Let $X$ be a Banach space with its dual $X^*$. Consider the mapping $f: X\rightarrow \mathbb{R}$ given by $$ f(x)=\frac{1}{2}\|x\|^2. $$ We have know that when $X$ is a real Hilbert space ($X=X^*$) then $f$ is strongly convex with modulus $\lambda=1$, i.e. $$ \alpha f(x)+(1-\alpha)f(y)\geq f(\alpha x+(1-\alpha)y)+\frac{\alpha(1-\alpha)}{2}\|x-y\|^2 $$ for all $x, y\in X$ and $\lambda\in [0, 1]$. Moreover, $f$ is Frechet differentiable and $$ \nabla_F f(x)=x\quad \forall x\in X. $$ I do not know besides Hilbert space, what kind of Banach spaces do we have two above properties (Frechet differentiability and strong convexity) of the function $f(x)$.
I would like to thank for all constructive comments, helping and pointing out the references related to this problem.
Note.
- If $X^*$ is strictly convex then $f(x)=\frac{1}{2}\|x\|^2$ is Gateaux differentiable and $$ \nabla_G f(x)=\{x^*\in X^*: \|x^*\|^2=\|x\|^2=\langle x^*, x\rangle\}; $$
- If $f(x)$ is strongly convex with constant $\lambda=1$ then $X$ is a Hilbert space.
The spaces with Fréchet differentiable norm are usually called Fréchet smooth spaces... which of course is just a name. Šmulian gave a useful characterization of this property: it holds if and only if for every unit vector $x$ and every sequence of unit functionals $f_n$ such that $f_n(x)\to 1$, the sequence $\{f_n\}$ is norm-convergent.
The stronger property of having uniformly continuous Fréchet derivative of squared norm admits a simpler characterization: it's equivalent to the space being uniformly smooth (hence, also equivalent to the dual space being uniformly convex).
The property of strong convexity amounts to the norm being uniformly convex with quadratic modulus of convexity. This is quite a bit stronger than simply being uniformly convex.
Examples of the spaces that satisfy both properties: $L^p$ and $\ell^p$ for $1<p\le 2$. (For $2<p<\infty$, the space is uniformly smooth and uniformly convex, but the modulus of convexity has power type $p$, not $2$.)
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