# Thread:

1. ## how to apply the Langrange equations in this case

In the proof to Theorem 7.3 from this paper on FNTFs, the authors invoke the so-called "Langrange equations." I assume they mean the Euler-Lagrange equations. (But maybe not...?) Unfortunately I'm not at all familiar with the Euler-Lagrange equations, and in reading what they are, I have no idea how to apply them in this case.

If anyone has some spare time and good will, can he/she please explain how to understand this?

The set-up:

Let $K=\mathbb{C}$ be the complex numbers and $S(K^d)$ the unit sphere in $K^d$ for some positive integer d. Let $\{x_n\}_{n=1}^N\subseteq S(K^d)$ be a fixed sequence in that unit sphere. Let $S=\{(a,b)\in\mathbb{R}^d\times\mathbb{R}^d:\lvert a\rvert^2+\lvert b\rvert^2=1\}$ be the unit sphere in $\mathbb{R}^d\times\mathbb{R}^d$, and define the function $\widetilde{FP}_l:S\to[0,\infty)$ by

$(a,b)\mapsto 2\sum_{n\neq l}(\langle a,a_n\rangle+\langle b,b_n\rangle)^2+(\langle b,a_n\rangle-\langle a,b_n\rangle)^2+1+\sum_{m\neq l}\sum_{n\neq l}|\langle x_m,x_n\rangle|^2,$

where the sums are otherwise over 1 through N, and l is some integer between 1 and N. Let $(a_l,b_l)\in S\subset\mathbb{R}^d\times\mathbb{R}^d$ be a local minimizer of $\widetilde{FP}_l$.

The problem:

Show that there exists a scalar $c\in\mathbb{R}$ such that both of the following equations hold:

(7.1) $\nabla_a\widetilde{FP}_l(a,b)|_{(a,b)=(a_l,b_l)}=c \nabla_a(\lvert a\rvert^2+\lvert b\rvert^2)|_{(a,b)=(a_l,b_l)};$

(7.2) $\nabla_b\widetilde{FP}_l(a,b)|_{(a,b)=(a_l,b_l)}=c \nabla_b(\lvert a\rvert^2+\lvert b\rvert^2)|_{(a,b)=(a_l,b_l)}.$

I assume that $\nabla_a,\nabla_b$ refer to the gradients on $a,b$, respectively. However I'm not sure about that.

If possible, I would like someone to show me in a textbook (I can get almost anything online or from my university library) what theorem to use, and what choices to make in applying the theorem. For instance, if a theorem calls for a function f, then what is a suitable choice of f in this case?

Thanks guys.
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