x orthogonal to y equivalent conditions Show that for vectors $ x,y \in H $ condition $ x \bot y $ is equivalent to:
a) $ \|x+ \lambda y\|=\|x- \lambda y\|   $
b) $ \|x+ \lambda y\| \geq\|x\|   $
For every $ \lambda $
To show that $ x \bot y \Rightarrow$ a I have:
$ \|x+ \lambda y\|^2=\langle x+ \lambda y, x+ \lambda y \rangle= \langle x, x+ \lambda y \rangle+ \langle \lambda y, x+ \lambda y \rangle= \overline{\langle x+ \lambda y, x\rangle} + \overline{\langle x+ \lambda y, \lambda y\rangle}= \overline{\langle x, x\rangle + \langle \lambda y, x\rangle }+ \overline{\langle x, \lambda y\rangle + \langle \lambda y, \lambda y\rangle }= \|x\|^2+ \overline{\lambda}\langle x,y \rangle+\lambda \langle y,x \rangle+ |\lambda|^2 \|y\|^2 $
Now because $\langle x,y \rangle=0$ we have
$ \|x+ \lambda y\|^2=  \|x\|^2 + |\lambda|^2 \|y\|^2$
And analogously
$ \|x- \lambda y\|^2=\langle x- \lambda y, x- \lambda y \rangle= \langle x, x- \lambda y \rangle- \langle \lambda y, x- \lambda y \rangle= \overline{\langle x- \lambda y, x\rangle} - \overline{\langle x- \lambda y, \lambda y\rangle}= \overline{\langle x, x\rangle - \langle \lambda y, x\rangle }- \overline{\langle x, \lambda y\rangle - \langle \lambda y, \lambda y\rangle }= \|x\|^2- \overline{\lambda}\langle x,y \rangle-\lambda \langle y,x \rangle+ |\lambda|^2 \|y\|^2 $
Taking $\langle x,y \rangle=0$ we have
$ \|x- \lambda y\|^2=  \|x\|^2 + |\lambda|^2 \|y\|^2$
So $ \|x+ \lambda y\|^2=  \|x- \lambda y\|^2$ and  $ \|x+ \lambda y\|=  \|x- \lambda y\|$
That shows that  $ x \bot y \Rightarrow a $ but how can I proof that  $ a \Rightarrow x \bot y $
I have the same problem with b. It is easy to show that $ x \bot y \Rightarrow b $:
Taking $\langle x,y \rangle=0$ we have
$  \|x+ \lambda y\|^2=  \|x\|^2 + |\lambda|^2 \|y\|^2 \geq  \|x\|^2$ so $  \|x+ \lambda y\| \geq \|x\|^2 $
How to prove that  $ b \Rightarrow x \bot y $?
 A: I am assuming that $H$ is a real Hilbert space but the complex case needs only obvious modifications.
Squaring both sides of a) we get $\|x\|^{2}+\lambda^{2}\|y\|^{2}+2\lambda \langle x, y \rangle=\|x\|^{2}+\lambda^{2}\|y\|^{2}-2\lambda \langle x, y \rangle$ which gives $\langle x, y \rangle=0$.
Similarly, b) gives $\lambda^{2}\|y\|^{2}+2\lambda \langle x, y \rangle \geq 0$.If this holds for all real $\lambda$ then we can divide by $\lambda$ and let $\lambda \to 0$ to see that $\langle x, y \rangle=0$
A: Note that $\overline{\lambda} \langle x,y\rangle + \lambda \langle y,x\rangle = 2\Re (\lambda\langle x,y\rangle)$ so from the alternative expressions for $\|x+\lambda y\|^2$ and $\|x-\lambda y\|^2$ which you have obtained you get that $\Re(\lambda \langle x,y\rangle) = 0$  for every $\lambda \in \mathbb{C}$.
Now, if you take $\lambda = \overline{\langle x,y\rangle}$, then you get that $\Re(|\langle x,y\rangle|^2) = 0$, hence $\langle x,y\rangle = 0$.
For the second point, after squaring we end up with $|\lambda|^2\|y\|^2 + 2\Re(\lambda\langle x,y\rangle) \geq 0$, we take $\lambda = \gamma \overline{\langle x,y\rangle}$ for $\gamma<0$.
Then we divide by $\gamma$ and take the limit $\gamma\to 0^-$, similarly to what Kavi Rama Murthy suggests in his answer. We obtain that $|\langle x,y\rangle|^2 \leq 0$.
