Can it be shown that the Banach algebra $(L^1(\mathbb{R}^n), \ast)$ does not have a unit element?

Question

Let $f, g \in L^1(\mathbb{R}^n)$ and define the convolution of $f$ with $g$ by $$ (f \ast g)(x) = \int_{\mathbb{R}^n} f(y)g(x-y)dy. $$ Then $(L^1(\mathbb{R}^n), \ast)$ is a Banach algebra. I have seen it stated in a few places that this Banach does not have a unit element. Is it possible to show this property? It seems to me that the dirac delta function could be a unit element here as it will pick out the original values of the function. E.g. $(f \ast \delta)(x) = f(x)$?

Answer 1

Let us assume towards a contradiction that $f $ is a convolutional unit for $L^1$.

By the convolution theorem, you have $\hat {g} = \widehat {f \ast g} = \hat {f} \hat {g} $ for arbitrary $g \in L^1$. But there is some $g $ (for example, a Gaussian) whose Fourier transform never vanishes. Thus, $\hat {f} \equiv 1$.

But the Riemann Lebesgue lemma shows that $\hat {f} $ vanishes at $\infty $. Thus, $L^1$ has no unit.

Answer 2

This is essentially the same as PhoemueX's answer, but skips the intermediate step of taking the Fourier transform.

Let $e_k(t) = 1_{[0,1]^n}(t) e^{i 2 \pi \langle k , t \rangle}$, $t \in \mathbb{R}^n, k \in \mathbb{Z}^n$. Note that $\|e_k\| = 1$.

Let $\phi_k(t) = \int e_k(t-x) g(x) dx $ and $\gamma(t) = \int_x 1_{[0,1]^n}(t-x) |g(x)| dx$. Note that $|\phi_k(t)| \le \gamma(t)$ and an application of Tonelli shows that $\int \gamma = \|g\|$.

Expanding $\phi_k(t) = e^{i 2 \pi \langle k , t \rangle} \int 1_{[0,1]^n}(t-x) e^{-i 2 \pi \langle k , x \rangle} g(x) dx$ and applying Riemann Lebesgue we get $\lim_{\|k\| \to \infty} \phi_k(t) = 0$, hence $\lim_{\|k\| \to \infty} \|\phi_k\| = 0$.

Suppose $g$ was a unit, then we would have $e_k = e_k * g = \phi_k$ which is a contradiction.