2
$\begingroup$

Let $a$ be a non-zero element of an Hilbert space $H$. I try to prove that for every $x\in H$, $$ d(x, \{a\}^{\perp})=\frac{\left|\langle x,a\rangle \right|}{\left\|a\right\|}. $$ So $d(x, \{a\}^{\perp})= \left\|x- p(x)\right\|$ where $p(x)$ is the orthogonal projection of $x$ on $\{a\}^{\perp}$, if I can show that $x- p(x)$ is parallel to $a$, then we conclude the proof by using the Cauchy-Schwarz inequality. In fact, writing $x=x- p(x) + p(x)$ it gives
$\left| \langle x,a\rangle \right|=\left|\langle x- p(x),a\rangle \right|$ because $p(x) \in \{a\}^{\perp}$.
By what argument can I say that $x- p(x)$ is parallel to $a$?

Thank's

  • 0
    $p$ is projection onto the subspace generated by $a$, isn't it?2011-12-16

1 Answers 1