0||\leq \inf_{g\in S}||f-g||$.
Let $n \in \mathbb{N}$ be s.t. $\min(||f-g_n||-\inf_{g\in S}||f-g||, \ ||g_n-g_0||) < \frac{\epsilon}{2}$. Then substitute into the first equation to get $||f-g_0||\leq \inf_{g\in S}||f-g|| + \epsilon$.
