Vrftsjtryk

Define the sequence $phi_n$ on the interval $[-pi,pi]$ (the torus if you will) by:
$$
phi_n = c_n (1 + cos x)^n
$$
Where $c_n$ is a normalizing constant to ensure that $int_T phi_n = 1$ for any choice of $n$. I.e $c_n = int_T (1 + cos x)^n$. Now, I want to prove this sequence is an approximation to the identity. The main property I must prove is that for any $delta$ greater than $0$, we have that:
$$
int_{{delta < |x| < pi}} phi_n(x) mathrm{d}x to 0
$$
I.e, if we shift a little off from $1$ and take a limit of integrals, all the mass becomes concentrated around $0$, as we would think happens in an approximation to the identity. My main strategy for the proof was to do the following. First, write:
$$
(1 + cos x)^n = 2^{n} cos^{2n} left(frac{x}{2}right)
$$
using trigonometric identities. Then we see that $cos^{2n} left(frac{x}{2}right) to 0$ pointwise for any $x neq 0$. Moreover, if we have for fixed $x_0$ that:

$$
left|cos^{2n} left(frac{x_0}{2}right)right|<epsilon
$$
Then we see that $x > x_0$ implies:
$$
left|cos^{2n} left(frac{x}{2}right)right|<epsilon
$$
Hence, given any $delta > 0$, we can take $N$ such that $n > N$ implies:
$$
left|cos^{2n} left(frac{delta}{2}right)right|<epsilon
$$
So that for any $x$ with $|x| > delta$ (exploiting symmetry of cosine) we have that:
$$
left|cos^{2n} left(frac{x}{2}right)right|<epsilon
$$
And hence:
$$
int_{{delta} < |x|<pi} cos^{2n} left(frac{x}{2}right) < 2(pi - delta)epsilon to 0
$$
So that the LHS becomes arbitrarily small and converges to $0$. However, I am having issues dealing with the normalizing factor. Namely, we have:
$$
int_{{delta < |x| < pi}} phi_n(x) mathrm{d}x = frac{1}{int_{[-pi,pi]} 2^{n} cos^{2n} left( frac{x}{2}right) mathrm{d}x} int_{{delta < |x| < pi}} 2^{n} cos^{2n} left( frac{x}{2}right)mathrm{d}x
$$
The $2^n$ powers cancel, so we have:
$$
frac{int_{{delta < |x| < pi}} cos^{2n} left( frac{x}{2}right)mathrm{d}x}{int_{[-pi,pi]} cos^{2n} left( frac{x}{2}right) mathrm{d}x}
$$
And the numerator of this expression tends to $0$, by my previous argument. However, how can I deal with the normalizing integrals? How do I make sure they do not go to $0$, and if they do (I am fairly sure they do), how can I rigorize the fact that the numerator converges to $0$ faster?

If you look at the graph of $f(x) = 1+cos x$ you can intuitively guess that with the powers $f^n$ the mass of the integral will "bunch up" around $x=0$. Points closer to 0 will shoot up faster than points further away (basically because if a>b then $a^n >>b^n$ as n grows large). Here's a way to make that precise:

Let $f(x)$ be positive and strictly decreasing on $[0,pi]$. We can compare $int_{[0,delta]} f^n(x) ,dx$ to $int_{[delta,pi]} f^n(x), dx$ by taking their ratio and showing
$$
frac{int_{[0,delta]} f^n(x) ,dx}{int_{[delta,pi]} f^n(x) ,dx} rightarrow +infty
$$

By choosing a fixed $0<epsilon<delta$ we have that the numerator is at least $int_{[0,epsilon]} f^n(x) ,dx$ and this is at least $epsilon f(epsilon)^n$ (both of these fact follow since $fgeq 0$ and $f$ is decreasing.
Similarly the denominator is less than $(pi-delta)f(delta)^n$. Since
$$
frac{epsilon f(epsilon)^n}{(pi-delta)f(delta)^n} rightarrow infty
$$
we have our result.

Now to apply this result to your problem, you can choose $f(x) = cos^2(frac{x}{2})$ as in your last line. You can invert the limit above going to $infty$ and do one more comparison:

$$
frac{int_{[delta,pi]} f^n(x) ,dx}{int_{[0,pi]} f^n(x) ,dx} leq frac{int_{[delta,pi]} f^n(x) ,dx}{int_{[0,delta]} f^n(x) ,dx} rightarrow 0
$$

You can compute the normalising factor explicitly: $$int_{-pi}^pi(1+cos x)^n,mathrm{d}x = 2^{-n}binom{2n}{n+1}cdot 2pisim 2^ncdotfrac{2pi}{sqrt{npi}}$$
where the last asymptotic is by Stirling's formula. On the other hand, for $delta<lvert xrvert<pi$, we have $cos^{2n}frac{x}{2}<cos^{2n}fracdelta2leqexp(-cndelta^2)$, some $c>0$. So
$$
frac{int_{delta<lvert xrvert<pi}(1+cos x)^n,mathrm{d}x}{int_{-pi}^pi(1+cos x)^n,mathrm{d}x}lesssimfrac{2^nexp(-cndelta^2)cdot 2pi}{2^ncdot{2pi}/{sqrt{npi}}}simexp(-cndelta^2)sqrt{npi}to 0
$$