Proving a linear transformation is continuous Correct me if I'm mistaken, but a linear transformation is continuous if it is bounded, correct?
 Let $T: (\mathbb{R^n}, \left\|\cdot\right\|_1) \rightarrow (\mathbb{R^m}, \left\|\cdot\right\|_1)$ be a linear transformation. Show that $T$ is continuous. Note that in this case, $$\|T\| = \sup_{\|x\|_1\leq 1} \|Tx\|_1.$$
Thus, we have $\|x\|_1 \leq 1$ so that $|\sum_{i=1}^{n} x_i| \leq 1$. Thus, $$\|T\| = \sup_{\|x\|_1\leq 1} \|Tx\|_1 = \sup_{\|x\|_1\leq 1} \left(\left|\sum_{i=1}^{n} t_{1i} x_i\right| + \dotsb + \left|\sum_{i=1}^{n} t_{mi} x_i\right|\right).$$
However, from this point we cannot say that $$\tag{**}\|T\| \leq \left(\left|\sum_{i=1}^{n} t_{1i}\right| + \dotsb + \left|\sum_{i=1}^{n} t_{mi}\right|\right) = \sum_{i=1}^{n} \sum_{j=1}^{m} t_{ji}.$$
This was my original idea, but I believe this is incorrect as $\|x\|_1 \leq 1 \nrightarrow (^{**})$, correct?
If this is true, how do I go about finishing the proof?
edit: I'm now seeing a statement in my textbook that says the following:
 If $S$ is a metric space, if $a_{11}, ..., a_{mn}$ are real continuous functions on $S$, and if, for each $p\in S$, $A_p$ is the linear transformation of $R^n$ into $R^m$ whose matrix has entries $a_{ij}(p)$, then the mapping $p \rightarrow A_p$ is a continuous mapping of $S$ into $L(R^n, R^m)$.
What exactly are $S$ and $p$?
 A: It's easier to think in terms of a basis for $\mathbb{R}^n$ and how $T$ acts on that basis rather than working component by component. 
Let $\{e_i: 1 \leq i \leq n \}$ be the standard basis for $\mathbb{R}^n$. Then for $\lambda_i \in \mathbb{R}$
$$\lvert \lvert T(\sum_{i=1}^n \lambda_i e_i) \rvert \rvert = \lvert \lvert \sum_{i=1}^n \lambda_i T(e_i) \rvert \rvert \leq \sum_{i=1}^n \lvert \lambda_i \rvert \cdot \lvert \lvert T(e_i) \rvert \rvert \leq \sum_{i=1}^n \lvert \lambda_i \rvert \max \lvert \lvert T(e_i) \rvert \rvert \\ = \max \lvert \lvert T(e_i) \rvert \rvert \cdot  \bigg (\sum_{i=1}^n \lvert \lambda_i \rvert \bigg )= \max \lvert \lvert T(e_i) \rvert \rvert \cdot \lvert \lvert \sum_{i=1}^n \lambda_i e_i \rvert \rvert_1$$ 
which shows that $T$ is continuous since if we express $x = \sum_{i=1}^n \lambda_i e_i$ and $\lvert \lvert x \rvert \rvert_1 \leq 1$ then the above gives $\lvert \lvert Tx \rvert \rvert \leq \max \lvert \lvert T(e_i) \rvert \rvert$ so we get that $T$ is bounded.
Edit: To answer the question added in your edit, A metric space $S$ is a set equipped with a function $d:S^2 \to [0,\infty)$ giving the distance between two points satisfying $d(x,y) = 0$ if and only if $x=y$, $d(x,y) = d(y,x)$ and the triangle inequality - $d(x,z) \leq d(x,y) + d(y,z)$. A norm induces a metric by setting $d(x,y) = \lvert \lvert x-y \rvert \rvert$ but not all metrics arise this way.
