Question related to Cauchy Schwarz Inequality

Question

If $x$ and $y$ are two vectors such that $\|y\|_2=1$ and $x$ is in the non-negative orthant then how to show that the condition $$x^Ty\leq1$$ leads to the condition that $$\|x\|_2\leq1.$$ I know that Cauchy Schwarz Inequality means that $$\|x^Ty\|_2\leq \|x\|_2\|y\|_2$$ but I still do not understand how in this case $$x^Ty=\|x^Ty\|_2.$$ Any help in this regard will be much appreciated. Thanks in advance.

Answer 1

If you need $x^\top y \leq 1$ to hold for all $y$ such that $\| y \| = 1$, try the following: take $y = \frac{x}{\| x \|}$. In that case

$$ x^\top y = \frac{x^\top x}{\| x \|} = \frac{\|x\|^2}{\|x\|} = \| x \| $$

So we know that $x^\top y \leq 1, \; \forall y : \|y\| =1 \Rightarrow \| x \| \leq 1$

Now, check that any $x$ that satisfies $\|x \| \leq 1$ satisfies $x^\top y \leq 1$: using the Cauchy-Schwarz inequality, you will get

$ x^\top y \leq \| x \| \| y \| \leq 1 \Rightarrow x^\top y \leq 1 $.

Answer 2

It is not true. For example, in $\mathbb R^2$ take $x=(2,0)$, $y=(0,1)$. Then $$ x^Ty=0,\ \ \|x\|_2=2. $$