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We know that Fejer Kernel: $(K_n)_{n=0}^{\infty}$ is an approximate identity of $L^1(T)$.

$K_n=\sum_{k=-n}^{n} \left(1-\frac{|k|}{n+1}\right)e_k$ , ($n\in Z_+$).

I am trying to use this in order to show that the series $(\|K_n\|_2^{-2}K^2_n)_{n=0}^{\infty}$ is an approximate identity of $L^1(T)$.

So, we want to show by definition that the above sequence satisfies the (three) conditions of an approximate identity.

  • We say that a sequence $(f_n)_{n=1}^{\infty}$ is an approximate identity of $L^1(T)$ if it satisfies:

  1. $\sup_{n\in \mathbb N} \|f_n\|_1 <\infty$.

  2. For all $n, \int_{T} f_n dm=1$.

  3. for $0< \delta\leq \pi$: $\lim_{n \to \infty} \int_{e^{it} : \delta \leq t \leq 2\pi-\delta} |f_n| dm = 0$.

An approximate identity is said to be positive if it consists of positive functions, we notice that in this case the 2nd condition leads to the 1st condition.

We also know that $K_n(e^{it})=\frac{1}{n+1} (\frac{\sin(( n+1)t/2)}{\sin(t/2)})^2$.

So we can express $K_n(e^{it})^2$, and that shows that $\|K_n\|^{-2} K^2_n$ is positive, thus the series is bounded that means $\sup_{n} \|K_n\|_2^{-2} K^2_n <\infty$.

We can also use this representation to show the third condition, however I am not quite sure about it!. I think we can use that $\|K_n\|^{-2}_2$ is a positive number (in terms of n), is that right? And how can the second condition: $\int_T \|K_n\|_2^{-2} K^2_n dm =1$ be proved?

Lam18373
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1 Answers1

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If $f\in L^2[0,2\pi],$ $f\ne 0,$ then

$$\int_0^{2\pi}\frac{f(t)^2}{\|f\|_2^2}\,dt = \frac{1}{\|f\|_2^2}\int_0^{2\pi}f(t)^2\,dt = \frac{1}{\|f\|_2^2}\|f\|_2^2 =1.$$

Apply this to $f=K_n^2\|K_n\|_2^{-2}$ and we obtain both conditions 1. and 2. in the definition of an $L^1$ approximate identity.

To obtain condition 3., note that by Jensen

$$1^2 = \left (\frac{1}{2\pi}\int_0^{2\pi}K_n(t)\,dt \right)^2 \le \frac{1}{2\pi}\int_0^{2\pi}K_n(t)^2\,dt = \|K_n\|_2^2.$$

Let $\delta>0.$ Then

$$\tag 1 \sup_{\delta<t<2\pi-\delta} |K_n(t)|^2\le \frac{1}{(n+1)^2}\frac{1}{\sin^4(\delta/2)}.$$

Since $\|K_n\|_2^{-2}\le 1,$ we see by $(1)$ that $K_n^2\|K_n\|_2^{-2}\to 0$ uniformly on $[\delta,2\pi-\delta].$ This shows condition 3. holds for $K_n^2\|K_n\|_2^{-2}.$

Therefore $K_n^2\|K_n\|_2^{-2}$ is an $L^1$ approximate identity.

zhw.
  • 107,943
  • Thanks, it makes sense! can you please state the jensen inequality ? @zhw. – Lam18373 Jul 08 '21 at 11:02
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    We don't necessarily need Jensen; the result follows from Holder:

    $$1=\frac{1}{2\pi}\int_0^{2\pi}K_n(t),dt \le \frac{1}{2\pi}(\int_0^{2\pi}K_n(t)^2,dt)^{1/2}(2\pi)^{1/2}.$$

    That will lead to $1\le|K_n|_2^2.$

    But Jensen's inequality is of great importance in analysis. See https://en.wikipedia.org/wiki/Jensen%27s_inequality#Measure-theoretic_and_probabilistic_form Jensen was an interesting mathematician; see https://mathshistory.st-andrews.ac.uk/Biographies/Jensen/ for a short biography.

     – zhw. Jul 08 '21 at 16:36
  • So here you're using jolder on $K_n\in L^2(T)$ and $\sqrt{2\pi} \in L^2(T)$? – Lam18373 Jul 08 '21 at 17:21
  • No, I am using $K_n=K_n\cdot 1,$ a useful trick. – zhw. Jul 08 '21 at 17:39