This is exercise 6, part b) in chapter 5 out of Folland's Real Analysis textbook.
Let $X$ be a finite dimensional vector space (over $\mathbb{R}$) with basis $\{e_i\}_{i = 1}^{n}$. Equip $X$ with the following norm: $||\sum_{j=1}^{n}a_je_j||_1 = \sum_{j=1}^{n}|a_j|$
Next, define the map $$p: \mathbb{R}^n \longrightarrow X$$ $$(a_1, ... ,a_n) \mapsto \sum_{j=1}^{n}a_je_j$$
I am interested in showing that the map $p$ is continuous when $\mathbb{R}^n$ is equipped with the usual Euclidean topology and $X$ is equipped with the topology induced by $|| \cdot ||_1$. I would appreciate if anyone could verify my proof.
Firstly, it is clear that the map $p$ is linear. Next, since both the domain and range of $p$ are metric spaces (induced by the norm), it suffices to show sequential continuity.
I will use superscripts to denote the sequence index and subscripts to denote the number component of the $n$-tuple.
Let $\{x^k\}$ be a sequence in $\mathbb{R}^n$ that converges to $x = (x_1, ..., x_n)$ in the Euclidean norm. It is then a standard fact that each component $x_j^k$ converges to $x_j$ in $\mathbb{R}$ for $1 \leq j \leq n$.
Thus, given $\epsilon > 0$, for each $j \in \{1, ..., n\}$, there is a natural number $N_j$ such that $|x_j^k - x_j| < \frac{\epsilon}{n}$ whenever $k \geq N_j$. Define $N:= \max\{N_j\}_{j=1}^n$.
Then for $k \geq N$, we have that $||p(x^k) - p(x) ||_1 = ||p(x^k - x)||_1$ (since $p$ is linear) = $||p((x_1^k - x_1, ... , x_n^k - x_n))||_1 = ||\sum_{j=1}^{n}(x_j^k - x_j)e_j ||_1 = \sum_{j=1}^{n}|x_j^k - x_j| < \sum_{j=1}^{n}\frac{\epsilon}{n} = \epsilon$
Thus, $p(x^k) \longrightarrow p(x)$ in $X$.
Does this check out? Also, since $p$ is linear it would also suffice to show that it is bounded for then continutity would follow. However, I could not find an easy way to show this.
Your proof of continuity is fine!
For showing boundedness, note that we must find a constant $C$ such that $C\|v\| \geq |p(v)|_1$ for all $v \in \mathbb R^n$.
For this,note that if $v = (v_1,...,v_n)$ then $\sum |v_i| = |p(v)|_1$ by the definition. Note that $||v|| = \sqrt{\sum v_i^2}$ is the Euclidean distance.
Now, the famous RMS-AM inequality tells us that for any real quantities $v_i$, we have $$\sqrt{\frac{\sum v_i^2}{n}} \geq \frac{\sum |v_i|}{n}$$
Which gives a constant $C =\sqrt n$.