Differentiation of $l^p(\mathbb{N})$ norm I was asked a problem, to show that $x\mapsto \Vert x\Vert_{l^p}^p=\sum_{k=0}^\infty \vert x_k\vert^p$ is differentiable at any point $x\in l^p(\mathbb{N})$ when $p\in]1,\infty[$ and twice differentiable when $p\in[2,\infty[$.
Where are talking Fréchet differentiability here ($l^p(\mathbb{N})$ being a Banach space). My guess is that it would mimic the way it works on $\mathbb{R}^n$ but I cannot seem to figure out how to properly write that out in this infinite dimension context.
Is writing 
$$
\Vert x+h\Vert_{l^p(\mathbb{N})}^p-\Vert x\Vert_{l^p(\mathbb{N})}^p
$$
for $h=(h_n)_{n\in\mathbb{N}}\in l^p(\mathbb{N})$ even relevant in the first place to answer the question ?
Thank you for your help !
 A: Use my comments and the fact that for $p\in (1,2]$ the function $\mathbb{R} \rightarrow \mathbb{R}, x\mapsto \vert x \vert^{p-1}$ is $(p-1)$-Hölder continuous (with constant equal to $1$). To obtain
$$ \vert \Vert x+h \Vert_p^p - \Vert x \Vert_p^p - A_x(h) \vert \leq p \Vert h \Vert_p^p.$$
Hence, we get
$$ \frac{\vert \Vert x+h \Vert_p^p - \Vert x \Vert_p^p - A_x(h) \vert}{\Vert h \Vert_p} \leq p \Vert h \Vert_p^{p-1} \rightarrow 0 $$
for $\Vert h \Vert_p \rightarrow 0$.
For $p\in (2,\infty)$ we can use the mean value theorem once more and get $\eta_j$ between $x_j$ and $\xi_j$ such that 
$$ \Vert x+h \Vert_p^p - \Vert x \Vert_p^p - A_x(h) = \sum_{j\in \mathbb{N}} f''(\eta_j)(\xi_j - x_j) \cdot h_j. $$
Thus, we get
$$ \vert \Vert x+h \Vert_p^p - \Vert x \Vert_p^p - A_x(h) \vert \leq \sum_{j\in \mathbb{N}} \vert f''(\eta_j) \vert \cdot \vert h_j \vert^2. $$
Therefore,
$$ \vert \Vert x+h \Vert_p^p - \Vert x \Vert_p^p - A_x(h) \vert \leq p(p-1)\sum_{j\in \mathbb{N}} (\vert x_j \vert + \vert h_j \vert)^{p-2}\cdot \vert h_j \vert^2. $$
Hölder gives us
$$ \vert \Vert x+h \Vert_p^p - \Vert x \Vert_p^p - A_x(h) \vert \leq p(p-1)\Vert (\vert x_j \vert + \vert h_j \vert)_{j\in \mathbb{N}}\Vert_p^{p-2} \cdot \Vert h \Vert_p^2. $$
For $\Vert h \Vert_p \leq \Vert x \Vert_p$ we get
$$ \vert \Vert x+h \Vert_p^p - \Vert x \Vert_p^p - A_x(h) \vert \leq p(p-1)(\Vert x \Vert_p + \Vert h \Vert_p)^{p-2} \cdot \Vert h \Vert_p^2
\leq p(p-1)2^{p-2} \Vert x \Vert_p^{p-2} \cdot \Vert h \Vert_p^2. $$
