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This is a very beautiful sample problem from ISI MStat PSB 2014 Problem 2 based on the use and properties of a function . Let's give it a try !!

Let \( a_{1}<a_{2}<\cdots<a_{m}\) and \(b_{1}<b_{2}<\cdots<b_{n}\) be real numbers such

that \(\sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right| \text { for all } x \in \mathbb{R} \)

Show that \(m=n\) and \(a_{j}=b_{j}\) for \(1 \leq j \leq n\)

Differentiability

Mod function

continuity

Let , \(\sum_{i=1}^{m}\left|a_{i}-x\right|=\sum_{j=1}^{n}\left|b_{j}-x\right|=f(x) \text { for all } x \in \mathbb{R} \)

Then , \( f(x)=\sum_{i=1}^{m}\left|a_{i}-x\right| \) is not differentiable at \( x=a_1,a_2, \cdots , a_m \) ---(1)

As we know the function \(|x-a_i|\) is not differentiable at \(x=a_i\) .

Again we have , \( f(x) = \sum_{j=1}^{n}\left|b_{j}-x\right| \) it also not differentiable at \( x= b_1,b_2, \cdots , b_n \) ----(2)

Hence from (1) we get f has m non-differentiable points and from (2) we get f has n non-differentiable points , which is possible only when m and n are equal .

And also the points where f is not differentiable must be same in both (1) and (2) .

As we have the restriction that \( a_{1}<a_{2}<\cdots<a_{m}\) and \(b_{1}<b_{2}<\cdots<b_{n}\) .

So , we have \(a_{j}=b_{j}\) for \(1 \leq j \leq n\) .

\(a<b \in \mathbb{R} .\) Let \(f:[a, b] \rightarrow[a, b]\) be a continuous and differentiable on (a,b) . Suppose that \(\left|f^{\prime}(x)\right| \leq \alpha<1\) for all \(x \in(a, b)\) for some \(\alpha .\) Then prove that there exists unique \(x \in[a, b]\) such that \(f(x)=x\)

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This problem has an absolutely elementary solution. For any $x < \min (a_1, b_1)$, the given condition implies that:

\[\sum_{i=1}^m a_i - mx= \sum_{j=1}^n b_j - nx.\]

i.e.,

\[(m-n)x= \sum_{i=1}^m a_i - \sum_{j=1}^n b_j.\]

If $m\neq n$, this linear equation has only one solution. But the above equation should be valid for any of the infinitely many $x<\min (a_1, b_1)$. Hence we must have $m=n$ and $\sum_{i=1}^na_i =\sum_{i=1}^n b_i$.\\

Next, if $a_1 \neq b_1$, assume $a_1 < b_1$, and choose $x$ such that $a_1<x< \min(a_2, b_1)$. The given condition implies that

\[x-a_1 + \sum_{i=2}^m a_i - (m-1)x = \sum_{i=1}^nb_i -nx.\]

This, coupled with the previous result, implies that $x=a_1$, which is a contradiction. Hence we have $a_1=b_1$. Proceeding similarly, it follows that $a_i=b_i, \forall 1\leq i \leq n$.