QUESTION: Show that equality holds in the Cauchy-Schwarz inequality |⟨x, y⟩| ≤ ∥x∥ ∥y∥ . for x, y if and only if x and y are linearly dependent.
I have written up this but I not sure if it is a full proof, I'm especially not sure if my (==>) proof is correct. Any help would be lovely.
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FOR (==>)
So, if y=0 then the result is trivial. Hence, assume y is non-zero.
Let z = x - $\frac{<x,y>}{<y,y>}y$
Then, by the linearity axiom:
< z,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 so hence, the equality holds if x and y are linearly dependent.
NOW FOR (<==)
Case 1: X and Y are linearly dependent if there exists a real number a such that x=ay
By direct substitution, the equality holds:
< ay, y > ≤ ||ay|| ||y||
a < y,y > ≤ a||y|| ||y||
$ay^2$ ≤ $ay^2$
clearly, $ay^2$ = $ay^2$ so equality holds when x and y are linearly dependent.
Case 2: X and Y aren't linearly dependent if there exists a vector, z, orthogonal to y, and a non-zero scaler a, such that x = ay+z
||x|| ||y|| = ||ay+z|| ||y|| = a||$y^2$|| + ||z|| ||y|| = a$y^2$ + ||zy||
and
|< x,y >| = |< ay+z , y >| = |a||(< y,y > + < z,y >)| = a$y^2$ (since z and y are orthogonal)
clearly $ay^2 < ay^2 + ||zy||$
hence, equality does not hold if x and y aren't linearly dependent.
Your proof is ok, but you need only what you call Case 1 and Case 2. Indeed, in case one you prove "linearly dependent implies equality" and in case two you prove "linearly independent implies inequality", which is equivalent to "equality implies linearly dependent". By the way, let me add another possible proof. Assume without loss of generality that $x\neq 0$ (if one of the two is zero the assertion is trivial). A way to prove Cauchy-Schwarz inequality is to notice that $$ 4\langle x,y \rangle^2 -4||x||^2 ||y||^2 $$ equals the discriminant of the quadratic inequality given by $||y+tx||^2\geq 0$. As this is always satisfied, the discriminant is non positive, which yields the desired inequality. Therefore, the equality holds if and only if the discriminant is zero, which in turn means that the quadratic polynomial has a zero. This is only possible if $y+tx=0$ for some $t$, which tells that $x,y$ are linearly dependent.