Proving that the convolution operator on $L^1$ does not have an identity element.

Question

consider the operator $*$ from $L^1(R^d) \to L^1(R^d)$ with

$$f*g = \int_{R^d}f(x-y)g(y)dy$$

I want to show this operation doesn't have an identity element, namely, there does not exist an element $I \in L^1(R^d)$ with $$f*I = f$$ I have argued as follows:

Assume there exists such an element, hence $$\int_{R^d}I(x-y)I(y)dy = I(x)$$ taking integral from both sides with respect to $x$ and using Fubini-Tonelli's theorem: $$\int_{R^d}I(x-y)dx\int_{R^d}I(y)dy = \int_{R^d}I(x)dx$$ using invariance properties of Lebesgue integral then yields: $$\int_{R^d}I(x)dx = 0 \ or \ 1$$ case1$\int_{R^d}I(x)dx = 0$: $$\int_{R^d}f(x-y)dx\int_{R^d}I(y)dy = \int_{R^d}f(x)dx$$ and this yields $\int_{R^d}f(x)dx = 0$ which is a contradiction since $f$ was arbitrary.
case2:$\int_{R^d}I(x)dx = 1$: $$\int_f(x-y)I(y)dy = \int_{R^d}f(x)I(y)dy$$ hence $$\int_{R^d}[f(x-y)-f(x)]I(y)dy = 0$$

And I'm stuck here. The best idea I have is to consider some $f$ and evaluate it at some point and get to a contradiction. Any ideas?

Answer 1

The Riemann-Lebesgue lemma implies that for $f\in L^1(\mathbb{R}^d)$ we have $$\lim_{\|\xi \|\to \infty }\hat{f}(\xi)=0$$ The Fourier transform of $g(x)=e^{-\|x\|^2}$ does not vanish. We have $$\widehat{f*g}=\hat{f}\,\widehat{g}$$ Assume by contradiction that $f*g=g.$ Then $\hat{f}=1,$ a contradiction.

The argument given in this post is more complicated in my opinion.

Answer 2

Suppose $e$ is an identity, then $e * 1_{[0,1]} = 1_{[0,1]}$. Then $1_{[0,1]} (y) = \int 1_{[0,1]}(y-x) e(x) dx$ ae. In particular, $\int 1_{[0,1]}(y-x) e(x) dx = \int_{y-1}^y e(x)dx= 1_{[0,1]}(y)$ for ae. $y$. Differentiating gives $e(y)=e(y-1)$ ae. and so $e$ is essentially $1$-periodic. Since $e \in L^1$, this gives $e=0$ which is a contradiction.