# The component of f perpindicular to g in the Schwarz inequality?

I was wondering why this equation for the perpendicular component of f can be written as such, in the Schwartz inequality written for two vectors f and g.

Isn't the second part of the right side saying that the parallel component of f is the magnitude of the parallel component times the vector g? How can this be equal to the parallel component of f?

• hint: what is $\langle g|f_\perp\rangle$? Jul 22, 2017 at 18:50
• It should be 0, but how is this related to the relation above? I see how applying the bra vector g to each side gets me an equality that satisfies the equation, but what is the geometric intuition behind why it's the magnitude of the parallel component of f on g multiplied by the g vector? Is there a geometric intuition? Jul 22, 2017 at 19:37

You should ask yourself: parallel to what? The second term is the projection of $f$ in the direction parallel to $g$, while $f_\perp$ is the projection of $f$ perpendicular to $g$. You can verify this by noting that
$$|f_{\parallel}\rangle = \frac{ \langle g |f \rangle }{ \langle g |g \rangle } | g \rangle$$
is parallel to $g$ and has magnitude $\langle g |f \rangle$.