Reverse Triangle Inequality/Normed Vector Space

Theorem

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.

Then:

$\forall x, y \in X: \norm {x - y} \ge \size {\norm x - \norm y}$

Proof

Let $d$ denote the metric induced by $\norm {\, \cdot \,}$, that is:

$\map d {x, y} = \norm {x - y}$

From Metric Induced by Norm is Metric we have that $d$ is indeed a metric.

Then, from the Reverse Triangle Inequality as applied to metric spaces:

$\forall x, y, z \in X: \size {\norm {x - z} - \norm {y - z} } \le \norm {x - y}$

Then:

$\forall x, y \in X: \size {\norm x - \norm y} = \size {\norm {x - 0} - \norm {y - 0} } \le \norm {x - y}$.

$\blacksquare$

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.2$: Normed and Banach spaces. Normed spaces