Show that there is no \(I\in L^1(\RR^n)\) such that \[ f*I=f \qquad\text{for every }f\in L^1(\RR^n). \]
Solution
1. If such an \(I\) existed, then\[ \widehat f(\xi)\widehat I(\xi)=\widehat f(\xi) \qquad\text{for every }f\in L^1(\RR^n) \]and every \(\xi\in\RR^n\).
Taking Fourier transforms in \(f*I=f\) and applying the convolution formula for Fourier transforms gives \[ \widehat{f*I}=\widehat f\,\widehat I, \] which is the stated identity.
2. The identity in step 1 forces \(\widehat I\equiv1\).
Choose the Gaussian \(f(x)=e^{-\pi\abs{x}^2}\). It belongs to \(L^1(\RR^n)\) and its Fourier transform is again a Gaussian, so \[ \widehat f(\xi)\ne0 \qquad\text{for every }\xi. \] Therefore \[ \widehat I(\xi)=1 \qquad\text{for every }\xi\in\RR^n. \]
3. This contradicts the Riemann–Lebesgue lemma.
Since \(I\in L^1(\RR^n)\), the Riemann--Lebesgue lemma gives \[ \widehat I(\xi)\longrightarrow0 \qquad(\abs{\xi}\to\infty), \] contradicting step 2.
4. Q.E.D.
The contradiction in step 3 proves that no such convolution identity element exists in \(L^1(\RR^n)\).