Real Analysis. Show that this $\lim$ doesn't depends on the norm. 
Let a sequence of linear transformations $A_{k} : \mathbb{R}^{m} \longrightarrow \mathbb{R}^{n}$ suppose that, for all $x \in \mathbb{R}^{m}$, theres exists $\displaystyle Ax = \lim_{k\to \infty}A_{k}x$. Prove that the transformation $A: \mathbb{R}^{m} \longrightarrow \mathbb{R}^{n}$ is linear and $\displaystyle \lim_{k\to \infty}A_{k} = A$ in any norm of $\mathcal{L}(\mathbb{R}^{m}, \mathbb{R}^{n})$.

I did the first part of the question, but I do not know that $\displaystyle \lim_{k\to \infty}A_{k} = A$ in any norm. I've already shown that any two norms in $\mathbb{R}^{n}$ are equivalent, but I do not know how that helps. Thanks for any hint.
 A: Fix norms on $\mathbb R^m$ and $\mathbb R^n$.  We will show that $A_n\to A$ in the norm $\|A\|=\max_{\|x\|\leq1}\|Ax\|$.  The basic idea is to use compactness of the unit ball of $\mathbb R^m$ to obtain a bound on $\|Ax-A_kx\|$ for arbitrary $x$ in the unit ball of $\mathbb R^m$, and hence on $\|A-A_n\|$.
Put $M=\max\{1,\|A\|,\|A_1\|,\|A_2\|,\ldots\}$, and fix $\varepsilon>0$.  For each $x\in\mathbb R^m$, there is some $N_x\in\mathbb N$ such that $\|Ax-A_nx\|<\varepsilon/3$ whenever $n\geq N_x$.  If $\|x-y\|<\varepsilon/3M$ and $n\geq N_x$, then 
$$\|Ay-A_ny\|\leq\|Ay-Ax\|+\|Ax-A_nx\|+\|A_nx-A_ny\|<\varepsilon.$$
Since the unit ball of $\mathbb R^m$ is compact, there exist $x_1,\ldots,x_k$ in the unit ball of $\mathbb R^m$ such that for each $x\in\mathbb R^m$ with $\|x\|\leq1$, there is some $j$ such that $\|x-x_j\|<\varepsilon/3M$.  Put $N=\max\{N_{x_1},\ldots,N_{x_k}\}$.   Then for $n\geq N$ and any $x\in\mathbb R^m$ with $\|x\|\leq1$, we have 
$$\|Ax-A_nx\|\leq\|Ax-Ax_j\|+\|Ax_j-A_nx_j\|+\|A_nx_j-A_nx\|<\varepsilon.$$
  Taking the supremum over $x$ yields $\|A-A_n\|\leq\varepsilon$, and we are done.
