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I am considering these functions: For $i=1,\ldots,n$, define $$f_i(\lambda_1,\dots,\lambda_n)=\frac{1}{(n-2)^2}\sum_{k,l\neq i}(\lambda_k-\lambda_l)^2-\frac{2n}{n-2}\sum_{k=1}^n\lambda_k^2.$$ Suppose that $$\lambda_1+\cdots+\lambda_n=R$$ where $R$ is a positive constant. I wonder what the minimum value of $f_i(\lambda_1,\ldots,\lambda_n)$ is.

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@Paul: For the first sum, do you mean $k \ne l$ instead of $k, l \ne i$? –  André Nicolas Jun 11 '11 at 5:15
    
Lagrange multiplier method might help: en.wikipedia.org/wiki/… –  Did Jun 11 '11 at 5:17
    
It's not clear over which indices the first sum goes... no "$i$" appears anywhere behind that sum (?)... But my first choice would be, as Didier Piau said, Lagrange multipliers, though this might get ugly. –  Patrick Da Silva Jun 11 '11 at 5:21
    
Dear user6312 and Patrick Da Silva: No, it's not $k\neq l$. I editted the question. Hope that it's clearer now. Thank you for your comment. –  Paul Jun 11 '11 at 5:50
    
@Patrick: the $i$ is in the index of the function. It would seem that there are $n$ functions at work, each one slightly different. –  Arturo Magidin Jun 11 '11 at 5:52

2 Answers 2

up vote 1 down vote accepted

We suppose of course that $n \ge 3$.

If the $\lambda_j$ are $\ge 0$, it is clear that if we can make the first sum as small as it could conceivably be, and the second as large as it could conceivably be, $f_i$ will be as small as possible.

The smallest conceivable value of the first sum is $0$, and it is reached for example when all the $\lambda_j$ other than $\lambda_i$ are $0$.

Now given any non-negative $\lambda_j$ summing to $R$, the maximum value of the second sum is $R^2$, reached when all $\lambda_j$ but one are $0$.

This is an easy general fact. Probably the easiest proof is the observation that $$\sum \lambda_j^2 \le \left(\sum \lambda_j\right)^2$$ (just expand the right-hand side, the "mixed" terms are non-negative). Clearly we have equality iff one $\lambda_j$ is $R$ and the rest are $0$.

By the way, but not relevant to your problem, the minimum of $\sum \lambda_j^2$ is reached when the $\lambda_j$ are equidistributed, meaning that $\lambda_j=R/n$ for all $j$.

Thus, if the $\lambda_j$ are $\ge 0$, we can simultaneously minimize the first sum and maximize the second by choosing $\lambda_i=R$ and $\lambda_j=0$ for $j \ne i$.

If the condition is that the $\lambda_j$ are positive, then there is no minimum, but your expression can be made arbitrarily close to $-2nR^2/(n-2)$ by choosing $\lambda_i$ very close to $R$ and the remaining $\lambda_j$ (say) equal and very close to $0$. Under the positivity constraint, although the minimum does not exist, the infimum does and is equal to $-2nR^2/(n-2)$.

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There is no minimum value if the $\lambda_k$ are unrestricted: choose $\lambda_k=-tR$ for every $k\ne i$ and $\lambda_i=(1+(n-1)t)R$ such that the sum is $R$, then the first sum in $f_i$ is zero and the value of $f_i$ goes to $-\infty$ when $t\to\pm\infty$.

If the $\lambda_k$ are restricted to be nonnegative, most probably the minimum value is realized on the boundary of the positive quadrant. To wit, Lagrange multipliers method yields a point such that $\lambda_k$ does not depend on $k$ for $k\ne i$ and a value of $f_i$ at that point which behaves like $-2R^2/n$ when $n\to+\infty$, while, at $\lambda_i=R$ and $\lambda_k=0$ for every $k\ne i$, the value of $f_i$ is $-2nR^2/(n-2)$.

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