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.
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1$\begingroup$ Any vector $x$ orthogonal to $y$ trivially satisfies $x^\top y = 0 \leq 1$ but $\| x \|$ can be as large as you want. However, if you want $x^\top y \leq 1$ for all $y$ such that $\| y \| = 1$, it is straightforward. $\endgroup$ – VHarisop May 9 '18 at 0:31
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$\begingroup$ @VHarisop yes I want it for all $y$. How to show that? What straight forward thing I am missing here? $\endgroup$ – Frank Moses May 9 '18 at 0:37
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1$\begingroup$ Take $y = \frac{x}{\| x \|}$ and see what $x^\top y \leq 1$ gives you in that case. $\endgroup$ – VHarisop May 9 '18 at 0:45
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$\begingroup$ @VHarisop why consider only $y=\frac{x}{\|x\|_2}$ why not any other $y=\frac{p}{\|p\|_2}$? $\endgroup$ – Frank Moses May 9 '18 at 0:49
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$\begingroup$ Because this is the "most pathological" example you can find in the unit sphere. After you've solved this example, all the others follow from Cauchy-Schwarz. $\endgroup$ – VHarisop May 9 '18 at 0:52
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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 $.
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$\begingroup$ Is it right to say that (I think this was what you meant from pathalogical case) "we should take $y=\frac{x}{\|x\|}$ because for any other $y$ of the form $y=\frac{p}{\|p\|}$ will always produce a smaller value than for $y=\frac{x}{\|x\|}$ case due to Cauchy Schwarz Inequality, which becomes equality only when $y$ is a linear combination of $x$"? $\endgroup$ – Frank Moses May 9 '18 at 1:09
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1$\begingroup$ @FrankMoses: Yes, this is essentially what you are doing by considering $y$ collinear to $x$. $\endgroup$ – VHarisop May 9 '18 at 1:30
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It is not true. For example, in $\mathbb R^2$ take $x=(2,0)$, $y=(0,1)$. Then $$ x^Ty=0,\ \ \|x\|_2=2. $$
