Why am I getting two different results while solving an inequality that involves L1 norm?

[zeeshas901]

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

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

[MATH]f(\boldsymbol{x},\boldsymbol{y})=\frac{1}{2}\sum\limits_{i=1}^{3}|y_{i}-x_{i}|.[/MATH]
I have approached the question in two ways.

Solution 1: Numerical Example

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

Solution 2: General Case
Secondly, I tried to prove this for the general case as follows.
Consider,

[MATH]f(\boldsymbol{a},\boldsymbol{b})\le f(\boldsymbol{u},\boldsymbol{v})[/MATH]
[MATH]\frac{1}{2}\sum\limits_{i=1}^{2}|\frac{v_{i}}{v_{1}+v_{2}}-\frac{u_{i}}{u_{1}+u_{2}}|\le\frac{1}{2}\sum\limits_{i=1}^{3}|v_{i}-u_{i}|[/MATH]
[MATH]\frac{1}{(v_{1}+v_{2})(u_{1}+u_{2})}\sum\limits_{i=1}^{2}|v_{i}-u_{i}|\le\sum\limits_{i=1}^{3}|v_{i}-u_{i}|[/MATH]
[MATH]\frac{1}{(v_{1}+v_{2})(u_{1}+u_{2})}\le|v_{3}-u_{3}|[/MATH]
[MATH] 1 \le |v_{3}-u_{3}| (v_{1}+v_{2})(u_{1}+u_{2})[/MATH]
This last express does not look "correct" to me as [MATH]u_{i}[/MATH] and [MATH]v_{i}[/MATH] are proportions so the product of the LHR in the last expression "cannot" be greater than "1" (Isn't it?). Also note [MATH]u_{1}+u_{2}+u_{3}=1[/MATH] and [MATH]v_{1}+v_{2}+v_{3}=1[/MATH].

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!

 

[Cubist]

[MATH]\frac{1}{2}\sum\limits_{i=1}^{2}|\frac{v_{i}}{v_{1}+v_{2}}-\frac{u_{i}}{u_{1}+u_{2}}|\le\frac{1}{2}\sum\limits_{i=1}^{3}|v_{i}-u_{i}|[/MATH]
[MATH]\frac{1}{(v_{1}+v_{2})(u_{1}+u_{2})}\sum\limits_{i=1}^{2}|v_{i}-u_{i}|\le\sum\limits_{i=1}^{3}|v_{i}-u_{i}|[/MATH]

I spotted a couple of things wrong with the above step.
Instead:-

[MATH]\frac{1}{2}\sum\limits_{i=1}^{2}\left|\frac{v_{i}(u_{1}+u_{2})}{(v_{1}+v_{2})(u_{1}+u_{2})}-\frac{u_{i}(v_{1}+v_{2})}{(u_{1}+u_{2}) (v_{1}+v_{2}) }\right|\le\frac{1}{2}\sum\limits_{i=1}^{3}|v_{i}-u_{i}|[/MATH]
[MATH]\frac{1}{|(v_{1}+v_{2})(u_{1}+u_{2})|}\sum\limits_{i=1}^{2}|v_{i}(u_{1}+u_{2}) - u_{i} (v_{1}+v_{2}) |\le\sum\limits_{i=1}^{3}|v_{i}-u_{i}|[/MATH]
Note the modulus around the extracted factor. See if you can proceed from here!

 

[zeeshas901]

I spotted a couple of things wrong with the above step.
Instead:-

[MATH]\frac{1}{2}\sum\limits_{i=1}^{2}\left|\frac{v_{i}(u_{1}+u_{2})}{(v_{1}+v_{2})(u_{1}+u_{2})}-\frac{u_{i}(v_{1}+v_{2})}{(u_{1}+u_{2}) (v_{1}+v_{2}) }\right|\le\frac{1}{2}\sum\limits_{i=1}^{3}|v_{i}-u_{i}|[/MATH]
[MATH]\frac{1}{|(v_{1}+v_{2})(u_{1}+u_{2})|}\sum\limits_{i=1}^{2}|v_{i}(u_{1}+u_{2}) - u_{i} (v_{1}+v_{2}) |\le\sum\limits_{i=1}^{3}|v_{i}-u_{i}|[/MATH]
Note the modulus around the extracted factor. See if you can proceed from here!

Thank you for pointing out the mistakes. I have got that the above inequality does not hold if [MATH]u_{1}=v_{2}[/MATH] and [MATH]u_{2}=v_{1}[/MATH] (of course, that is not the only case).
However, I am wondering how can I prove that for n-dimensions? May you assist in the general case, please? That is:
(i) [MATH]n-[/MATH]dimensional proportion vectors [MATH]\boldsymbol{u}[/MATH] and [MATH]\boldsymbol{v}[/MATH](ii) m-dimensional proportion vectors [MATH]\boldsymbol{a}[/MATH] and [MATH]\boldsymbol{b}[/MATH] , where [MATH]m[/MATH] elements are subset of [MATH]n[/MATH] elements (as defined in the above question).

 

[Cubist]

(ii) m-dimensional proportion vectors [MATH]\boldsymbol{a}[/MATH] and [MATH]\boldsymbol{b}[/MATH] , where [MATH]m[/MATH] elements are subset of [MATH]n[/MATH] elements (as defined in the above question).

Post#1 only defines vectors a & b for case m=2. Is this perhaps the general definition for each element of a and b:-

[math] \mathit{a_i}=\frac{\mathit{u_i}}{\sum_{j=1}^{m}{\mathit{u_j}}} , \mathit{b_i}=\frac{\mathit{v_i}}{\sum_{j=1}^{m}{\mathit{v_j}}}[/math]
Have you tried starting this proof? Why not start in a similar way to the "m=2" proof above? If you get stuck then post your work. Hopefully myself, or another member, might be able to help.

--

BTW: I have not fully checked your numerical example in post#1, but f(u,v) ≠ 1/10 using your definition of f(x,y).

[math] f(u,v) = \frac{1}{2} \left(\frac{\left|7-5\right|}{10}+\frac{\left|2-3\right|}{10}+\frac{\left|1-2\right|}{10}\right) = \frac{1}{5} [/math]
Maybe it would help if you could post the full, original, question.