Continuity of a function $\varphi : \ell_1 \to \mathbb{R}$ defined by $\varphi \left (\left ( x_n \right )_{n=1}^{\infty} \right )=\sum x_n^2$ 
Show that the function $\varphi: \ell_1 \to \mathbb{R}$ defined by $$\varphi \left (\left ( x_n \right )_{n=1}^{\infty} \right )=\sum_{n=1}^\infty x_n^2$$
  is continuous

I've tried proving using the $\varepsilon-\delta$ definition, but I seem to fail finding a proper $\delta$. I have tried bounding $d\left(\varphi\left(x_{n}\right),\varphi\left(a_{n}\right)\right)$ with $\sum\limits _{n=0}^{\infty}\left|x_{n}-a_{n}\right|\left|x_{n}+a_{n}\right|$ for some $\delta$ (where $a_n$ is the series in the $\delta$-ball around $x_n$) but it seems to be a dead end. Is this the right approach, or should I try a different bound?
 A: First observe that $l^1\subset l^2.$ Thus $\varphi$ is well defined.
Fix $x\in l^1.$ Then for any $y$ with $\|y-x\|_1 < 1,$ we have
$$\tag 1 |\varphi(y) - \varphi (x)| = |\sum_{n=1}^{\infty}(y_n^2 -x_n^2)| \le \sum_{n=1}^{\infty}|y_n -x_n||x_n+y_n|.$$
Now each $|x_n+y_n|$ is bounded above by $\|x\|_1 + \|y\|_1 \le 2\|x\|_1 + 1.$ Thus the right side of $(1)$ is bounded above by
$$(2\|x\|_1 +1)\sum_{n=1}^{\infty}|y_n -x_n| = (2\|x\|_1 +1)\|y-x\|_1.$$
This shows $\varphi$ is continuous at $x$ as desired.
A: Let $x,y \in l^2$
Then $y_0=(y_n)$ and $\epsilon>0$.
We have that $x_n \to 0$ and $y_n \to 0$ from the convergence of the series thus $x_n,y_n$ are bounded.
So exist $M,N>0$ such that $|y_n| \leq N,\forall n \in \Bbb{N}$
Thus for $\delta<1$ we have that $||x||_1 \leq ||y_0||_1 +1$
$$|\phi(x)-\phi(y)|=|\sum_{n=1}^{\infty}x_n^2-y_n^2| \leq \sum_{n=1}^{\infty}|x_n-y_n||x_n+y_n| \leq \sum_{n=1}^{\infty}|x_n-y_n|(2||y_0||_1+1)=(2||y_0||_1+1)||x-y_0||_1$$
Take $\delta<\min\{1,\frac{\epsilon}{2||y_0||_1+1}\}$  and you are done.
So your function is  continuous.
