Equivalent relations on norms (With inner product)

Question

For $x,y\in X.$ Where $X$ is an inner product space, Prove that the following are equivalent:

1. $||x+y||=|||x||+||y||$
2. $<x,y>=||x||.||y||$
3. $x=\lambda y$ or $y=\lambda x$ for some $\lambda\geq0$

My attempt
To prove $1\rightarrow2$, I tried to apply the lemma $||x+y||=||x||^2+2Re<x,y>+||y||^2$, But then I end up with $Re<x,y>=||x||.||y||$

For $2\rightarrow3$ I couldn't see a correct path.

For $3\rightarrow1$ It was able to obtain it by direct substitution of $x=\lambda y$ (Or $y=\lambda x $)in $1$

I would appreciate a help

Answer 1

(1) $\Rightarrow$ (2): to continue, by Cauchy-Schwarz you have $$\| x \| \cdot \|y \|= \operatorname{Re}\langle x,y \rangle \leq | \langle x,y \rangle | \leq \| x \| \cdot \|y \|$$

This implies $\operatorname{Re}\langle x,y \rangle = | \langle x,y \rangle |$. Now, if $\langle x,y \rangle =a+ib$ you get that $$a=\sqrt{a^2+b^2}$$ from where you get immediately that $b=0$.

Deduce from here that $\langle x, y\rangle = \operatorname{Re}\langle x,y\rangle = \|x \| \cdot \|y \|$.

(2) $\Rightarrow$ (3) You have equality in Cauchy-Schwarz, thus....

Added The C-S says that $$| \langle x, y \rangle | \leq \| x \| \cdot \|y \|$$ with equality if and only if $x,y$ are linearly dependent.

Since you have equality, it follows that $x,y$ are linearly dependent, meaning either $x= \lambda y$ or $y = \lambda x$.

Now use $\langle x, y \rangle =\| x \| \cdot \|y \|$ to deduce that $\lambda \geq 0$.

Answer 2

For $2$ implies $3$: assume $x\neq 0$ without loss of generality. The assumption implies that $$\left\|y-\frac{\langle x,y \rangle}{\langle x,x \rangle}x\right\|^2=0.$$ (just expand, using $\|\cdot\|^2=\langle\cdot,\cdot\rangle$.)