How to prove $1-L_{\infty} ~ Norm$ is non-convex (concave) function?

Question

I know that $L_\infty$ norm is convex and the 1-convex function will lead to a concave function.

My attempt to shows this as: \begin{align} \|(1 - \lambda)x + \lambda y\|_\infty & \le (1 - \lambda)\|x\|_\infty + \lambda\|y\|_\infty & & \text{Condition of convexity}\\ -\|(1 - \lambda)x + \lambda y\|_\infty & \geq -(1 - \lambda)\|x\|_\infty -\lambda\|y\|_\infty\\ 1-\|(1 - \lambda)x + \lambda y\|_\infty & \geq 1-(1 - \lambda)\|x\|_\infty -\lambda\|y\|_\infty & & \text{Condition of convcavity} \end{align}

Now, if this is correct, what is the convex overestimation of $1-\|x\|_\infty$?

Answer 1

To complete the argument, note that \begin{align*} 1-(1 - \lambda)\|x\|_\infty -\lambda\|y\|_\infty &= (1 - \lambda) - (1 - \lambda)\|x\|_\infty + \lambda - \lambda\|y\|_\infty \\ &= (1 - \lambda)(1 - \|x\|_\infty) + \lambda(1 - \|y\|_\infty), \end{align*} finishing the proof of concavity.