estimate on a convolution Let $\psi$ be a non-negative Schwartz function on $\mathbb{R}$ such that supp$\hat{\psi}$ is contained in $[-0.1, 0.1]$ and $\hat{\psi}(0)=1$. Define $\psi_k(x)=2^k\psi(2^kx)$ for any integer $k$. Let $I_k$ be an interval of length $2^{-k}$ and $1_{I_k}$ be the indicator function  of $I_k$. Define $f(x)=1_{I_k}*\psi_k(x)$. Prove that $$|f(x)|\le\frac{C_N}{(1+2^kdist(x, I_k))^N}$$
I can prove this if I assume $\psi$ is radial and decreasing (in this case no information of Fourier transform is needed), but I do not know how to use $\hat{\psi_k}$ or its support (which is an interval of length $0.2\times2^k$) in the proof of this estimate. Any help is greatly appreciated.
 A: Perhaps I am missing something and have made an error, but I don't see why you need to the information on the support of $\hat{\psi}$ to prove the estimate for $f$; that $\psi(x)\lesssim_{N} (1+\left|x\right|)^{-N}$, for any $N>0$, together with the information on $I_{k}$ is sufficient.
Write $I_{k}=[a-2^{-k-1},a+2^{-k-1}]$. First, I claim that it suffices to consider the case $a=0$. Write $I_{k}'=[-2^{-k-1},2^{-k-1}]$. Making the change of variable $z+a=y$, we see that
$$f(x)=\int_{\mathbb{R}}\mathrm{1}_{I_{k}}(y)\psi_{k}(x-y)dy=\int_{\mathbb{R}}\mathrm{1}_{I_{k}}(z+a)\psi_{k}(x-(z+a))dz=\int_{\mathbb{R}}\mathrm{1}_{I_{k}'}(z)\psi_{k}((x-a)-z)dz$$
So if we have the estimate for the case $a=0$, then using the fact that $\text{dist}(x-a,I_{k}')=\text{dist}(x,I_{k})$ (you can check this), we have
$$\left|f(x)\right|\lesssim_{N}(1+2^{k}\text{dist}(x-a,I_{k}'))^{-N}=(1+2^{k}\text{dist}(x,I_{k}))^{-N}$$
So it suffices to assume that $a=0$. Observe that
$$\int_{\mathbb{R}}\mathrm{1}_{I_{k}}(y)\psi_{k}(x-y)dy=\int_{\mathbb{R}}\mathrm{1}_{I_{k}}(y)2^{k}\psi(2^{k}(x-y))dy=\int_{\mathbb{R}}\mathrm{1}_{I_{k}}(2^{-k}z)\psi(2^{k}x-z)dz=\int_{-1/2}^{1/2}\psi(2^{k}x-z)dz, \tag{1}$$
where we make the change of variable $z=2^{k}y$. Since $\psi$ is Schwartz, the RHS in (1) is
$$\leq C_{N}\int_{-1/2}^{1/2}(1+\left|2^{k}x-z\right|)^{-N}dz \tag{2}$$
for any $N>0$. We consider now three case.
First case $x\geq 2^{-k-1}$. Then the integrand in (2) is majorized by $(1+\left|2^{k}x-1/2\right|)^{-N}$, and since $\text{dist}(x,I_{k})=\left|x-2^{-k-1}\right|$, we get
$$\left|f(x)\right|\leq C_{N}(1+2^{k}\text{dist}(x,I_{k}))^{-N}$$
Second case $x\leq -2^{-k-1}$. The integrand is majorized by $(1+\left|2^{k}x+1/2\right|)^{-N}$, and since $\text{dist}(x,I_{k})=\left|x+2^{-k-1}\right|$, we get
$$\left|f(x)\right|\leq C_{N}(1+2^{k}\text{dist}(x,I_{k}))^{-N}$$
Third case $-2^{-k-1}<x<2^{-k-1}$. The integrand is majorized by $1$, and since $\text{dist}(x,I_{k})=0$, we get
$$\left|f(x)\right|\leq 1=(1+2^{k}\text{dist}(x,I_{k}))^{-N}$$
Taking $C_{N}'=\max\left\{1,C_{N}\right\}$ completes the proof.
