Proving a sequence of functions is an approximation to the identity

Question

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?

Answer 1

You can compute the normalising factor explicitly: $$\int_{-\pi}^\pi(1+\cos x)^n\,\mathrm{d}x = 2^{-n}\binom{2n}{n+1}\cdot 2\pi\sim 2^n\cdot\frac{2\pi}{\sqrt{n\pi}}$$ where the last asymptotic is by Stirling's formula. On the other hand, for $\delta<\lvert x\rvert<\pi$, we have $\cos^{2n}\frac{x}{2}<\cos^{2n}\frac\delta2\leq\exp(-cn\delta^2)$, some $c>0$. So $$ \frac{\int_{\delta<\lvert x\rvert<\pi}(1+\cos x)^n\,\mathrm{d}x}{\int_{-\pi}^\pi(1+\cos x)^n\,\mathrm{d}x}\lesssim\frac{2^n\exp(-cn\delta^2)\cdot 2\pi}{2^n\cdot{2\pi}/{\sqrt{n\pi}}}\sim\exp(-cn\delta^2)\sqrt{n\pi}\to 0 $$

Answer 2

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 $f\geq 0$ 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:

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