Can we conclude that $\langle x, y \rangle = 0\ $?

Question

Let $V$ be a complex inner product space. Let $x,y \in V$ be such that for any $\lambda \in \mathbb C$ we have the following inequality $:$

$$|\lambda|^2 \|y\|^2 + 2\ \mathfrak R \left (\lambda \langle x, y \rangle \right ) \geq 0.$$

From here can we conclude that $\langle x, y \rangle = 0\ $?

Actually I get this while working with a problem in functional analysis but I failed to justify it properly. Would anybody please shed some light on it? Any suggestion regarding this will be greatly appreciated.

Thanks in advance.

Answer 1

Let $\langle x, y \rangle \neq 0.$ Then there exists $s \gt 0$ and $\phi \in [0,2\pi)$ such that $\langle x, y \rangle = s\ e^{i \phi}.$ Let us take $r,\theta \in \Bbb R$ in such a way that $\|y\| \lt \sqrt {\frac {2s} {r}}$ and $\theta + \phi = \frac {3 \pi} {2}.$ Let $\lambda = r e^{i \theta}.$ Then from the above inequality it follows that $r^2 \|y\|^2 - 2rs \geq 0.$ In other words $\|y\|^2 \geq \frac {2s} {r},$ a contradiction.

Answer 2

Hint: Suppose $\langle x,y\rangle\neq 0$. Then you can choose $\lambda=-\frac{\epsilon}{\langle x,y\rangle}$ for an arbitrary (real) $\epsilon$. And then think about what happens if $\epsilon$ is a very small (positive) number.