# Characteristics of Birkhoff-James Orthogonality

## Theorem

Let $\struct {V, \norm {\,\cdot\,} }$ be a normed linear space.

Let $x, y \in V$.

Then $x$ and $y$ are Birkhoff-James orthogonal if and only if either:

$(1): \quad x = 0$

or:

$(2): \quad$ there exists a continuous functional $f$ on $\struct {V, \norm {\,\cdot\,} }$ such that:
$\norm f = 1$
$\map f x = \norm x$
$\map f y = 0$

## Proof

### Necessary Condition

Let $x \perp_B y$.

Let $V' \subset V$ be the subspace spanned by $x$ and $y$.

Define $\overline f$ on $V'$ as:

$\map {\overline f} {a x + b y} = a \norm x$

for $a$ and $b$ scalars.

Clearly, $\overline f$ is linear and:

$\map {\overline f} x = \norm x$
$\map {\overline f} y = 0$

Further:

 $\ds \norm {a x + b y}$ $=$ $\ds \size a \norm {x + \dfrac b a y}$ $\ds$ $\ge$ $\ds \size a \norm x$ $\ds$ $=$ $\ds \size {\map {\overline f} {a x + b y} }$

proving that $\overline f$ is a bounded functional of norm $1$.

Now by Hahn-Banach Theorem, $\overline f$ can be extended to a functional $f$ on $V$ such that $\norm f = \norm {\overline f} = 1$

This proves the necessity.

$\Box$

### Sufficient Condition

Let such a functional $f$ on $V$ exist.

Then for any scalar $\lambda$:

$\norm {x + \lambda y} \ge \size {\map f {x + \lambda y} } = \norm x$

establishing the sufficiency.

$\blacksquare$