In an inner product space X, if ||λx + (1-λ)y|| = ||x|| for all λ ∈ [0, 1] then x = y

Question

I tried to prove that, in an inner product space $X$, if $\|λx+(1-λ)y\|$ = $\|x\|$ for all $λ$ $∈$ $[0, 1]$ and for $x$ and $y$ from $X$, then $x$ $=$ $y$.

Is it correct to plugin particular value of $λ$ to get the result? Or is there any simple way to prove it?

Answer 1

Square both sides to get $$\lambda^2 \|x-y\|^2 + \|y\|^2 + \lambda \langle x-y, y\rangle = \|x\|^2$$ for all $\lambda \in [0,1]$. Taking $\lambda = 0$ in the original condition shows that $\|x\|=\|y\|$, so two terms above can be canceled. Thus for all $\lambda > 0$ we have $$\lambda \|x-y\| = \langle y - x, y\rangle.$$

Answer 2

Consider $\lambda = 0, {1\over2}$

$\|y\|=\|{x+y\over 2}\|=\|x\|$

$\|{x+y\over 2}\|=\sqrt{{\|x\|^2\over4}+{\|y\|^2\over4}+{\langle x,y\rangle\over2}}=\sqrt{{\|x\|^2\over2}+{\langle x,y\rangle\over2}}\leq \|x\|$ with equality if and only if $x=y$

Answer 3

Let $f(\lambda) = {1 \over 2} \|\lambda x + (1-\lambda) y \|^2$. Since $f$ is constant for $\lambda \in [0,1]$, we must have $f'(\lambda) = 0$ for $\lambda \in (0,1)$.

Note that $f'(\lambda) = \lambda \|x-y\|^2 -\|y\|^2 + \langle y, x \rangle$.

It follows that $\|x-y\| = 0$ or equivalently, $x=y$.