Regarding equivalent conditions of Frechet differentiability

Question

Gateaux and Frechet differentiability in a Banach space are defined as below. Can you tell below how (ii) implies (i). The rest is easy.

Answer 1

Let me denote $f(x):=\|x\|$ and the limit $$ f'(x;h) = \lim_{t\to0} \frac1t{f(x+th)-f(x)} \quad x\in S_X,h\in X. $$ Due to (ii), this limit exists for all such $x,t$.

First, let me show that $h\mapsto f'(x;h)$ is a linear map. Let $h\in X$, $s\in \mathbb R\setminus\{0\}$ be given. Then $$ f'(x; sh) = \lim_{t\to0} \frac1t{f(x+sth)-f(x)} = s \lim_{t\to0} \frac1{st}{f(x+sth)-f(x)} = s f'(x;h). $$ Let now $h_1,h_2$ be given. Then by convexity (or triangle inequality) $$ \|x + \frac12t(h_1+h_2)\| \le \frac12 \|x+th_1\| + \frac12 \|x+th_2\|. $$ Subtracting $\|x\|$ on both sides, dividing by $t>0$, passing to the limit $t\searrow0$ yields $$ f'(x,\frac12(h_1+h_2))\le \frac12 f'(x,h_1+h_2)+\frac12 f'(x,h_1+h_2), $$ which implies $f'(x;h_1+h_2) \le f'(x;h_1)+f'(x;h_2)$. Replacing $h_i$ by $-h_i$ implies the reverse inequality, hence $f'(x;h_1+h_2) = f'(x;h_1)+f'(x;h_2)$. This proves Gateaux differentiability of the norm.

The uniformity of the limit then gives the required $o$-property to obtain Frechet differentiability.