Uniform smoothness and twice-differentiability of norms

Question

To get to the simplest case, consider a norm $\|\cdot\|$ over $R^n$ that is uniformly smooth of power-type 2, that is, there is a constant $C$ such that $$\frac{\|x+y\| + \|x - y\|}{2} \le 1 + C \|y\|^2$$ for all $x$ with $\|x\| = 1$ and for all $y$.

Question: Does this guarantee that $\|\cdot\|$ has a second-order Taylor expansion on $R^n \setminus \{0\}$, that is, there is a vector $g$ and a symmetric matrix $A$ such that $$\|x + y\| = \|x\| + \langle g, y \rangle + \frac{1}{2} \langle Ay, y \rangle + o(\|y\|^2)$$ for all $x \neq 0$. (Apparently this is a weaker requirement than twice-differentiability of $\|\cdot\|$ on $R^n \setminus \{0\}$)

It is easy to see that $\|\cdot\|$ is differentiable on $R^n \setminus \{0\}$, and a classic result of Alexandrov guarantees that the above second-order Taylor expansion holds for any convex function on almost every point $x$. It is also known that the norm of any separable Banach space can be approximated arbitrarily well by a power-type 2 norm that is twice differentiable on $R^n \setminus {0}$ (see Lemma 2.6 here). But I wonder if the original norm itself has a second-order Taylor expansion.

Answer 1

Assuming you meant uniform smoothness instead of uniform convexity, and that the inequality $$ \| x + y \| + \|x - y\| \leq 2 + C \|y\|^2 $$ is exactly what you intended, then you have a counterexample on $\mathbb{R}^3$ with

$$\|(x,y,z)\| = \sqrt{x^2 + \sqrt{y^4 + z^4}} $$

Supposing the Taylor expansion exists, then we have $$ \|(1,y,z)\| = 1 + \frac12 \langle A(y,z), (y,z)\rangle + o(y^2 + z^2) $$ This requires $$ 1 + \sqrt{y^4 + z^4} \approx 1 + \langle A(y,z), (y,z) \rangle $$ or $$ y^4 + z^4 = \langle A(y,z), (y,z)\rangle^2 $$ which is impossible.