This is related to this other question, I mean, the linked question comes to my mind trying to solve the following exercise:
Show that any non-trivial ideal of $(L_1,*)$ is dense.
Here $(L_1,*)$ is the space $L_1(\Bbb R^n,\Bbb R)$ and $*$ stay for convolution. I know that $(L_1,+,*)$ is a Banach algebra without unity. I dont know the concept of "ideal" related to a Banach algebra so I assumed that it is asking for a set $I\subset L_1$ such that $f*\varphi\in I$ for any pair $(f,\varphi)\in L_1\times I$.
Correct me if my interpretation was wrong.
FIRST APPROACH: I know that if $g\in L_1$ and $\|g\|_1=1$ then $g_\epsilon(x):=\epsilon^{-n}g(x/\epsilon)$ defines a kernel $\{g_\epsilon:\epsilon>0\}$ that approximates the unity, that is, I know that
$$\lim_{\epsilon\to 0} g_\epsilon* f=\|g\|_1 f=f\tag1$$
in $L_1$ for any chosen $f\in L_1$.
Then my first idea was construct some approximation to the unity from a function $f*\varphi\in I$ (where $(f,\varphi)\in L_1\times I$), that is, I want to show that if $f*\varphi\in I$ then $(f*\varphi)_\epsilon\in I$ for any $\epsilon>0$, from here it would be easy to see that
$$h_\epsilon:=\frac{(f*\varphi)_\epsilon}{\|f*\varphi\|_1}\in I\tag2$$
and $\lim_{\epsilon\to 0}h_\epsilon*r=r$ for any chosen $r\in L_1$, what would imply that $I$ is dense in $L_1$. Then note that
$$(f*\varphi)_\epsilon(x)=\epsilon^{-n}\int f(x/\epsilon-y)\varphi(y)\, dy=\epsilon^{-n}\int f(x-y+K)\varphi(y)\, dy\tag3$$
for $K:= x/\epsilon-x$. Then also note that $\epsilon^{-n} f(\,\cdot+K)\in L_1$ for any chosen $K\in\Bbb R^n$, so an heuristic argument make me think from $(3)$ that $(f*\varphi)_\epsilon\in I$. However, this heuristic argument seems wrong, because we cannot fix $K$ as a constant and say that $g(x-y):=f(x/\epsilon-y)$ belongs to $L_1$ because $g$ is not well-defined.
SECOND APPROACH:
Note that
$$(f*\varphi)_\epsilon(x)=\epsilon^{-n}\int f(x/\epsilon-y)\varphi(y)\, dy=\epsilon^{-n}\int \tau_{x/\epsilon}\check f(y)\varphi(y)\, dy\\ =\epsilon^{-n}\int(\tau_{x/\epsilon}\check f\cdot \varphi)(y)\, dy =\epsilon^{-2n}\int\tau_{x/\epsilon}\check f(y/\epsilon)\varphi(y/\epsilon)\, dy \\=\epsilon^{-2n}\int f\left(\frac{x-y}\epsilon\right)\varphi(y/\epsilon)=(f_\epsilon*\varphi_\epsilon)(x)\tag4$$
where $\tau_a f(x)=f(x+a)$ and $\check f(x)= f(-x)$. Then it would be enough to show that if $\varphi\in I$ then $\varphi_\epsilon\in I$, but this doesn't seems feasible.
THIRD APPROACH:
Let any smoothing kernel $\{\varphi_\epsilon:\epsilon>0\}\subset L_1$ (by example the Gaussian kernel) and some $\psi\in I\setminus\{0\}$. Then, by the definition of ideal, we knows that $\varphi_\epsilon*\psi\in I$ for all $\epsilon>0$, thus by $(1)$ we can see that $\psi$ is a limit point of $I$, and because this holds for every function on the ideal then we conclude that $I$ is closed and perfect.
Then, if $I$ is dense, it must be the case that $I=L_1$. However Im again stuck here, that is, I dont know how to show that $I=L_1$.
Moreover: if it would be true that $I=L_1$ then $I$ is, indeed, a trivial ideal of $L_1$, contradicting the existence of non-trivial ideals, so the statement to be proved will be a vacuous truth.
Some help will be appreciated, thank you.