# Prove that a projector is continuous using Cauchy-Schwarz's inequality

Given a vector $a$ in an Euclidean Space with $a\cdot a = 1$ ($\cdot$ = scalar product), then $P(b) = (a \cdot b)a$ defines the orthogonal projection $P$ on vector $a$.

How do you show that $P$ is continuous using Cauchy-Schwarz's inequality?

This is really simple: $$Px - Px_0 = (a\cdot x)a - (a\cdot x_0)a = \left(a\cdot x - a\cdot x_0\right)a = \big(a\cdot(x-x_0)\big)a.$$ Hence, $$\|Px - Px_0\| = |a\cdot(x-x_0)|\|a\| = |a\cdot(x-x_0)|.$$ Now, use Cauchy-Schwarz.
• It even shows that $P$ is Lipschitz continuous. I will not explain this because I do not regard Math Stack Exchange as a service for complete solutions. You will have to do a little work yourself, but I can assist. So, what's your problem in seeing from the line above that $P$ is continuous? What is your definition of continuity? Mar 6 '16 at 17:10
• If you use CS on the above to get an inequality, that inequality will making showing continuity with the $\epsilon-\delta$ definition very easy. Mar 6 '16 at 18:19