# Annihilator of a set M in inner product space X

• May 31st 2009, 06:47 AM
frater_cp
Annihilator of a set M in inner product space X
Show that the annihilator (M perpendicular) of a non-empty set M in an inner product space X, is a closed subspace of X.
• May 31st 2009, 08:32 AM
TheEmptySet
Quote:

Originally Posted by frater_cp
Show that the annihilator (M perpendicular) of a non-empty set M in an inner product space X, is a closed subspace of X.

Let $x,y \in M \perp$

Note to be in $M \perp$ $ = = 0, \forall m \in M$

Now lets show that it is closed. let $\alpha, \beta \in \mathbb{R}$

$<\alpha x+\beta y,m>=<\alpha x,m> +<\beta y,m>=\alpha +\beta =\alpha \cdot 0 +\beta \cdot 0 =0$
• May 31st 2009, 10:02 AM
Rebesques
I think the question demands "show it is topologically closed".

To that end, define for fixed $x\in M$ the function $f_x(y)=\langle x,y \rangle, \ y\in X$. This function is continuous (a fundamental property of the inner product), so $f_x^{-1}(\{0\})=\{y: f_x(y)=0\}$ is a closed set. This means

$M^{\perp}=\{y: \ \langle x,y \rangle=0, \ \forall x\in M\}=\{y: y\in f_x^{-1}(\{0\}), \ \forall x\in M\}=\cap_{x\in M}f_x^{-1}(\{0\})
$

is closed as the intersection of a family of closed sets.
• May 31st 2009, 10:38 AM
frater_cp