Minkowski Functional of Convex Absorbing Set is Sublinear

Theorem

Let $\GF \in \set {\R, \C}$.

Let $X$ be a vector space over $\GF$.

Let $p$ be a seminorm on $X$.

Let $A$ be a convex absorbing set.

Let $\mu_A$ be the Minkowski functional of $A$.

Then for each $x, y \in X$ we have:

$\map {\mu_A} {x + y} \le \map {\mu_A} x + \map {\mu_A} y$

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Proof

Let $\epsilon > 0$.

By the definition of infimum, we can pick $t > 0$ such that:

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$\map {\mu_A} x \le t \le \map {\mu_A} x + \epsilon$

and $t^{-1} x \in A$.

We can also pick $s > 0$ such that:

$\map {\mu_A} y \le s \le \map {\mu_A} y + \epsilon$

and $s^{-1} y \in A$.

We have:

$\ds \frac t {s + t} + \frac s {s + t} = 1$

So, since $A$ is convex, we have:

$\ds \frac t {s + t} \paren {\frac x t} + \frac s {s + t} \paren {\frac y s} = \frac {x + y} {s + t} \in A$

So, we have:

$\map {\mu_A} {x + y} \le s + t \le \map {\mu_A} x + \map {\mu_A} y + 2 \epsilon$

Since this inequality holds for all $\epsilon > 0$, we have:

$\map {\mu_A} {x + y} \le \map {\mu_A} x + \map {\mu_A} y$

$\blacksquare$

Sources

• 1991: Walter Rudin: Functional Analysis (2nd ed.) ... (previous) ... (next): $1.34$: Theorem