I want to prove that a ball for infinity norm is convex:
$$ B_\infty=\{x\in\mathbb R^n : \|x\|_\infty\le1\} $$
I came up with this proof and appreciate it if someone can help to verify if this is correct: \begin{align} \|x\|_\infty&=\|(1-\lambda)x+\lambda y\|_\infty\\&=\max_j|(1-\lambda)x|+\max_j|\lambda y|\\&=(1-\lambda)\max_j|x|+\lambda\max_j|y|\\&=(1-\lambda)+\lambda=1 \end{align}
P.S. This is a homework. Thanks!
Suppose $x, y \in B_{\infty}$. Let $x_i$ and $y_i$ denote components of $x$ and $y$. Then for any $\lambda \in [0, 1]$, \begin{align} \|(1 - \lambda)x + \lambda y\|_\infty & = \max_i |(1 - \lambda)x_i + \lambda y_i| & & \text{by definition of $\|\cdot\|_\infty$}\\ & \le \max_i (1 - \lambda)|x_i| + \max_i \lambda|y_i| & & \text{by $|a + b| \le |a| + |b|$} \\ & = (1 - \lambda)\|x\|_\infty + \lambda\|y\|_\infty & & \text{by definition of $\|\cdot\|_\infty$}\\ & \le (1 - \lambda) + \lambda & & \text{because $x \in B_\infty$ and $y \in B_\infty$}\\ & \le 1 \\ \therefore (1 - \lambda)x + \lambda y & \in B_{\infty} \end{align} Therefore, $B_{\infty}$ is convex.