If $T(X) = AX $, $A \subset \mathbb{R^{m \times 1}}, \forall X \in \mathbb{R^{n}},$ Then $T$ is continuous.
My thoughts:
I have proved (or I was helped to prove ) it using the facts that every linear operator is continuous and that every linear operator is bounded in a finite dimensional space.
Now, I want to prove this statement using more elementary tools, like $\epsilon - \delta$ definition, could anyone help me in doing this please?
I'm assuming that $A \in \mathbb{R}^{m\times n}$ so $T : \mathbb{R}^n \to \mathbb{R}^m$ otherwise it doesn't make much sense.
For $x\in \mathbb{R}^n$ we have
$$\|Tx\|_2^2 = \|Ax\|_2^2= \left\|\left(\sum_{j=1}^n A_{ij}x_j\right)_{i=1}^m\right\|^2 = \sum_{i=1}^m \left(\sum_{j=1}^n A_{ij}x_j\right)^2$$ Now apply Cauchy-Schwarz inequality to obtain:
$$\|Tx\|_2^2 \le \sum_{i=1}^m \left(\sum_{j=1}^n A_{ij}^2\right)\left(\sum_{j=1}^n x_j^2\right) = \left(\sum_{i=1}^m\sum_{j=1}^n A_{ij}^2\right) \|x\|_2^2$$
where $\sum_{i=1}^m\sum_{j=1}^n A_{ij}^2$ is a fixed nonnegative constant.
Now we have $$\|Tx-Ty\|_2^2 = \|T(x-y)\|_2^2 \le \left(\sum_{i=1}^n\sum_{j=1}^n A_{ij}^2\right)\|x-y\|_2^2$$
so $T$ is continuous.