The Ideal of integrable real functions with vanishing Fourrier transform is not closed in $L^1(\mathbb{R})$ Consider $(L^1(\mathbb{R}),+,\ast)$ with the convolution operator, and the Banach ring $(L^1(\mathbb{R}), ||.||)$. I am looking for an example of a Banach ring with non closed ideal, with respect to the norm ||.||. more precisely, i'm trying to show that : $I = \lbrace f \in L^1(\mathbb{R}) : \exists \epsilon >0, \hat{f} = 0$ in $]- \epsilon, \epsilon[ \rbrace $ is not closed in $L^1(\mathbb{R})$ I can see that it is an ideal since the convolution in $L^1(\mathbb{R})$ corresponds to pointwise multiplication of the Fourier transforms, any help for the last one would be appreciated!

Well clearly you want to find $f_n\in I$ with $||f_n-f||_1\to0$ but $f\
otin I$. A hint regarding one way to construct such a sequence: Say $f\in L^1$ and $g(t)=e^{ict}f(t)$. How are $\hat f$ and $\hat g$ related?

Ok, a solution, since the OP has given up. We know that if $\psi\in C^2_c$ then there exists $f\in L^1$ with $\psi=\hat f$. So we can choose $f\in L^1$ such that $\hat f(\xi)=0$ for all $\xi<0$ while $\hat f(\xi)>0$ for all $\xi\in(0,1)$. Let $f_n(x)=e^{ix/n}f(x)$. Then $f_n\in I$, $||f-f_n||_1\to0$ and $f\
otin I$.