$f$ well defined, continuous, not bounded and maps to $l^2$ If $X$ is a infinite-dimensional Banach space, $u_0 \in X$ and $r>0$, from Riesz theorem we know that the sphere $$\overline B(u_0,r)= \{v \in X | \vert\vert v-u_o\vert\vert \leq r\}$$ is not compact. Therefore, the image of a continuous function on $\overline B(u_0,r)$ is not necessarily bounded.
Now consider the space $X=l^2$ of the square summable sequences $u=(u_1,u_2,...)$ where $$\vert\vert u \vert\vert _{l^2}:=(\sum_{k=1}^\infty \vert u_k \vert^2)^\frac{1}{2} < \infty$$
and the function $f$ on the unit ball $\overline B(0,1) \subset l^2$ defined by $$f(u):=(\sum_{k=1}^\infty \frac{2^{-k}}{1+3^{-k}-u_k},0,0,...).
$$
What is the approach when proving that $f$ is well defined, continuous and maps to $l^2$ and that the image of $f$ is not bounded?
 A: Let's consider the functional 
$$ \phi\colon B_{\ell^2} \to \def\C{\mathbf C}\C, \qquad u \mapsto \sum_{k=1}^\infty \frac{2^{-k}}{1 + 3^{-k} - u_k} $$
on the closed unit ball $B_{\ell^2}$ of $\ell^2$. 
Note that $f$ is concatenation of $\phi$ with the first coordinate inclusion $\C \to \ell^2$ which is isometric. Hence it suffices to prove, that $\phi$ is well-defined, continuous and unbounded.
First note that $$\def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|}\abs{u_k}\le \norm u_2 \le 1 < 1 + 3^{-k}$$
for each $k$, hence the quotient is defined. As $u_k \to 0$, there is some $K(u)$ such that $\abs{u_k}\le \frac 12$, this gives 
$$ \abs{1 + 3^{-k} - u_k}\ge 1 + 3^{-k} - \frac 12 \ge \frac 12 $$
for $k \ge K$, this gives 
$$ \abs{\phi(u)} \le \abs{\sum_{k=1}^K \frac{2^{-k}}{1 + 3^{-k} - u_k}} + \sum_{k=K+1}^\infty 2^{1-k} < \infty $$
Hence, $\phi$ is well defined. To prove that $\phi$ is continuous, let $u \in B_{\ell^2}$ and $\def\eps{\varepsilon}\eps > 0$ be given. Choose $K$ such that $\sum_{k=K+1}^\infty 2^{2-k}< \frac \eps 2$. Now, note that $\phi^K : u \mapsto \sum_{k=1}^K \frac{2^{-k}}{1 + 3^{-k} - u_k}$ is continuous on $B_{\ell^2}$, as a finite sum of continuous maps. Hence, there is $\delta > 0$ such that $\abs{\phi^K(u)-\phi^K(v)} < \frac{\eps}2$ for all $y\in B_{\ell^2}$ with $\norm{u-v} < \delta$. Now let $v \in \ell^2$ with $\norm{u-v}_2< \delta$, then 
\begin{align*}
  \abs{\phi(u) - \phi(v)} &= \abs{\sum_{k=1}^\infty\left( \frac{2^{-k}}{1 + 3^{-k} - u_k} - \frac{2^{-k}}{1 + 3^{-k}- v_k}\right)}\\
   &\le \abs{\phi^K(u) - \phi^K(v)} + \sum_{k=K+1}^\infty 2^{2-k}\\
  &< \frac \eps 2 + \frac \eps 2\\
  &= \eps.
\end{align*}
To see that $\phi$ is unbounded, recall that 
\begin{align*}
  \phi(e_n) &= \sum_{k=1}^\infty \frac{2^{-k}}{1 + 3^{-k} - \delta_{nk}}\\
     &\ge \frac{2^{-n}}{1 + 3^{-n} - 1}\\
     &=  \left(\frac 32\right)^n 
\end{align*}
