Let $A: J\subseteq \mathbb{R} \to \mathbb{K}^{n \times n}$ be a
function, and denote $A = (A_{ij}(x))$. Equip $J$ with the euclidean
metric, and the space of $n\times n$ matrices with the operatornorm
$\Vert A \Vert_{op} = \sup_{\Vert y\Vert = 1, y \in \mathbb{K}^n}
\Vert Ay\Vert$ where $\mathbb{K} \in \{\mathbb{R}, \mathbb{C}\}$ and
$\Vert \cdot \Vert$ is the usual norm on $\mathbb{K}^n$Prove that $A$ is continuous if and only if every component function
of $A$ is continuous, i.e., for every $i$ and $j$, the function
$A_{ij}: J \to \mathbb{K}$ is continuous.
My attempt:
$\Rightarrow$
Let $1 \leq i,j \leq n$; $k,l \in J$
$$|a_{ij}(k) – a_{ij}(l)| \leq \Vert A_i(k) – A_i(l)\Vert = \Vert A(k)e_i – A(l)e_i\Vert \leq \sup \Vert(A(k)-A(l))y\Vert = \Vert A(k) – A(l) \Vert_{op}$$
where $A_i$ denotes the i'th column of the matrix and $e_i = (0, \dots ,i, \dots, 0)$
$\Leftarrow$
Let $k \in J$ be fixed, $\epsilon > 0$. Choose $\delta$ such that for $|k-l| < \delta$, we have $|a_{ij}(k)- a_{ij}(l)| < \epsilon$ for every $i$ and $j$. Then, for $l \in J$ with $|k-l| < \delta$:
$$\Vert A(k) – A(l) \Vert_{op} = \sup \Vert (A(k)-A(l))y \Vert$$
$$ = \sup\left\Vert \left(\sum_{i=1}^n(A_{1i}(k)-A_{1i}(l))y_i, \dots ,\sum_{i=1}^n(A_{1n}(k)-A_{1n}(l))y_i\right)\right\Vert$$
$$ = \sup \sqrt{\sum_{j=1}^n \left(\sum_{i=1}^n (A_{ji}(k) – A_{ji}(l))y_i\right)^2}$$
$$ \leq \sup \sum_{j=1}^n \left(\sum_{i=1}^n (A_{ji}(k) – A_{ji}(l))y_i\right)$$
$$< \sup \sum_{j=1}^n \left(\sum_{i=1}^n \epsilon y_i\right)$$
$$= \epsilon \sup \sum_{j=1} \sum_{i=1}y_i = \epsilon \sup \sum_{j=1} \Vert y \Vert_1$$
Using the fact that the usual norm and the $1$-norm are equivalent, there exists $C > 0$ such that $\Vert y \Vert_1 \leq C \Vert y \Vert$ and the expression becomes smaller than
$$\epsilon Cn \sup \Vert y \Vert = \epsilon C n$$
and hence the function is continuous.
Is this correct?
Best Answer
A matrix space, as set, is simply the cartesian product $\Bbb K\times\cdots\times\Bbb K$ ($n\times n$ times). As in finite dimension all the norms are equivalent, using the 1-norm $$\|(x_{i,j})\| = \sum_{i,j}|x_{i,j}|$$ is very easy to prove that a function $$Y\longrightarrow{\Bbb K}^{n\times n}$$ is continuous iff the composition with the canonical projection over each factor is continuous.
(In fact, this is the universal property of the product topology, but general topology isn't required)