# For what $p \in [1; +\infty)$ is $l_p$ a FD-space?

Suppose $$(V, \|\cdot\|_V)$$ and $$(W, \|\cdot\|_W)$$ are two Banach spaces and $$f: V \to W$$ is some function. We call a bounded linear operator $$A \in B(V, W)$$ Fréchet derivative of $$f$$ in $$x \in V$$ iff

$$\lim_{h \to 0} \frac{\|f(x + h) - f(x) - Ah\|_W}{\|h\|_V} = 0$$

We call a $$f$$ Fréchet differentiable in $$x$$ iff there exists a Fréchet derivative of $$f$$ in $$x$$.

We call a Banach space $$(V, \|v\|)$$ FD-space iff $$f: V \to \mathbb{R}, v \mapsto \|v\|_V$$ is Fréchet differentiable $$\forall x \in V \setminus \{0\}$$.

Now let’s define the collection of Banach spaces $$l_p := (\{(a_n)_{n \in \mathbb{N}} \subset \mathbb{R}| \sum_{n = 1}^\infty |a_n|^p < \infty\}, \|\cdot \|_p := (\sum_{n = 1}^\infty |a_n|^p)^{\frac{1}{p}}\}$$ for $$p \in (1; +\infty)$$.

For what $$p$$ is $$l_p$$ a FD-space?

If $$n \in \mathbb{N}$$ and $$p = 2n$$, then it definitely is.

Proof:

$$(h_k)_{k \in \mathbb{N}} \mapsto 2n\sum_{k = 1}^\infty a_k^{2n-1}h_k$$ is the Fréchet derivative of $$\sum_{k = 1}^\infty a_k^{2n}$$

However I do not know how to deal with the other possible values of $$p$$.

• Yes it should hold. Maybe try the derivative $(h_k)_{k \in \mathbb{N}} \mapsto p \sum_{k=1}^\infty |a_k|^{p-2} a_k h_k$? It would be the only one that would make sense because the derivative of $f: \mathbb{R} \to \mathbb{R}: x \mapsto |x|^p$ is $x \mapsto p |x|^{p-2} x$ everywhere except zero. Commented Dec 21, 2019 at 17:22
• @Demophilus, your argument does indeed work for $p > 1$. For $p = 1$ the situation is, however, a bit different... Commented Dec 21, 2019 at 17:57

For $$p > 1$$ $$l_p$$ is indeed a FC-space, as $$(h_k)_{k \in \mathbb{N}} \mapsto p\sum_{k = 1}^\infty |a_k|^{p-2}a_kh_k$$ is derivative of $$(a_k)_{k \in \mathbb{N}} \mapsto \sum_{k = 1}^\infty |a_k|^p$$ everywhere except zero.
However, $$l_1$$ is not an FC-space, as $$\lim_{h \to 0} \frac{|\|x + h\|_1 - \|x\|_1 - f^*(h)|}{\|h\|_1}$$ does not exist for any $$f^* \in l_1^*$$ if at least one coordinate of $$x$$ is zero.