$A$ is continuous w.r.t. $||\cdot||_{L(\Bbb{K}^n,\Bbb{K}^m)}$, iff $a_{\mu\nu}$ is continuous for all $\mu$ and $\nu$ Let $m,n\in\Bbb{N}$, let $X$ be a metric space, let $a_{\mu\nu}:X\to\Bbb{K}$ for $\mu\in\{1,\dots,m\}$ and $\nu\in\{1,\dots,n\}$, as well as $$A:\begin{cases}X\to\Bbb{K}^{m\times n},\\x\mapsto(a_{\mu\nu}(x))_{\mu=1,\dots,m,\\ \nu=1,\dots,n.}\end{cases}$$Also let $L(\Bbb{K}^n,\Bbb{K}^m)=\{A:\Bbb{K}^n\to\Bbb{K}^m,\ A\text{ is linear}\}=\Bbb{K}^{m\times n}$ be a vectorspace and $||A||_{L(\Bbb{K}^n,\Bbb{K}^m)}=\sup_{0\neq x\in\Bbb{K}^n}\frac{||Ax||}{||x||}$ be  the operator norm on $L(\Bbb{K}^n,\Bbb{K}^m)$.
I want to show that $A$ is continuous w.r.t. $||\cdot||_{L(\Bbb{K}^n,\Bbb{K}^m)}$, iff $a_{\mu\nu}$ is continuous for all $\mu$ and $\nu$, but I don't  know where to start. I would appreciate any help, thanks.
 A: I proved your theorem for $\mathbb{K}=\mathbb{R}$ because I wasn't sure what norm is in $\mathbb{K}^n$ for arbitrary field.
Assume we are given a metric space $X$ and functions $a_{\mu\nu}:X\rightarrow \mathbb{R}$ for any $\mu\in\{1,\ldots,m\},\nu\in\{1,\ldots,n\}$ and let $A:X\rightarrow \mathbb{R}^{m\times n}$ be
$$A(x)=\left[\begin{array}{ccc}a_{11}(x) & \ldots & a_{1n}(x) \\ \vdots & &\vdots\\ a_{m1}(x)& \ldots & a_{mn}(x)\end{array}\right]$$
In $\mathbb{R}^{m\times n}$ we have the operator norm
$$\|B\|=\sup_{0\neq t\in\mathbb{R}^n}\frac{\|Bt\|}{\|t\|}$$
First observe that a function $\xi:\mathbb{R}^{m\times n}\rightarrow L(\mathbb{R}^n,\mathbb{R}^m)$ given by the formula
$$\xi(B)(t)=Bt$$ is an isomorphism, where in $L(\mathbb{R}^n,\mathbb{R}^m)$ we have standard operator norm.
Also for any $\Lambda\in L(\mathbb{R}^n,\mathbb{R}^m)$
$$\xi^{-1}(\Lambda)=\left[\begin{array}{ccc}\Lambda(e_1)_1 & \ldots & \Lambda(e_n)_1 \\ \vdots & &\vdots\\ \Lambda(e_1)_m& \ldots & \Lambda(e_n)_m\end{array}\right]$$
, where $e_1,\ldots,e_n$ form standard basis of $\mathbb{R}^n$.
For any $\nu\in\{1,\ldots,n\}$ a function $T_{\nu}:L(\mathbb{R}^n,\mathbb{R}^m)\rightarrow\mathbb{R}^m$ given by the formula
$$T_{\nu}(\Lambda)=\Lambda(e_{\nu})$$
is continuous.
Now assume that $A$ is continuous and fix $\mu\in\{1,\ldots,m\},\nu\in\{1,\ldots,n\}$. All we need to do is to observe that
$$a_{\mu\nu}=\pi_{\mu}\circ T_{\nu}\circ \xi\circ A$$
where $\pi_{\mu}:\mathbb{R}^m\rightarrow\mathbb{R}$ is the standard projection. Then $a_{\mu\nu}$ is continuous as a composition of continuous functions.
Conversely, assume that all $a_{\mu\nu}$ are continuous.
For any $\mu\in\{1,\ldots m\}$ let $\alpha_{\mu}:X\rightarrow\mathbb{R}^n$ be given by the formula $$\alpha_{\mu}(x)=\left(a_{\mu 1}(x),\ldots,a_{\mu n}(x)\right)$$
Then $\alpha_{\mu}$ are continuous functions because their projections are all continuous.
For any $x,y\in X$ we have
$$\|A(x)-A(y)\|=\sup_{0\neq t\in\mathbb{R}^n}\frac{\|(A(x)-A(y))t\|}{\|t\|}=\sup_{0\neq t\in\mathbb{R}^n}\frac{\|A(x)t-A(y)t\|}{\|t\|}=\sup_{0\neq t\in\mathbb{R}^n}\frac{\sqrt{\sum_{\mu=1}^{m}(A(x)t-A(y)t)_{\mu}^2}}{\|t\|}=\sup_{0\neq t\in\mathbb{R}^n}\frac{\sqrt{\sum_{\mu=1}^{m}\left((A(x)t)_{\mu}-(A(y)t)_{\mu}\right)^2}}{\|t\|}=\sup_{0\neq t\in\mathbb{R}^n}\frac{\sqrt{\sum_{\mu=1}^{m}\left(\langle\alpha_{\mu}(x)| t\rangle-\langle\alpha_{\mu}(y)| t\rangle\right)^2}}{\|t\|}=\sup_{0\neq t\in\mathbb{R}^n}\frac{\sqrt{\sum_{\mu=1}^{m}\langle\alpha_{\mu}(x)-\alpha_{\mu}(y)| t\rangle^2}}{\|t\|}\leq \sup_{0\neq t\in\mathbb{R}^n}\frac{\sqrt{\sum_{\mu=1}^{m}\|\alpha_{\mu}(x)-\alpha_{\mu}(y)\|^2 \|t\|^2}}{\|t\|} =\sqrt{\sum_{\mu=1}^{m}\|\alpha_{\mu}(x)-\alpha_{\mu}(y)\|^2}$$
Fix $x\in X$ and $\varepsilon>0$. For any $\mu\in\{1,\ldots,m\}$ from continuity of $\alpha_{\mu}$ at $x$ there exists $\delta_{\mu}>0$ such that for $y\in X$
$$\rho(x,y)<\delta_{\mu} \implies \|\alpha_{\mu}(x)-\alpha_{\mu}(y)\|<\frac{\varepsilon}{\sqrt{m}}$$
Let $\delta=\min\{\delta_1,\ldots,\delta_m\}$. It is easy to verify that for $y\in X$
$$\rho(x,y)<\delta \implies \|A(x)-A(y)\|<\varepsilon$$
