# Is norm of $\mathcal{C}^2(X,\mathbb{R})$?

Given a Banach space $\mathcal{X}$, consider $\phi(u)=\|u\|_{\mathcal{X}}^2,$ is $\phi\in\mathcal{C}^2(\mathcal{X},\mathbb{R})$?

For a Hilbert space $\mathcal{H}$, we have $\left\langle\phi'(u),v\right\rangle=2(u,v)_{\mathcal{H}},\forall v\in \mathcal{H}$, but now what is the gateaux derivative of $\phi$ in a Banach space?

Usually, when we calculate the frechet derivative, we first guess the gateaux derivative, and then prove that the gateaux derivative is continuous, but now I even can't guess the gateaux derivative of $\phi$.

• Answer below is correct, but maybe $R^2$ with supremum norm would be an easier counterexample. Try differentiating that norm at $(1,1)$ Commented Dec 25, 2017 at 9:09
• You're right~So what condition should be added to $\mathcal{X}$ or $\phi$ to grarantee that $\phi\in\mathcal{C}^2(\mathcal{X},\mathbb{R})$? Commented Dec 30, 2017 at 5:30
• The differentiability of the norm is a property of interest in itself, and there is no shorter equivalent characterisation (especially if you want to have exactly C^2 differentiability) as far as I know. Google 'differentiable' or 'smooth' norms to find out when does have equivalent norms that are differentiable (as in the other answer) Commented Jan 2, 2018 at 10:47
• ok,thx for answering~ Commented Jan 13, 2018 at 4:02

In general, it is not necessary to hold $\phi\in C^2(X,\mathbb R).$ For example, $X=C[a,b]$ with the $L^\infty$ norm.
However, if $X$ is reflexive, then there exists a differentiable norm which is equivalent to the original norm of $X$.