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Is it true that if $\|\cdot\|:\mathbb{R}^n \to \mathbb{R}$ is an arbitrary norm then $\|\cdot\|^2$ is $\mathcal{C}^\infty$ function ?


If the norm comes from an inner product $\langle\cdot,\cdot\rangle$, i.e. $$\|x\|^2 = \langle x , x \rangle; $$ Then it is easy to see that $\| \cdot\|^2$ is a $\mathcal{C}^\infty$ function, because, if $\{v_1,...,v_n\}$ is an orthonomal basis with respect to the inner product $\langle\cdot, \cdot\rangle$, and $A$ $\in$ $M_n(\mathbb{R})$ is the change basis matrix of $\{e_1,...,e_n\}$ to $\{v_1, ..., v_n\}$ then

$$\|x\|^2=\langle x, x\rangle = (A\cdot x)^{\text{T}}G (A\cdot x)$$

where $G$ is a matrix such that $[G]_{ij} = [\langle v_i, v_j\rangle]$. Using the above formula is easy to conclude that $\|\cdot \|^2$ is a $\mathcal{C}^\infty$ function.

But when $\|\cdot \|$ is an arbitrary norm I just was able to conclude that $\| \cdot \|^2$ is a continuous function.

Can anyone help me?

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I'd say the answer is no, thinking of "square" balls with corners, such as the case with $\|\cdot\|_\infty$ in $\Bbb R^2$. We have $$\lim_{h\to 0^+} \frac{\|(1+h,1)\|_\infty^2-1}{h} = \lim_{h\to 0^+}\frac{(1+h)^2-1}{h} = 2$$and$$\lim_{h\to 0^-} \frac{\|(1+h,1)\|_\infty^2-1}{h} = \lim_{h\to 0^-}\frac{1-1}{h} = 0,$$so that $\dfrac{\partial\|\cdot\|_\infty^2}{\partial x}(1,1)$ does not exist.

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    $\begingroup$ Perfect answer, thx. $\endgroup$ – Matheus Manzatto Jul 1 '18 at 2:44
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Consider the taxicab norm squared $$ \Vert x \Vert^2_T = \left(\sum_{k=1}^n |x_k| \right)^2$$

Now the partial derivative with respect to the first component is $$ 2x_1 + 2\frac{x_1}{|x_1|} \left(\sum_{k=2}^n |x_k| \right)$$

which is clearly discontinuous.

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