Let $\phi: S^{k-1}\to E\setminus Q$ be a continuous map, here


* $E$ is an infinite dimensional Hilbert space
* $Q$ is a compact set in $E$


I need extend $\phi$ to $B^{k}$ such that
$\phi: B^{k}\to E\setminus Q$ is still continuous.


If $Q := \{u\in E:\|u\|\_E=1\}\setminus E\_{k+1}~ \text{where}~ E\_{k+1} ~\text{is any $k+1$-dimension subspace of} ~E$ (not compact), that's Ok, how about other $Q$?



extend a continuous map on sphere to ball such that the image is out of a compact set
Extending $\phi$ to the ball is equivalent to proving that $\phi:S^{k-1}\to H\setminus Q$ is nulhomotopic.


To prove this, you can consider the map $F:\phi(S^{k-1})\times Q\to E;(x,y)\to\frac{y-x}{|y-x|}$, which has compact image, so there is some vector $v$ with $|v|=1$ outside the image of $F$. So you can homotope $\phi(S^{k-1})$ to $\phi(S^{k-1})+kv$ with $k$ as big as you want without intersecting $Q$ at any point. If you take $k$ big enough, $\phi(S^{k-1})+kv$ and $Q$ will be separated by a hyperplane (perpendicular to $v$, for example), so you can just linearly homotope $\phi(S^{k-1})+kv$ to a point in its side of the hyperplane without intersecting $Q$.


