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.


$(h_k)_{k \in \mathbb{N}} \mapsto 2n\sum_{k = 1}^\infty a_k^{2n-1}h_k$ 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 $p$.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Demophilus
    Commented Dec 21, 2019 at 17:22
  • $\begingroup$ @Demophilus, your argument does indeed work for $p > 1$. For $p = 1$ the situation is, however, a bit different... $\endgroup$ Commented Dec 21, 2019 at 17:57

1 Answer 1


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.


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