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Why is $g = \sum_{k=1}^n $ $\sum_{l=1}^n cos(g_k-g_l)$ always non-negative?

g is the magnitude square of the complex-valued function, f, defined below, so it is , for sure, non-negative, but I have such a hard time accepting its non-negativity by looking at the current form of g. Is there a way to re-write g in such a way that it would be easy to see its non-negativity property?

$f= \sum_{k=1}^n e^{ig_k}$ is a complex-valued function where g_k is a real-valued function of k.

Then the magnitude square of the function $\ |f|^2 = \sum_{k=1}^n e^{ig_k} \sum_{l=1}^n e^{-ig_l}$ = $\sum_{k=1}^n $ $\sum_{l=1}^n e^{ig_k} e^{-ig_l}$ = $\sum_{k=1}^n $ $\sum_{l=1}^n cos(g_k-g_l)$

Therefore $\sum_{k=1}^n $ $\sum_{l=1}^n cos(g_k-g_l)$ >= 0 as it is the magnitude of a complex number.

erfaun
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    $\sum_{l=1}^n e^{ig_k} e^{-ig_k}$ should be $\sum_{l=1}^n e^{ig_k} e^{-ig_l}$ (last subscript is typed incorrectly.) – saulspatz Jun 29 '19 at 23:39
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    You have given a neat proof (modulo a typo pointed out by saulspatz) so where is the confusion? – Kavi Rama Murthy Jun 29 '19 at 23:42
  • $e^{ig_k} e^{-ig_l} \ne \cos(g_k - g_l)$, or are you doing something more complicated in that last step that I'm not seeing? – JonathanZ Jun 30 '19 at 01:43
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    @JonathanZ OP is taking the real part, since the overall sum is a real number already. – anon Jun 30 '19 at 03:08
  • @saulspatz Thank you for pointing out the typo...just fixed it. – erfaun Jun 30 '19 at 12:07
  • @JonathanZ just use Euler's formula and write the complex exponential in terms of sine and cosine. The imaginary parts cancel each others out and what is left is what I have here. and it makes sense because you expect the magnitude to be a real number. – erfaun Jun 30 '19 at 12:12

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Using the angle-addition formula, we have $$\cos(g_k - g_{\ell}) = \cos(g_k) \cos(g_{\ell}) + \sin(g_k) \sin(g_{\ell})$$ and thus $$\sum_{k, \ell} \cos(g_k - g_{\ell}) = \sum_{k, \ell} \cos(g_k) \cos(g_{\ell}) + \sum_{k, \ell} \sin(g_k) \sin(g_{\ell})$$ $$ = \left(\sum_{k} \cos(g_k) \right)^2 + \left(\sum_{k} \sin(g_k) \right)^2 \ge 0$$ Note that the equality condition $\sum_{k} \cos(g_k) = \sum_{k} \sin(g_k) = 0$ is precisely equivalent to $f = \sum_k e^{ig_k} = 0$.

Sameer Kailasa
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