Could someone please help me with this proof?
I have written up this but I not sure if it is a full proof, since it is an if and only if statement. Could someone read and inform me where I can go from here if there is more to prove?
So, if y=0 then the result is trivial. Hence, assume y is non-zero.
Let z = x - $x_y$ = x - $\frac{<x,y>}{<y,y>}y$ , y
= < x, y > - $\frac{< x,y >}{< y,y >}$ < y,y >
= 0
Thus, z (the projection of x) is a vector orthogonal to y and hence, if x and y are linearly dependent then the equality holds.
I am unsure whether this satisfies ==> direction of the proof or <== or both (i'm assuming it proves ==>, though).
I tried to further my proof by saying:
for (<==) let y=ax (ie. x and y are linearly independent) for some real a.
so, |< x,y >| = |< x,ax >| = ||x||.||ax||
now, |a| |< x,x >| = ||x||.||ax|| ≥ 0
if < x,x > = 0 then the inequality clearly holds.
but now I am unsure of how to expand on if < x,x > doesn't equal 0.
Following your idea:
$1).\ x$ and $y$ are linearly dependent, the there is an $a\in \mathbb C$ such that $x=ay$. A direct substitution shows that equality holds in Cauchy Schwartz.
$2).$ If $x$ and $y$ are not dependent, then there must be a vector $z\perp y$ and a non-zero scalar $a$ such that $x=ay+z, $ in which case $\|x\|^2\cdot \|y\|^2=(a^2\|y\|^2+\|z\|^2)\cdot \|y\|^2=a^2\|y\|^4+\|z\|^2\cdot \|y\|^2$ whereas $|\langle x,y\rangle|^2=a^2\|y\|^4$, so strict inequality holds.