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

Question

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

Answer 1

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)

Answer 2

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