Reverse Triangle Inequality/Seminormed Vector Space

Theorem

Let $\struct {K, +, \circ}$ be a division ring with norm $\norm {\,\cdot\,}_K$.

Let $X$ be a vector space over $K$.

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

Then:

$\forall x, y \in X : \size {\map p x - \map p y} \le \map p {x - y}$

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Proof

We have:

\(\ds \map p x - \map p y\)

\(=\)

\(\ds \map p {x - y + y} - \map p y\)

\(\ds \)

\(\le\)

\(\ds \map p {x - y} + \map p y - \map p y\)

Seminorm Axiom $\text N 3$: Triangle Inequality

\(\ds \)

\(=\)

\(\ds \map p {x - y}\)

We also have:

\(\ds \map p y - \map p x\)

\(=\)

\(\ds \map p {y - x + x} - \map p x\)

\(\ds \)

\(\le\)

\(\ds \map p {y - x} + \map p x - \map p x\)

Seminorm Axiom $\text N 3$: Triangle Inequality

\(\ds \)

\(=\)

\(\ds \map p {y - x}\)

\(\ds \)

\(=\)

\(\ds \norm {-1}_K \map p {x - y}\)

Seminorm Axiom $\text N 2$: Positive Homogeneity

\(\ds \)

\(=\)

\(\ds \map p {x - y}\)

Norm of Negative of Unity of Division Ring

Therefore:

$-\map p {x - y} \le \map p x - \map p y \le \map p {x - y}$

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In view of Definition of Absolute Value, the claim is proved.

$\blacksquare$