$x^Ty \le 1$ for all $y$ with $||y||_2 = 1$ iff $||x||_2 \le 1$

Question

$x^Ty \le 1$ for all $y$ with $||y||_2 = 1$ iff $||x||_2 \le 1$

I can prove $\Leftarrow$ but I can't see a way to prove $\Rightarrow$.

I start with Cauchy's inequality and have $x^Ty \le ||x||_2 ||y||_2 \le ||x||_2$ since $||y||=1$ but I can't figure a way to show that $||x||_2 \le 1$ from this.

Anyone have any ideas?

Answer 1

Let $y = \frac{x}{\|x\|_2}$, so $\|y\|_2 =1$. Then $$x^T y = \frac{x^T x}{\|x\|_2}=\frac{\|x\|_2^2}{\|x\|_2}=\|x\|_2,$$ giving the required result.

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A little idea of why this is the natural approach to take: Our goal is to show that $\|x\|_2 \leq 1$, given that $\forall y, \|y\|_2 = 1 \implies x^T y \leq 1$. Since we know some statement is true for any $y$, it makes sense to try it at a particular $y$. A natural choice for $y$ is then just $x$, bearing in mind that $\|x\|_2 = x^T x$. But, $x$ might not have the required magnitude. So, just scale it to make it work.