norms simplification. I encountered this in a book. $||.||$ is the norm of a real normed space.
$$\lim_{\lambda\rightarrow\pm 0}\frac{||x+\lambda y||^2-||x||^2}{2\lambda}=||x||.\lim_{\lambda\rightarrow\pm 0}\frac{||x+\lambda y||-||x||}{\lambda}.$$ I could not understand the derivation. Any help would be appreciated.
After a little googling, I found that the limit on the left side exists due to the convexity of the function $$f(t)=\|x+ty\|^2,\quad t\in \mathbb{R}$$ and the lemma (e.g., Kuczma (1985)) below:
Lemma: Let $I\subset \mathbb{R}$ be an open interval and let $f:I\rightarrow \mathbb{R}$ be convex. Then for every $x\in I$ there exist the right and left side derivatives.
The existence of the limit on the right side is the existence of the Gateaux differential of the norm functional.
How do I equate these two expressions? Any kind of help would be appreciated. Thanks in advance!
 A: This inequality can't be true in general. Take the normed vector space to be $\mathbb{R}$ with the standard norm $|.|$, and $x,y,\lambda$ to be $1$.
Your left hand side then gives $1.5$ and your right gives $1$.
A: $\lambda$ =|0 .Your equality is by obvious algebra  equivalent to 
(||x+$\lambda$y||-||x||)^2 =0 which is equivalent to  ||x+$\lambda$y||=||x|| which is clearly not true in general.Please look in your book and see what additional conditions are given 
A: For brevity, let the LHS and RHS expressions be $L$ and $R.$
(1). If $x\ne 0$ then $L=MR$ where $M=(\|x\|+\|x+\lambda y\|)/2\|x\|.$
We have $\lim_{\lambda\to 0}M=1$ because  $\|x\|-|\lambda|\cdot \|y\|\leq \|x+\lambda y\|\leq \|x\|+|\lambda|\cdot \|y\|.$
So either $\lim_{\lambda\to 0}L=\lim_{\lambda \to 0}R$ or neither side has a limit when $\lambda \to 0.$
(2). If $x=y=0$ them $L=R=0.$
(3). If $x=0 \ne y$ then $|L|=|\lambda|\cdot \|y\|^2/2$ which $\to 0$ as $\lambda \to 0,$ and $|R|=0.$
Examples. In $\mathbb R^2$ let $x=(1,0)$ and $y=(0,1) .$ 
(a). If $\|(u,v)\|=\max (|u|,|v|)$ then $L=R=0$ when $|\lambda|\leq 1.$ 
(b).If $\|(u,v)\|=|u|+|v|$ then $R=1$ when $\lambda >0$ but  $R=-1$ when $-1<\lambda <0.$
