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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?

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  • $\begingroup$ Much faster: in finite dimension all the norms are equivalent and endow ${\Bbb K}^{n\times n}$ with the product topology. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 2 '18 at 16:43
  • $\begingroup$ What is the product topology for a matrix space? $\endgroup$ – user370967 Apr 2 '18 at 16:50
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    $\begingroup$ A matrix space, as set, is simply the product $\Bbb K\times\cdots\times\Bbb K$ ($n \times n$ times). In the product topology a function $Y\longrightarrow$ (product space) is continuous iff the composition with the canonical projection over each factor is continuous (en.wikipedia.org/wiki/Product_topology#Properties). $\endgroup$ – Martín-Blas Pérez Pinilla Apr 2 '18 at 17:00
  • $\begingroup$ Make this an answer and I'll accept it. $\endgroup$ – user370967 Apr 2 '18 at 17:04
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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)

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