why two ans for single inequality in L1 norm?


why two ans for single inequality in L1 norm?

Postby zeeshas901 » Thu Mar 05, 2020 1:23 pm


Can anyone assist regarding the following a scalar measure of the difference between to vectors, please?

Thank you!

Let $\boldsymbol{u}=[u_{1}~~u_{2}~~u_{3}]$ and $\boldsymbol{v}=[v_{1}~~v_{2}~~v_{3}]$ be two known row vectors of proportions.~Suppose two row vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ are defined as follows:
Does the following equation holds?
$$f(\boldsymbol{a},\boldsymbol{b})\le f(\boldsymbol{u},\boldsymbol{v}),$$
where the function for $\boldsymbol{x}$ and $\boldsymbol{y}$ is defined as:

I have approached the question in two ways.

Firstly, I assumed some values for $\boldsymbol{u}=[5/10~~3/10~~2/10]$ and $\boldsymbol{v}=[7/10~~2/10~~1/10]$. Then, I computed $f(\boldsymbol{u},\boldsymbol{v})=1/10$ and $f(\boldsymbol{a},\boldsymbol{b})=11/72.$ Here, clearly $f(\boldsymbol{a},\boldsymbol{b}) < f(\boldsymbol{u},\boldsymbol{v})$ holds.

Secondly, I tried to prove this for the general case as follows. Consider,
$$f(\boldsymbol{a},\boldsymbol{b})\le f(\boldsymbol{u},\boldsymbol{v})$$
$$ 1 \le |v_{3}-u_{3}| (v_{1}+v_{2})(u_{1}+u_{2})$$

This last express does not look "correct" to me as $u_{i}$ and $v_{i}$ are proportions so the product of the LHR in the last expression "cannot" be greater than "1" (Isn't it?). Also note $u_{1}+u_{2}+u_{3}=1$ and $v_{1}+v_{2}+v_{3}=1$.

I am wondering where have I make a mistake in the second or first method? If both are correct, then why I am getting a different answer? Moreover, how may I correct if there is any mistake it?

Thank you!
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