Gateaux and Frechet derivatives

Question

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?

Answer 1

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 $\|h\|\to0.$

Answer 2

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 $x=0$.