# Are all FD-spaces strictly convex?

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 $$\forall x \in V \setminus \{0\}$$.

We call a Banach space $$(V, \|v\|)$$ strictly convex, iff $$\forall x \neq y \in V, \lambda \in (0,1)$$ if $$\|x\|=\|y\|=1$$, then $$x + \lambda(y-x) < 1$$.

Are all FD-spaces strictly convex?

Currently I know two classes of examples of FD-spaces and both of them satisfy this property:

All Hilbert spaces are FD-spaces.

Proof:

One can manually check, that $$h \mapsto \frac{h}{2\sqrt{x_0}}$$ is a Fréchet derivative for $$x \mapsto \sqrt{|x|}$$ in $$x_0 \neq 0$$. One can also manually check, that $$h \mapsto 2\langle v, h \rangle_V$$ is a Fréchet derivative for $$x \mapsto \langle x, x \rangle_V$$ in all $$v \in V$$. And it is a well known fact, that the composition of Fréchet derivatives of two functions is a Fréchet derivative of their composition. Thus, as $$\|v\|_V = \sqrt{\langle v, v \rangle_V}$$, we have, that $$h \mapsto \ \frac{\langle v, h \rangle_V}{\|v\|_V}$$ is a Fréchet derivative of $$\|v\|_V$$ in all $$v \in V \setminus \{0\}$$.

All Hilbert spaces are strictly convex

Proof:

If $$\langle x, x\rangle = 1$$ and $$\langle y, y \rangle = 1$$, then $$\langle x + \lambda(y-x), x + \lambda(y-x) \rangle = (1-\lambda)^2 + \lambda^2 + 2(1-\lambda)\lambda \langle x, y \rangle < (1-\lambda)^2 + \lambda^2 + 2(1-\lambda)\lambda = 1$$

Suppose $$(X, \Omega, \mu)$$ is a measurable space, $$n \in \mathbb{N}$$. Then $$L_{2n}(X, \Omega, \mu)$$ is FD-space.

Proof:

One can manually check, that $$h \mapsto 2n\int_X f^{2n-1}hd\mu$$ is the Fréchet derivative for $$\int_X f^{2n}d\mu$$.

Suppose $$(X, \Omega, \mu)$$ is a measurable space, $$n \in \mathbb{N}$$. Then $$L_{2n}(X, \Omega, \mu)$$ is strictly convex.

Proof:

Suppose $$\int_X f^{2n}d\mu = \int_X g^{2n}d\mu = 1$$ and $$f \neq g$$. Then $$\int_X (\lambda f + (1 - \lambda)g)^{2n}d\mu < \sum_{i=0}^2n C_n^i\lambda^i(1 - \lambda)^{2n - i}$$

However I do not know how to prove this statement in general.

Note however that not all strictly convex spaces are FD-spaces. An example of a strictly convex space that is not a FD-space is $$(\mathbb{R}^2, \|(x,y)\| := \sqrt{ \max(x^2 + 2y^2, \ 2x^2 + y^2 )})$$.

In $$V = \mathbb{R}^2$$, consider the set $$B$$ of points $$p$$ with $$\lVert p\rVert_{\infty}\leqslant 1$$, such that either (at least) one of the components is $$\leqslant 1/2$$ in absolute value, or the Euclidean distance to one of the four points $$(\pm 1/2, \pm 1/2)$$ is $$\leqslant 1/2$$. Thus we round the corners, and do so that the boundary curve is $$C^1$$. Let $$\lVert\,\cdot\,\rVert$$ be the Minkowski functional of $$B$$. Then $$(V,\lVert\,\cdot\,\rVert)$$ is not strictly convex, but since $$\partial B$$ is a $$C^1$$ curve, it is an FD-space.
Consider a point $$(x,y)$$ in the right half-plane with $$\lvert y\rvert < \frac{1}{2} x$$. Then $$\lVert (x,y)\rVert = x$$, which is continuously differentiable in that angular wedge. Consider next a point in the right half-plane with $$\frac{1}{2} x < y < 2x$$. Let $$t = \lVert (x,y)\rVert$$. Then \begin{aligned} &&(t^{-1}x - 1/2)^2 + (t^{-1}y - 1/2)^2 &= 1/4 \\ &\iff& (x - t/2)^2 + (y - t/2)^2 &= t^2/4 \\ &\iff& \frac{t^2}{4} - (x+y)t + x^2 + y^2 &= 0 \\ &\iff& (t/2 - (x+y))^2 &= 2xy \\ &\iff& 2(x+y) \pm 2\sqrt{2xy} &= t\,. \end{aligned} Looking at $$x = y = 1/2$$ shows that the correct sign is $$-$$, so $$\lVert (x,y)\rVert = 2(x+y) - 2\sqrt{2xy}$$ in the wedge $$\frac{1}{2}x < y < 2x$$ of the right half-plane. This is also continuously differentiable.
It remains to consider the line $$y = \frac{1}{2} x$$, where we must check that the derivatives fit together. Then the differentiability on all of $$\mathbb{R}^2 \setminus \{(0,0)\}$$ follows by symmetry. On the wedge $$\lvert y\rvert < \frac{1}{2} x$$ the derivative is constant, its matrix is $$\begin{bmatrix}1& 0\end{bmatrix}$$. On the wedge $$\frac{1}{2} x < y < 2x$$ the Jordan matrix is $$\begin{bmatrix} 2 - \sqrt{\frac{2y}{x}} & 2 - \sqrt{\frac{2x}{y}}\end{bmatrix}\,.$$ On the line $$y = \frac{1}{2}x$$ these coincide, hence the norm is continuously differentiable everywhere except at $$(0,0)$$.