Let $A: \Bbb{R}^{+}\to M_{n\times n}(\Bbb{R})$, then $A(t)$ is continuous if and only if $a_{ij}(t)$ is continuous, $\forall\;i,j=1,\cdots,n$

Question

Suppose \begin{align}A: \Bbb{R}^{+}\to M_{n\times n}(\Bbb{R})\end{align} \begin{align}t\mapsto A(t),\end{align} where $A(t)=\big(a_{ij}(t)\big).$

I want to prove that the following are equivalent.

$i.$ $A(t)$ is continuous;

$ii.$ $a_{ij}(t)$ is continuous, $\forall\;i,j=1,\cdots,n$.

MY TRIAL

Assume that $a_{ij}(t)$ is continuous $\forall\;i,j=1,\cdots,n$. Since $M_{n\times n}(\Bbb{R})$ is a finite dimensional vector space, then all norms are equivalent. Take

\begin{align}\Vert A(t)\Vert=\sum^{n}_{i,j=1}|a_{ij}(t)|\end{align} Since $A(t)$ is a finite sum of continuous functions, then it is continuous.

Now, I need to prove that converse but no way. Please, can anyone help me? Kindly help me check if my proof is as well, correct!

Answer 1

The last step of your proof for $(ii)\Rightarrow (i)$ does not sound quite right to me. In your last equality, it is $\|A(t)\|$ which is a sum of continuous functions, and hence continuous. In general, this does not imply that $A(t)$ is also continuous.

Hint: consider this equality instead $$\|A(t)-A(t_0)\|=\sum_{i,j=1}^{n}|a_{ij}(t)-a_{ij}(t_0)| ,\qquad \forall t,t_0\in \mathbb{R}^+$$ and take the limit for $t\to t_0$ (starting from the side of the equality where the limit actually exists).