A question about approximate unit and Convolutions

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Let $n\in Z_+$. Define $U_n:=2K_{2n+1}-K_n$ where $(K_n)$ is Fejer kernel in $L^1(T)$.

a. Prove that $(U_n)_{n=0}^{\infty}$ is an approximate unit of $L^1(T)$.

b. Prove that $\hat {U_n}(k)=1$ for all $n\in Z_+, k\in Z$ such that $|k|\leq n+1$.

c. Prove that for every $f\in L^1(T)$:

  1. $lim_{n\to \infty} U_n*f=f$ in $L^1(T)$.

  2. $\hat {U_n*f}(k)=\hat{f}(k)$ , $\forall k\in Z, n\in Z_+$ such that $|k|\leq n+1$.

I am not sure how I am supposed to doparts a and b correctly (I havn't done a similar thing before).

For c,

  1. It can be deduced by a theorem about approximate units, since $(U_n)$ is a convolution by a, and $f\in L^1(T)$.

  2. It can be seen by an identity of Fourier coefficients: $\hat {U_n*f}(k)=\hat{U_n}(k)\hat{f}(k)=$(by b)$=1*\hat{f}(k)$.

I would be glad if tou tell me or guide me.

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I think that part b) follows from the following computations, where $F$ is the Féjer Kernel, in your notation, $F = K$. By Proposition 3.1.7 from the Grafakos book we have $$ \hat{F}_{N}(m)=\left\{\begin{array}{lr} 1-\frac{|m|}{N+1} \quad \text { if }|m| \leq N \\ 0 \quad \text { c.c } \end{array}\right. $$ and $$ \hat{F}_{2 N+1}(m)=\left\{\begin{array}{l} 1-\frac{|m|}{2 N+2} \text { if }|m| \leq 2 N+1 \\ 0 \quad \text { c.c } \end{array}\right. $$ If $|m|=N+1$ we have $$ \hat{U}_{N}(m)=2 \hat{F}_{2 N+1}(m)-\hat{F}_{N}(m)=2\left(1-\frac{|m|}{2 N+2}\right)=2\left(1-\frac{N+1}{2(N+1)}\right)=2\left(1-\frac{1}{2}\right)=1. $$ and if $|m|<N+1$ we have $$ \hat{U}_{N}(m)=2 \hat{F}_{2 N+1}(m)-\hat{F}_{N}(m)=2-\frac{2|m|}{2 N+2}-1+\frac{|m|}{N+1}=1-\frac{|m|}{N+1}+\frac{|m|}{N+1}=1. $$

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You can check the book "Classical Fourier Analysis" from Grafakos (Third Edition, 2014) and I think that the following results will help you.

a) The Proposition 3.1.10 show us that the Fejer Kernel is an approximate identity. Using this you can conclude that $U_n$ is an approximate identity too.

b) Use Proposition 3.1.7 from this book.

Note that the "hard work" is done in this propositions. I hope that this was helpful. By the way, this is the exercise $3.1.4$.