# Suppose that $0 \notin \mathbb B(x_0, R)$. Then $\langle x_0,x \rangle$ is strictly positive on $\mathbb B(x_0, R)$

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

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 $$\|x_0\| \ge R$$, $$\|x\|^2 -2 \langle x, x_0 \rangle + \|x_0\|^2 . Hence $$\langle x_0,x \rangle = \langle x, x_0 \rangle > (1/2) \|x_0\|^2 > 0$$. This completes the proof.

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

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)
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 $$\frac{1}{2} ||x_0||$$.