Pre-Hilbert Spaces properties of norm

Question

How can I prove that in a pre-Hilbert space

$$||\lambda x + (1- \lambda) y|| = ||x||, \forall \lambda \in [0,1] \Rightarrow x = y$$.

Also, is this property also true in normed spaces?

Answer 1

Inner product spaces satisfy the parallelogram identity: $$||x+y||^{2}+||x-y||^{2}=2(||x||^{2}+||y||^{2})$$ Note that, taking $\lambda = 0$, we get $||y|| = ||x||$ and taking $\lambda = 1/2$ we get $||x+y||=2||x||$. Thus, $||y||^{2}=||x||^{2}$ and $||x+y||^{2}=4||x||^{2}$. Thus, using the parallelogram identity, we get: $$||x-y||^{2} = 2(||x||^{2}+||x||^{2}) - 4||x||^{2} = 0 \Rightarrow x = y$$ since $||x|| = 0$ iff $x = 0$. To see how this fail to be true in normed spaces, take $\mathbb{R}^{2}$ with, say, the max norm.

Answer 2

Letting $\lambda=0$ gives $\|x\|=\|y\|$. So $\|\lambda x+(1-\lambda)y\|=\|x\|$ gives $$ \lambda^2 \|x\|^2+2\lambda(1-\lambda)x\cdot y+(1-\lambda)^2\|y\|^2=\|x\|^2. $$ Putting $\|y\|=\|x\|$ in it, one has $$ 2\lambda(1-\lambda)x\cdot y+(1-\lambda)^2\|x\|^2+(\lambda^2-1)\|x\|^2=0 $$ or $$ \lambda(1-\lambda)(x\cdot y-\|x\|^2)=0. $$ So $\|x\|^2=x\cdot y$. Thus $$ \|x-y\|^2=\|x\|^2-2x\cdot y+\|y\|^2=2(\|x\|^2-x\cdot y)=0 $$ which implies $x=y$.