1

I am interested in the characterization of a real Banach space $X$ with norm $\|\cdot\|$ of which the square $\|\cdot\|^2$ is continuously Fréchet differentiable, that is, the derivative $\nabla \|\bar x\|^2 \in X^*$ exists and is continuous (with respect to the dual norm) in $\bar x$. This is true in a Hilbert space with a norm induced by the inner product. (Example and counter-example: $\|\cdot\|_2$ and $\|\cdot\|_\infty$, respectively, in $\mathbb R^n$.)

Question: Is there a Banach space with a continuously (Fréchet) differentiable squared norm which is not a Hilbert space?

The answer would be negative if the bivariate mapping $(x,y) \mapsto \nabla\|x\|^2(y)$ defines an inner product on $X$; it's linear in $y$ (since the derivative is a linear mapping) but is it always linear in $x$?

tcunis
  • 13
  • It seems to me that this boils down to differentiability of $f:x\mapsto |x|$ everywhere except for $x = 0$ (which I think there is literature for), and additionally condition that $|x|\cdot f'(x)\to 0$ as $x\to 0$. – Jakobian Mar 25 '25 at 02:49

1 Answers1

1

Yes, those are precisely the Fréchet smooth spaces. A Banach space $X$ is Fréchet smooth if the limit $\lim_{t\to0} \frac{\|x+ty\|-\|x\|}t $ exists for each $x,y \in S_X$ and the convergence is uniform in $y$.

Typical examples are Banach spaces $X$ with (locally) uniformly convex dual $X^*$ (see Section 5.6 in [1] for more examples). In particular, this includes Hilbert spaces and most classical reflexive function spaces such as $L^p$, $W^{k,p}$ e.t.c.

Fréchet smooth spaces can be given an equivalent characterization via the duality mapping $J \colon X \to 2^{X^*}$ which is given by $$J(x)=\{x^* \in X^* \colon \langle x^*,x \rangle = \|x\|^2, \, \|x^*\|=\|x\| \} .$$

Theorem. [Theorem 5.6.3 in [1] or Proposition 2.1 in [2].] A real Banach space $X$ is Fréchet smooth if and only if the duality mapping $J \colon X \to 2^{X^*}$ is single-valued (i.e. $J(x)=\{j(x)\}$) and norm-to-norm continuous (i.e. $x\mapsto j(x)$ is continuous from $X$ to $X^*$).

Now this is related to the differentiability of the function $f(x)=\frac 12 \|x\|^2$ by the fact that $J(x)$ is the subdifferential $\partial f(x)$ of $f$ at $x$ $$\partial f(x) := \{x^* \in X^* \colon \langle x^*,y-x \rangle \le f(y)-f(x), \quad \forall y \in X\} = J(x) $$ (Proposition 6.1.23 in [3]] and

Theorem (see this or this) Let $U$ be open and $f\colon U \subset X \to \mathbb R$ convex continuous. Then $f$ is Fréchet differentiable at $x_0 \in U$ if and only if $\partial f$ is single-valued and continuous at $x_0$, i.e., $\partial f(x_0)=\{x^*_0\}$ and $$ \text{ whenever $x_n \in U$ and $x^*_n \in \partial f(x_n)$ and } x_{n} \rightarrow x_0 \text{ then } x^*_{n} \to x^*_0 .$$ Moreover, $\partial f(x_0)= \{f'(x_0)\}$ and if $f$ is Fréchet differentiable everywhere then $f$ is $C^1$.

Applying the above to $f(x)=\frac 12 \|x\|^2$ and noting that $\partial f =J$ gives that $f$ is $C^1$ if and only if the duality mapping $J$ is single-valued and continuous, i.e. if and only if $X$ is Fréchet smooth.

Finally note that when $H$ is a Hilbert space then $J(x)=\{x\}$ is the identity on $H$ so that $(f'(x), h) = (x,h)$ as usual.

[1] Megginson, Robert E. An introduction to Banach space theory. Vol. 183. Springer Science & Business Media, 2012.

[2] Kim, Tae Hwa, and Ximing Yin. "REMARKS ON DIFFERENTIABILITY OF THE NORM AND UNIFORMLY CONVEX SETS." JOURNAL OF NONLINEAR AND CONVEX ANALYSIS 16.9 (2015): 1737-1745.

[3] Papageorgiou, Nikolaos S., and Patrick Winkert. Applied Nonlinear Functional Analysis: An Introduction. Walter de Gruyter GmbH & Co KG, 2018.

Evangelopoulos Foivos
  • 4,429