How to prove that the map $f: M_n(\Bbb R) \to M_n(\Bbb R)$ defined by $A \mapsto AA^t$ is continuous? This map has appeared several times on this site such as here, here and here. I thought of writing out a proof for the continuity of this map. Identify $M_n(\Bbb R)$ with $\Bbb R^{n^2}$.
I tried using sequential way to prove the continuity here.
So let $X_n$ be a converging sequence in $M_n(\Bbb R)$ and $||\cdot||$ be max-norm norm on $M_n(\Bbb R)$ i.e. $||A||=\max {(a_{ij})}.$ Then $\forall \; \delta \gt 0 \; \exists N \in \Bbb N$ such that $||X_n - X|| \lt \delta \; \forall \; n \ge N.$ This also means that $\forall \; \delta \gt 0 \; \exists N \in \Bbb N$ such that $||X_n^t - X^t|| \lt \delta \; \forall \; n \ge N.$
Consider $||f(X_n)-f(X)||=||X_n X_n^t-XX^t||=||X_n X_n^t - XX^t + X_nX^t - X_n X^t|| \le ||X_n X_n^t - X_n X^t||+
||X_nX^t-XX^t||= ||X_n(X_n^t-X^t)||+||X^t(X_n-X)||.$
But I am stuck at the last step. What can I do from here?
Also I am intereseted in knowing other ways to show this.
EDIT : I have found that as per the norm that I have used, $||AB|| \not \le ||A||||B||.$ Take $$A = \left[ \begin{matrix}
        1 & 2 \\
        3 & 4 \\
       \end{matrix}\right]$$ and $$B = \left[ \begin{matrix}
        4 & 5 \\
        6 & 7 \\
       \end{matrix}\right].$$ So my way will not work.
This is because I was thinking whether $||X_n(X_n^t-X^t)|| \le ||X_n||||X_n^t-X^t||$ could be done. But my purpose is defeated.

Unfortunately your norm $||A||=\max {|a_{ij}|}$ is not submultiplikative. Since all norms on $M_n(\Bbb R)$ are equivalent, we can choose a norm which is submultiplikative.
Let $ ||⋅|| $ such a norm.
Then we have
$||X_n(X_n^t-X^t)||+||X^t(X_n-X)|| \le ||X_n|| \cdot||X_n^t-X^t||+||X^t|| \cdot ||X_n-X||$.
Can you take it from here ? Observe that $(||X_n||)$  is bounded.
