How to prove a map between two spaces of real sequences $f : l^1 \to l^2 $ is well-defined and continuous the question is whether the following statement is ture or false, and justify it.
Here is the statement

The map $f : l^1\to l^2$ given by 
    $f(x_0, x_1, x_2,...)= (x_0, x_1, x_2,...) $ is well-defined and continuous.

[Recall that: 


*

*$l^1$ is  the space of real sequences $(x_n)$ such that $\sum_0^\infty|x_n|$ converges with norm $||(x_n)||_1 =\sum_0^\infty|x_n|$.

*$l^2$ is  the space of real sequences $(x_n)$ such that $\sum_0^\infty x_n^2$ converges with norm $||(x_n)||_1 =\sqrt{\sum_0^\infty x_n^2}$.]
[topological properties of standard metric spaces may be quoted without proof]
I know to prove well-defined is to prove something like $f(a)=f(b)$ then $a=b$. am I correct?
Also, I am little confused about proving continuity of this real sequences map. Could someone please give me some hints on this? (does it related to homeomorphism)
 Thanks a lot. 
 A: To prove that $f$ is well-defined, you want to show that if you pick $(x_n) \in l^1$, then $f\big((x_n) \big) \in l^2$, that is, the map actually does what it says it will do in terms of mapping an element from its domain to it's codomain.
Theorem: Suppose $1\le p_1<p_2\le\infty.$ Prove that $\ell^{p_1}\subseteq \ell^{p_2}$ by proving that $\|a\|_{p_2}\le\|a\|_{p_1}$ for any $a\in \ell^{p_1}.$
Proof: Define $b_n := \frac{a_n}{\|a\|_{p_1}}$ and observe that $\vert b_n \vert \leq 1$ for all $n \in \mathbb N$. Notice the following chain of equivalent statements:
                    $$\begin{align*}
      \vert b_n \vert^{p_2} &\leq \vert b_n \vert^{p_1} \\
      \sum_{n = 1}^\infty \vert b_n \vert^{p_2} & \leq \sum_{n = 1}^\infty \vert b_n \vert^{p_1} \\
      \sum_{n = 1}^\infty \bigg \vert \frac{a_n}{\|a\|_{p_1}} \bigg \vert^{p_2} & \leq \sum_{n = 1}^\infty \bigg \vert \frac{a_n}{\|a\|_{p_1}} \bigg \vert^{p_1} \\
      \Bigg(\sum_{n = 1}^\infty \bigg \vert \frac{a_n}{\|a\|_{p_1}} \bigg \vert^{p_2}\Bigg)^{1/p_1} & \leq \Bigg( \sum_{n = 1}^\infty \bigg \vert \frac{a_n}{\|a\|_{p_1}} \bigg \vert^{p_1} \Bigg)^{1/p_1}\\
      \frac{1}{\big(\|a\|_{p_1}\big)^{p_2/p_1}} \Bigg(\sum_{n = 1}^\infty \vert a_n \vert^{p_2} \Bigg)^{1/p_1} & \leq 1\\
      \Bigg(\sum_{n = 1}^\infty \vert a_n \vert^{p_2} \Bigg)^{1/p_1} & \leq \big(\|a\|_{p_1}\big)^{p_2/p_1}\\
      \sum_{n = 1}^\infty \vert a_n \vert^{p_2}  & \leq \big(\|a\|_{p_1}\big)^{p_2}\\
      \Bigg(\sum_{n = 1}^\infty \vert a_n \vert^{p_2} \Bigg)^{1/p_2} & \leq \|a\|_{p_1}\\
       \|a\|_{p_2}& \leq \|a\|_{p_1}.
     \end{align*}$$
Consider the case when $p_2 = \infty$, notice that $\vert a_n \vert \leq \bigg(\sum_{n = 1}^\infty \vert a_n \vert^{p_1} \bigg)^{1/p_1}$ for all $n \in \mathbb N$, and thus $\|a_n\|_{p_2} = \|a_n\|_\infty = \sup_{n \in \mathbb N} \vert a_n \vert \leq \bigg(\sum_{n = 1}^\infty \vert a_n \vert^{p_1} \bigg)^{1/p_1} = \|a_n\|_{p_1}$. $\square$
By applying this theorem to $p_1 = 1$ and $p_2 = 2$, we get that $f$ is well-defined.
To show that $f$ is continuous, let $\varepsilon > 0$, then if $\|(x_n) - (y_n) \|_1 < \delta = \varepsilon$, again by the second part of the theorem, we get that $\|f\big( (x_n)\big) - f\big((y_n) \big) \|_2 = \|(x_n) - (y_n)\|_2 \leq \|(x_n) - (y_n)\|_1 < \varepsilon$. Conclude by definition that $f$ is continuous.
