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

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- May 31st 2009, 06:47 AMfrater_cpAnnihilator 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 AMTheEmptySet

Let $\displaystyle x,y \in M \perp$

Note to be in $\displaystyle M \perp$ $\displaystyle <x,m> =<y,m> = 0, \forall m \in M$

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

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

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

$\displaystyle 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 AMfrater_cp
Thank you for your replies.