I have been investigating under which conditions a subset $T$ of a Banach space $X$ induces a gauge norm via $$ ||f|| := \inf \{ \lambda > 0 \, : \, f \in \lambda T \}. $$ So far I have assumed:
- $T$ absorbs $X$,
- $T$ is symmetric ($x\in T \implies -x \in T$),
- $T$ is bounded,
- $T$ is convex,
and was able to show positive definiteness and homogeinity.
To show the triangle inequality I am certain I need to make use of the convexity. However, I have not been able to derive the $\Delta \neq$ under these assumptions. Any hint on the proof would be highly appreciated.
Edit: A proof can be found in 'Introduction to functional analysis' by R. Meise and D. Vogt. Let $x \in tT$ and $y \in sT$. We have $$ \frac{1}{t+s}(x+y) = \frac{t}{t+s}\frac{x}{t} + \frac{s}{t+s}\frac{y}{s}. $$ Since $x/t,y/s \in T$. Using convexity of $T$ and the above identity we obtain $x+y \in (t+s)T$. This yields $||x+y|| \leq t + s$, which implies $$ ||x+y|| \leq ||x|| + ||y||.$$
Concerning the duplication: The referenced question replaces condition 1. for $0\in T$ and $T$ open. This implies that $T$ absorbs $X$ but the reverse is not true (the closed unit ball in $X$ is absorbing but not open). It is easy to show that under the weaker assumption of 1. the norm equivalence still holds.
