A norm is strictly convex on a given normed space $X$ if, for every $x,y \text{ with } x \neq y \in X$ of norm 1, we have that $||x + y|| < 2$.
$||.||_{1} = \sum_{j}^{n}|x_j|$
Say we take the set $N = \{x \in \mathbb{C}^{n} s.t. ||x||=1 \text{ and } n \geq 2\}$ and take $x,y \in N$
So $||x||_1 = \sum_{j}^{n}|a_j + b_ji| = \sum_{j}^{n}[(a_j)^2 + (b_j)^2]^{1/2} = 1$.
And we would have $||x + y||= \sum_{j}^{n}|(a_j + b_ji) + (a’_j + b’_ji)|$.
After performing the sum, we’d end up with $||x + y||= \sum_{j}^{n}|(a_j + a’_j) +(b_ji + b’ _ji)|$
And then taking the modulus: $||x + y|| = \sum_{j}^{n}[(a’’_j)^2 +(b’’_j)^2]^{1/2}$
Can we show that this sum is strictly less than 2?
As for the second part $||.||_{\infty} = \sup_{j}|x_j|$ for $j=1,2,...,n$
And the sum would be $||x+y||= sup_j|x_j + y_j| = sup_j[(a’’j)^2 + (b’’j)^2]^{1/2}$
Then again, how is this actually less than 2.
Firstly, strictly convex means that for $x,y\in X$ with $x\neq y$ such that $\|x\|=\|y\|=1$, we have $\|x+y\|<2$.
Secondly, this does not hold for $\mathbb C^n$ with $n\geq 2$ and the $1$-norm. Indeed, consider $n=2$, $x=(1,0)$, $y=(0,1)$. Then $\|x\|_1=\|y\|_1=1$, while $\|x+y\|_1=2$.
Thirdily, this does not hold for $\mathbb C^n$ with $n\geq 2$ and the $\infty$-norm. Again, consider $n=2$, but with $x=(1,0)$ and $y=(1,1)$. Then $\|x\|_\infty=\|y\|_\infty=1$, while $\|x+y\|_\infty=2$.