Gateaux and Frechet derivatives Let $X$ be a reflexive Banach space. We know then that the norm is Fréchet differentiable outside zero (actually there exists an equivalent one that is differentiable).
Now consider $f \colon X \to \mathbb{R},\ f(x)=|x|^2$.
Then its Fréchet derivative in $x_0$ is $2x_0$ so it is an element of $X$. But shouldn't the Fréchet derivative of $f$ (and also the Gateaux derivative) be an operator in $L(X,\mathbb{R})$ and so a functional?
What am I missing?
 A: Let's look at $F(x)=\|x\|^2$ on a Hilbert space $X$ over $\mathbb{R}$.
The Gateau derivative is a functional $dF:X\times X\to\mathbb{R}$ defined by
$$
dF(x; h)
:= \left.\frac{d}{d\lambda}F(x+\lambda h)\right|_{\lambda=0}
.
$$
Here,
$$
F(x+\lambda h)
= (x+\lambda h)^2
= \langle x+\lambda h, x+\lambda h\rangle
= \langle x, x \rangle + \langle x, \lambda h \rangle + \langle \lambda h, x\rangle + \langle \lambda h, \lambda h \rangle
\\
= \langle x, x \rangle + \lambda \langle x, h \rangle + \lambda \langle h, x\rangle + \lambda^2 \langle h, h \rangle
$$
so
$$
dF(x; h)
= \langle x, h \rangle + \langle h, x\rangle
= 2 \langle x, h \rangle
= \langle 2x, h \rangle
.
$$
The Fréchet derivative is a linear operator $X\to\mathbb{R}$, which in this case is $h \mapsto \langle 2x, h\rangle$ since
$$
\frac{| \|x+h\|^2 - \|x\|^2 - \langle 2x, h\rangle |}{\|h\|}
= \frac{| (\langle x, x\rangle + \langle x, h\rangle + \langle h, x\rangle + \langle h,h\rangle) - \langle x,x\rangle - \langle 2x, h\rangle |}{\|h\|}
\\
= \frac{\langle h, h\rangle}{\|h\|} 
= \frac{\|h\|^2}{\|h\|} 
\to 0
$$
as $\|h\|\to0.$
A: Consider $(\mathbb{R}^2,||.||_1)$ where $||(x,y)||_1=|x|+|y|$.
Then you have
$$||(x,1)||_1^2=x^2+2|x|+1$$
which is obviously non differentiable at $x=0$.
