Let $H$ a pre-hilbert complex space (complex space with inner product). Prove that $\|\cdot\|$ and $\|\cdot\|^2$ aren't differentiable (Fréchet) in $H$. I don't know how to use the fact that $H$ is a complex space. Does anyone know?
I already proved that for normed space the norm isn't differentiable in $x=0$. By contradiction, supose that the norm is differentiable in $x=0$. By definition, exists a bounded linear operator $T$ such that $$ \|0+h\| = \|0\|+T(h)+e(h) \quad \mbox{and}\quad \|0-h\| = \|0\|+T(-h)+e(h). $$ Hence, $\|h\|=e(h)$. Therefore, $\|\cdot\|$ isn't differentiable in $x=0$.