Let $C = \{x \in \ell_2 \text{ with } \|x\|_{\ell_{2}} \leq 1\}$ and let $F: C \rightarrow C$ be defined as $x \mapsto \{\sqrt{1-\|x\|^{2}},x_{1},x_{2},...\}$ show that F is continuous
Attempts so far
I've been trying to find a delta such that the epsilon delta definition of continuity works out, without much luck so far. Clearly we have $\|F(x)\|=1$ $\forall x \in \ell_2$ (haven't been able to use this yet but i thought it might be useful).
$\|F(x)-F(y)\|=\bigg|\sqrt{1-\|x\|^{2}}-\sqrt{1-\|y\|^{2}}\bigg|^2 + \|x-y\|$, if i could just set $X=\frac{x}{\|x\|}$ and $Y=\frac{y}{\|y\|}$ then we'd have $\|F(x)-F(y)\|= \|x-y\|$ so $\delta =\epsilon$ would work, but I don't think this is possible in this case as I think it misses out a few elements.
Since $0\leq\sqrt{1-\|x\|}\leq 1$ we have $\|F(x)-F(y)\|\leq 1 + \|x-y\|=\|F(x)\|+\|x-y\|$ but this doesn't give me a delta thats always greater than 0.
I've also tried rationalising $\bigg|\sqrt{1-\|x\|^{2}}-\sqrt{1-\|y\|^{2}}\bigg|^2 $ and a few other things without much luck so any help would be appreciated.
The function $g:[0,1] \to \Bbb R$ defined by $$g(t)=\sqrt{1-t^2}$$ is continuous on a compact set, thus it is uniformly continuous. Thus, $$\forall \varepsilon >0, \ \ \ \exists \delta >0: \ \ |t-u|< \delta \Rightarrow |g(t)-g(u)| < \varepsilon \ \ \ \ \ (*)$$
Now, fix $0 < \varepsilon <1$, and fix $0< \delta < \varepsilon$ satisfying $(*)$ above.
If $||x-y||_2 < \delta$, then $$|||x||_2-||y||_2| \le ||x-y||_2 < \delta$$ thus, by $(*)$ $$\left| \sqrt{1-||x||_2^2} -\sqrt{1-||y||_2^2}\right|^2 = |g(||x||_2) - g(||y||_2)|^2 < \varepsilon^2 < \varepsilon$$ Finally, $$||F(x)-F(y)||_2 = \left| \sqrt{1-||x||_2^2} -\sqrt{1-||y||_2^2}\right|^2 + ||x-y||_2 < \varepsilon + \delta \le 2 \varepsilon$$
This show that $F$ is uniformly continuous on $C$.