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

$\endgroup$
  • $\begingroup$ $p$ is projection onto the subspace generated by $a$, isn't it? $\endgroup$ – Dylan Moreland Dec 16 '11 at 5:12
3
$\begingroup$

I think you mean "perpendicular to", not "parallel to", in the question.

It is clear from the formula for $p(x)$ in terms of $x$, $a$, and the inner product, namely $$ p(x) = \frac{\langle x, a\rangle}{\langle a, a\rangle} a, $$ that $x - p(x)$ is perpendicular to $a$. Just do a short calculation with the formula: $$ \begin{align} \langle x - p(x), a\rangle & = \langle x, a\rangle - \langle p(x), a\rangle \\ & = \langle x ,a\rangle - \left\langle \frac{\langle x, a\rangle}{\langle a, a\rangle} a, a \right\rangle \\ & = \langle x,a\rangle - \frac{\langle x, a\rangle}{\langle a, a\rangle} \langle a, a\rangle \\ & = \langle x, a\rangle - \langle x, a\rangle \\ & = 0. \end{align} $$ So perhaps your question is really why $p(x)$ is given by that formula. To answer that we'd need to know whatever definition you had of $p(x)$. In any case, for what you want, it's enough to just take the above formula as the definition of $p(x)$. To give the argument:

Fix any $y$ in $\{a\}^{\perp}$. The above calculation shows that $x - p(x)$ is perpendicular to $a$, and so is $y$, so the difference $x - p(x) - y$ is also perpendicular to $a$, and hence to $p(x)$ (as $p(x)$ is a scalar multiple of $a$ by definition). So by the Pythagorean theorem $$ \begin{align} d(x,y)^2 & = \|x - y\|^2 \\ & = \|x - p(x) - y + p(x)\|^2\\ & = \|x - p(x) - y\|^2 + \|p(x)\|^2 \\ & \geq \|p(x)\|^2 \end{align} $$ and taking square roots one deduces $$ d(x,y) \geq \|p(x)\| = \left\|\frac{\langle x,a\rangle}{\|a\|^2} a\right \| = \frac{|\langle x,a\rangle|}{\|a\|^2} \|a\| = \frac{|\langle x,a\rangle|}{\|a\|}. $$ The argument given for this inequality also shows that equality holds if and only if $\|x - p(x) - y\|^2 = 0$, ie, if and only if $y = x - p(x)$ (which we know is possible, as we proved $x - p(x)$ was indeed in $\{a\}^{\perp}$ ). This establishes $$ d(x,\{a\}^{\perp}) = \frac{|\langle x,a\rangle|}{\|a\|}. $$

$\endgroup$
  • $\begingroup$ It is clear as crystal. I apologize for the mistake parallel/perpenducular. I made a drawing which I was misleading. Thank you very much. $\endgroup$ – Zbigniew Dec 16 '11 at 8:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.