I am trying to prove that the functions $\mathbf{f}:\mathbb{M}_{n\times n}\to \mathbb{M}_{n\times n},\ \mathbf{f}(A)=A^2,\ \mathbf{f}(A)=A^T A$ are differentiable.
What I have done:
We consider $\mathbf{D}(f_1)(A):H\mapsto AH+HA$ and $\mathbf{D}(f_2)(A):H\mapsto A^TH+H^TA$. These are linear transformations (*) from $\mathbb{M}_{n\times n}$ to $\mathbb{M}_{n\times n}$ and we want to prove that $$\lim\limits_{H\to [0]}\frac{1}{|H|}\left( (f_1(A+H)-f_1(A)) - (AH+HA) \right)=[0]$$ and $$\lim\limits_{H\to [0]}\frac{1}{|H|}\left( (f_2(A+H)-f_2(A)) - (A^TH+H^TA) \right)=[0].$$ Since $f_1(A)=A^2,$ we have $$|f_1(A+H)-f_1(A) - (AH+HA)|=|A^2+AH+HA+H^2-A^2-AH-HA|=|H^2|;$$ this gives $$\lim\limits_{H\to [0]}\frac{|H^2|}{|H|}\leq\lim\limits_{H\to [0]}\frac{|H||H|}{|H|}=0$$ so $f_1$ is differentiable with derivative $AH+HA$.\
Similarly, since $f_2(A)=A^TH+H^TA,$ we have $|f_2(A+H)-f_2(A)-(A^TH+H^TA)|=|(A+H)^T(A+H)-A^TA-A^TH-H^TA|=|(A^T+H^T)(A+H)-A^TA-A^TH-H^TA|=|A^TA+A^TH+H^TA+H^TH-A^TA-A^TH-H^TA|=|H^TH|$; this gives $$\lim\limits_{H\to [0]}\frac{|H^TH|}{|H|}\leq\lim\limits_{H\to [0]}\frac{|H^T||H|}{|H|}=\lim\limits_{H\to [0]} |H^T|=0$$ so $f_2$ is differentiable with derivative $A^TH+H^TA.$
- $f_1(H_1+H_2)=A(H_1+H_2)+(H_1+H_2)A=AH_1+AH_2+H_1A+H_2A=(AH_1+H_1A)+(AH_2+H_2A)=f_1(H_1)+f_2(H_2),$
$f_1(cH))=A(cH)+(cH)A=cAH+cHA=c(AH+HA)=cf_1(H),$
$f_2(H_1+H_2)=A^T(H_1+H_2)+(H_1+H_2)^TA=A^TH_1+A^TH_2+H_1^TA+H_2^TA=(A^TH_1+H_1^TA)+(A^TH_2+H_2^TA)=f_2(H_1)+f_2(H_2),$
$f_2(cH)=A^T(cH)+(cH)^TA=cA^TH+cH^TA=c(A^TH+H^TA)=cf_2(H)$ for all $H,H_1,H_2\in\mathbb{M}_{n\times n},\ c\in\mathbb{R}$
Is this correct? Thanks for the feedback.
$f_1(A+H)-f_1(A) -(AH+HA)= H^2$, so $\| f_1(A+H)-f_1(A) -(AH+HA) \| \le \|H\|^2$.
$f_2(A+H)-f_2(A) -(A^TH+H^TA)= H^T H$, so $\| f_2(A+H)-f_2(A) -(A^TH+H^TA) \| \le \|H\|^2$.
Given any $\epsilon>0$ you can choose $\|H\| < \epsilon$ from which it follows that $f_1,f_2$ are (Frechet) differentiable.