Let $x,y\in\mathbb{R}^n$ be such that $||x+ty||\geq||x||$ $\forall \ t \in \mathbb{R}$. Show that $\vec{x} \cdot \vec{y} = 0$

Question

Let x,y$\in\mathbb{R}^n$ be such that $||x+ty||\geq||x||$
$\forall $ $t \in \mathbb{R}$

Show that x.y=0

Approach.

I expanded the above inequality then I get

$-2t$ x.y $\leq$ $t^2||y||^2$

$\forall $ $t \in \mathbb{R}$

Now If v=0 then x.y=0 AND if v$\neq0$ then let $t=\frac{x.y}{||v||^2}$

Putting into the simplified inequality we get $-2$x.y $\leq$ x.y

Hence $3$x.y$\leq0$

Which implies x.y$=0$

Is it correct? Correct me if not.

Someone told me that it also holds for general vector space too.How do I prove it.I am unable to do anuthing except expanding the inequality.

Thanks

Answer 1

We want to prove that $$ \|x\| \leq \|x + ty\| \quad \forall t \in \mathbb{R} \implies \langle x, y \rangle = 0. $$ We prove the contrapositive, namely $$ \langle x, y \rangle \neq 0 \implies \text{there exists } t \in \mathbb{R} \text{ such that } \|x\| \leq \|x + ty\| \text{ is false}. $$ As you did, rewrite the inequality as $$ 0 \leq 2t \langle x, y \rangle + t^2 \|y\|^2. $$ Since we're assuming $\langle x, y \rangle \neq 0$ for the contrapositive, and we only need to find one value of $t \in \mathbb{R}$ which violates the given inequality, we may assume $t \neq 0$ and divide by $t \langle x, y \rangle \neq 0$ to get $$ 0 \leq 2 + t \frac{\|y\|^2}{\langle x, y \rangle}. $$ Now, if $\langle x, y \rangle > 0$, the inequality is clearly false for $t \ll 0$. Similarly, if $\langle x, y \rangle < 0$, the inequality is clearly false for $t \gg 0$.

Answer 2

Let $$ f(t)=\|x+ty\|^2-\|x\|^2=t^2\|y\|^2+2tx\cdot y $$ Since $f(0)=0$, if $f(t)\ge0$ for all $t$ in a neighborhood of $0$, we must have $f'(0)=0$.

This means $2x\cdot y=0$.

Answer 3

For $||y||= 0$ there is nothing to show. So, let $||y|| \neq 0$.

According to our assumption we have $\forall t \in \mathbb{R}$: $$||x + ty||^2 = ||x||^2+ t^2||y||^2 + 2t (x\cdot y) \geq ||x||^2 $$ Rearranging and square completion gives: $$(t||y||- \frac{x\cdot y}{||y||} )^2 \geq \frac{(x\cdot y)^2}{||y||^2} \stackrel{t=\frac{x\cdot y}{||y||^2}}{\longrightarrow}x \cdot y = 0$$