Given a convex absorbing set $A$ in a vector space $X$, the Minkowski functional of $A$ at $x$ is defined to be $$ \mu_{A}(x) = \left\{ t > 0 : \frac{x}{t} \in A \right\}. $$

One property of this functional is that it satisfies is subadditivity, $$ \mu_{A}(x + y) \leq \mu_{A}(x) + \mu_{A}(y) $$ for all $x, y \in X$.

In all the proofs I've seen for this, the final inequality is always $$ \mu_{A}(x + y) \leq \mu_{A}(x) + \mu_{A}(y) + 2\varepsilon \hspace{20pt}\forall\varepsilon > 0, $$ and the statement that follows is "This implies the desired inequality". Of course, I can see why this is the case (take $\varepsilon \to 0$), but how does one show that formally? What mathematical steps prove the implication?


We want to prove $$B\leq C+2\epsilon \quad\forall \epsilon>0 \Rightarrow B\leq C,$$ where $B=\mu_{A}(x+y)$ and $C=\mu_{A}(x)+\mu_{A}(y)$.

The proof proceeds by contradiction: suppose $B>C$. Then $\epsilon_0:=\frac{B-C}{4}>0$, so it is a valid choice for $\epsilon$. Hence we obtain $$B\leq C+2\epsilon_0=\frac{B+C}{2}<\frac{B+B}{2}=B,$$ an obvious contradiction.

  • $\begingroup$ Wonderful, thank you! $\endgroup$ – Bill Wallis Apr 12 '17 at 12:42

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