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 $(\mathbb{R}^2, \|(x,y)\| := \sqrt{ \max(x^2 + 2y^2, \ 2x^2 + y^2 )})$.
 A: 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.
The first assertion is obvious, for the second we need a little work.
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 $(0,0)$.
Thus an FD-space need not be strictly convex.
