We know $L^1(\mathbb{R})$ is a Banach algebra whose product is defined to be convolution: $$f*g(x)=\int f(x-t)g(t)dt.$$ In fact, it is a $*$-Banach algebra (Don't confused with the convolution), with $f^*(x)=\overline{f(-x)}$. The Fourier transform $$\mathcal{F}: L^1(\mathbb{R})\to C_0(\mathbb R)$$ is a continuous algebraic homomorphism with ${\cal F}(f^*)=\overline{f}$.
I want prove $\cal F$ is not onto. I know there are some methods, see this post for example.
Observe that if $\cal F$ is onto, then $\cal F$ is an algebraic isomorphism and preserve the involution. However, $C_0(\mathbb{R})$ is a $C^*$-algebra while $L^1(\mathbb{R})$ is not.
It is possible to use this observation to get a new proof?
Feel free to use non-elementary results in $C^*$-algebra and Banach-* algebra theory.
PS: maybe it will be a little bit cyclic argument, since one way to prove $L^1(\mathbb{R})$ not satisfying the $C^*$ identity is to check the convolution of two indicator functions. Anyway, let's admit the fact that $L^1(\mathbb{R})$ is not a $C^*$-algebra.
Should $\cal F$ be surjective, it would be invertible and hence $\cal F^{-1}$ would be a *-homomorphism defined on the $C^*$-algebra $C_0(\mathbb R)$, hence contractive. As a result $\cal F$ would be isometric, from where one would deduce that $L^1(\mathbb R)$ is a $C^*$-algebra, a contradiction.