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


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.