minimum of $-(\cos k_1 + \cos k_2 + \cos k_3)$

Question

I want the minimum value of the function

$$- (\cos k_1 + \cos k_2 + \cos k_3)$$

under the constraint of

$$ k_1 + k_2 +k_3 = K , $$

where the constant $K \in [-\pi, \pi]$. I guess the minimum is achieved at $k_1 = k_2 = k_3 = K/3$, but find it difficult to prove it.

The problem is that $-\cos x $ is not completley convex on the real axis. Here we have three terms. It is conjectured that if we have $N\geq 3 $ terms, the minimum is always achieved by $k_1 = k_2 = \ldots = k_N = K/N $.

This problem is relevant to solid state physics. I want to minimize the kinetic energy of a few electrons in a lattice.

Answer 1

I confirm your guess for the case $N=3$.
By your hint in the comments,

$K$ is a constant in $[-\pi, \pi]$

follows, $k_{1..3}$ are almost free. That is why I disbelieved your minimum if $k_{1..3}=K/N$ assumption.
My reasoning, with $k_1=k_2=0$ the "terms" $\cos(k_1)$ and $\cos(k_2)$ have a maximum impact on the result, only $\cos(k_3)$ with $k_3=K$ is what it is. But, at $k_n=0$ the effect of a "little change" may be lower than it is for $k_3=K$. Hence I modified your function $- (\cos k_1 + \cos k_2 + \cos k_3)$ to the ansatz $$-\left(2\cos\left(e\right)+\cos\left(K-2e\right)\right)$$ The first derivative to $e$ set to $0$ and solved for $e$ yields four solutions: $$\left\{e=\frac{4m\pi+2\pi+K}{3}\mathrm{,}\\e=\frac{4m\pi+K}{3}\mathrm{,}\\e=4n\pi+\pi+K\mathrm{,}\\e=4n\pi-\pi+K\right\}$$ with $m$ and $n$ arbitrary integers including $0$ and negative values, I opted for $m=n=0$. Next I checked the sign of the second derivative with this solutions, if it results a minimum or not. It depends on $K$ but not for solution 2 -- in respect of the given range of $K$. Here the plot of the $2^{nd}$ derivative with the solutions 1..3:

But, I'd like also to see the results of the solutions, green is the second shown above --

Contrary to my expectations it turns out, a minimum is achieved by $\displaystyle e=\frac{K}{3}=k_1=k_2$ and $k_3=K-2e$ what is -- well, youknowit.

Something missing? -- Yes, same again with an ansatz where $k_1\ne k_2$.

Answer 2

Some thoughts.

Fact 1. Let $k_1, k_2, k_3 \in \mathbb{R}$ with $k_1 + k_2 + k_3 \in [-\pi, \pi]$. Then $-\cos k_1 - \cos k_2 - \cos k_3 \ge -3\cos \frac{k_1 + k_2 + k_3}{3}$.

Fact 2. Let $u, v \in [-\pi, \pi]$ and $n\ge 3$. Then $-n\cos \frac{u}{n} - \cos (u - v) + (n + 1)\cos \frac{v}{n + 1}\ge 0$.

Fact 3. Let $k_1, k_2, \cdots, k_n \in \mathbb{R}$ ($n \ge 3$) with $k_1 + k_2 + \cdots + k_n \in [-\pi, \pi]$. Then $$-\cos k_1 - \cos k_2 - \cdots - \cos k_n \ge - n\cos \frac{k_1 + k_2 + \cdots + k_n}{n}.$$

Proof of Fact 3.

We use the Mathematical Induction.

When $n= 3$, by Fact 1, the statement is true.

Assume that the statement is true for $n$ ($n\ge 3$).

For $n + 1$, WLOG, assume that $k_1 + k_2 + \cdots + k_n \in [-\pi, \pi]$.
(Note: Otherwise, there exists $m\in \mathbb{Z}$ such that $k_1 + k_2 + \cdots + k_n + 2m\pi \in [-\pi, \pi]$. Let $k_1' = k_1 + 2m\pi, k_{n+1}' = k_{n+1} - 2m\pi$. Then $-\cos k_1 - \cos k_2 - \cdots - \cos k_{n+1} = -\cos k_1' - \cos k_2 - \cdots - \cos k_n - \cos k_{n+1}'$, and $k_1' + k_2 + \cdots + k_n + k_{n+1}' = k_1 + k_2 + \cdots + k_{n+1}$, and $k_1' + k_2 + \cdots + k_n \in [-\pi, \pi]$.)

By the inductive hypothesis, we have $$-\cos k_1 - \cos k_2 - \cdots - \cos k_n \ge - n\cos \frac{k_1 + k_2 + \cdots + k_n}{n}.$$ Thus, we have $$-\cos k_1 - \cos k_2 - \cdots - \cos k_{n+1} \ge - n\cos \frac{k_1 + k_2 + \cdots + k_n}{n} - \cos k_{n+1}.$$

By Fact 2, we have $$- n\cos \frac{k_1 + k_2 + \cdots + k_n}{n} - \cos k_{n+1} \ge -(n + 1)\cos \frac{k_1 + k_2 + \cdots + k_{n+1}}{n + 1}.$$ (Note: Let $u := k_1 + k_2 + \cdots + k_n$ and $v := k_1 + k_2 + \cdots + k_n + k_{n+1}$.)

Thus, we have $$-\cos k_1 - \cos k_2 - \cdots - \cos k_{n+1} \ge -(n + 1)\cos \frac{k_1 + k_2 + \cdots + k_{n+1}}{n + 1}.$$ Thus, the statement is true for $n + 1$.