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Assume that $f \in L^1(\mathbb{R})$, compute $ f*g$ with $ g(x)=e^{2 i \pi x}$


Thm1 let $f \in L^1(R)$ and $g\in C^p(R)$. Assuming $g^k$ is bounded for $k=0,1,\dots ,p$

1) $f*g \in C^p(R)$

2) $(f*g)^k=f*g^k$ for $k=1,\dots ,p$

The composition $$ (f*g) = \int f(t-\tau)y (\tau)d \tau$$

def of $L^1(R)$ is space meausrable where

$$ \int |f(t)|dt < \infty $$

know that $$ e^{i2 \pi x} = \cos(2 \pi x )+ i\sin(2 \pi x)$$

convolution is commutative $ f*g =g*f.$


$$\begin{aligned} (f*g) &= \int e^{i 2 \pi (t-\tau) } f(\tau) d \tau = ? \end{aligned} $$

want to say it is $(e^{i2 \pi} *f)$ distorting the first theorem but x is not an integer . Feel like doing countour integration in complex analysis but idk

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$$(f*g)(t) = \int_{\Bbb R} e^{i 2 \pi (t-\tau) } f(\tau) d \tau = e^{i 2 \pi t } \int_{\Bbb R} e^{-i 2 \pi\tau } f(\tau) d \tau = e^{i 2 \pi t }\hat f(1) $$

where the Fourier transform is given by $$\hat f(x)= \int_{\Bbb R} e^{-i 2 \pi x\tau } f(\tau) d \tau$$

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