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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?

Thanks for your help.

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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.

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  • $\begingroup$ how does this show that P is continuous? $\endgroup$ Mar 6 '16 at 16:54
  • $\begingroup$ 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? $\endgroup$ Mar 6 '16 at 17:10
  • $\begingroup$ If you use CS on the above to get an inequality, that inequality will making showing continuity with the $\epsilon-\delta$ definition very easy. $\endgroup$
    – abnry
    Mar 6 '16 at 18:19
  • $\begingroup$ To me continuity in the 1st and second dimension implies that for a point c, if I take the lim of my function and make it tend towards c it the limit would be f(c). I visualize it as drawing a straight line without taking my pen of the paper. I have a difficult time geometrically interpreting continuity for higher dimensions in Rn. $\endgroup$ Mar 6 '16 at 19:08
  • $\begingroup$ So for every arbitrarly small e > 0 there exists an arbitrarly small d > 0 such that || x -a || < e implies || f(x) - f(a)|| < d $\endgroup$ Mar 6 '16 at 19:11

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