$x \perp y$ if and only if $\Vert x + \alpha y \Vert \ge \Vert x \Vert$ for all scalars $\alpha$

Question

Here's Prob. 8 in the Problems after Sec. 3.2 in Introductory Functional Analysis With Applications by Erwine Kreyszig:

Show that in an inner product space, $x \perp y$ if and only if $\Vert x + \alpha y \Vert \ge \Vert x \Vert$ for all scalars $\alpha$.

If $x \perp y$, then $\langle x, y \rangle = 0$; so for any scalar $\alpha$, we have $$ \begin{align*} \Vert x + \alpha y \Vert^2 &= \langle x + \alpha y, x + \alpha y \rangle \\ &= \Vert x \Vert^2 + 2 \Re \bar{\alpha} \langle x, y \rangle + \vert \alpha \vert^2 \ \Vert y \Vert^2 \\ &= \Vert x \Vert^2 + \vert \alpha \vert^2 \ \Vert y \Vert^2 \\ &\ge \Vert x \Vert^2. \end{align*} $$ So $$ \Vert x + \alpha y \Vert \geq \Vert x \Vert. $$

Am I right?

Now how to prove the converse? It is my guess that we will have to put in a particular value for $\alpha$.

Answer 1

Assumig \begin{align*} \Vert x \Vert^2 + 2 \Re \bar{\alpha} \langle x, y \rangle + \vert \alpha \vert^2 \ \Vert y \Vert^2 \ge \Vert x \Vert^2, \end{align*} we get \begin{align*} 2 \Re \bar{\alpha} \langle x, y \rangle + \vert \alpha \vert^2 \ \Vert y \Vert^2 \ge 0 \end{align*} and \begin{align} \Re \bar{\alpha} \langle x, y \rangle \ge -\frac{1}{2} \vert \alpha \vert^2 \ \Vert y \Vert^2. \end{align}

Now take $\alpha = 2\varepsilon / \Vert y \Vert^2 > 0$, and you can cancel $\alpha$ from both sides to get \begin{align*} \Re \langle x, y \rangle \ge -\frac{1}{2} \alpha \Vert y \Vert^2 =-\varepsilon. \end{align*}

Since this holds for all $\varepsilon$, we get $\Re \langle x, y \rangle \ge 0$. Similarly, taking $\alpha = - 2\varepsilon / \Vert y \Vert^2$, we get $\Re \langle x, y \rangle \le 0$, and hence $\Re \langle x, y \rangle = 0$.

In the same manner, taking $\alpha = \pm i 2\varepsilon / \Vert y \Vert^2$, one can show that $\Im \langle x, y \rangle =0$.

A slightly shorter argument along the same lines, is as follows. Assume $\langle x, y \rangle = |\langle x, y \rangle| e^{i\theta}$. Taking $\alpha :=e^{i\theta} 2\varepsilon / \Vert y \Vert^2$, gives \begin{align*} \Re \bar{\alpha} \langle x, y \rangle = \Re \big(e^{-i\theta} |\alpha| |\langle x, y \rangle| e^{i\theta}\big) = |\alpha| |\langle x, y \rangle| \end{align*} from which we get $|\langle x, y \rangle| \le 0$ and similarly taking $\alpha :=e^{i(\theta-\pi)} 2\varepsilon / \Vert y \Vert^2$, gives $|\langle x, y \rangle| \ge 0$.

Answer 2

For the converse, if $\langle x,y \rangle \ne 0$ then in line 2 of your argument you can choose $\alpha \in \mathbb{R}$ to make the r.h.s. smaller than $||x||^2$, since for any constants $a \ne 0,b$ (in this case $a=2\langle x,y\rangle$ and $b=||y||$) you can choose $\alpha$ such that $a\alpha + b\alpha^2 < 0$. (the parabola $y=ax+bx^2$ intersects but isn't tangent to the x-axis.)

Answer 3

If $(x,y) \ne 0$ then $\|y\|\ne 0$, and the following is an orthogonal decomposition: $$ x = \left[x-\frac{(x,y)}{(y,y)}y\right]+\frac{(x,y)}{(y,y)}y. $$ Hence, $$ \|x\|^{2} = \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2}+\frac{|(x,y)|^{2}}{\|y\|^{2}} > \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2}. $$

Answer 4

$\boldsymbol{\langle x,y\rangle=0\implies\|x+\alpha y\|^2\ge\|x\|^2}$

$$ \begin{align} \|x+\alpha y\|^2 &=\|x\|^2+2\mathrm{Re}\left(\langle x,\alpha y\rangle\right)+|\alpha|^2\|y\|^2\\ &=\|x\|^2+2\mathrm{Re}\left(\overline{\alpha}\langle x,y\rangle\right)+|\alpha|^2\|y\|^2\\ &=\|x\|^2+|\alpha|^2\|y\|^2\\ &\ge\|x\|^2 \end{align} $$

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$\boldsymbol{\|x+\alpha y\|^2\ge\|x\|^2\implies\langle x,y\rangle=0}$

Let $\alpha=t\langle x,y\rangle$ for $t\in\mathbb{R}$, so that $\langle x,\alpha y\rangle=t|\langle x,y\rangle|^2\in\mathbb{R}$.

Suppose $\langle x,y\rangle\ne0$. For $-\frac2{\|y\|^2}\lt t\lt0$, we have $$ \begin{align} \|x+\alpha y\|^2 &=\|x\|^2+2\mathrm{Re}\left(\langle x,\alpha y\rangle\right)+|\alpha|^2\|y\|^2\\ &=\|x\|^2+2t|\langle x,y\rangle|^2+t^2|\langle x,y\rangle|^2\|y\|^2\\ &=\|x\|^2+t|\langle x,y\rangle|^2\left(2+t\|y\|^2\right)\\ &\lt\|x\|^2 \end{align} $$ Therefore, $\langle x,y\rangle=0$.