0
$\begingroup$

Good evening, I'm trying to solve this exercise about Hilbert space:

enter image description here

Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!


My attempt:

We have the norm induced by this inner product is $\|x\| = \sqrt{\langle x, x \rangle}$. Moreover, $\mathbb B(x_0, R) = \{x \in H \mid \|x-x_0\| < R\}$. It follows from $0 \notin \mathbb B(x_0, R)$ that $\|x_0\| \ge R$.

For all $x \in \mathbb B(x_0, R)$, we have $\|x-x_0\| < R$ and so $\|x-x_0\|^2 < R^2$. Hence $\langle x-x_0, x-x_0 \rangle < R^2$ or $\|x\|^2 -2 \langle x, x_0 \rangle + \|x_0\|^2 < R^2$. Because $\|x_0\| \ge R$, $\|x\|^2 -2 \langle x, x_0 \rangle + \|x_0\|^2 <R^2 \le \|x_0\|^2$. Hence $\langle x_0,x \rangle = \langle x, x_0 \rangle > (1/2) \|x_0\|^2 > 0$. This completes the proof.

$\endgroup$
  • $\begingroup$ Thank you so much @BrianMoehring! Could you please write your comment as an answer and I will accept it to close this question. $\endgroup$ – LAD Oct 8 '19 at 21:23
1
$\begingroup$

The last line should be $ \langle x, x_0 \rangle >(1/2)\|x\|^2$. Otherwise it looks fine (assuming the inner product in your hilbert space is real)

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Everything makes perfect sense. However it seems that $\phi(x) = \langle x,x_0\rangle \geq \frac{1}{2}(||x_0||^2-R^2)$ rather than $\frac{1}{2} ||x_0||$.

| cite | improve this answer | |
$\endgroup$

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.