Jtdylktuy

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 $Isubset L_1$ such that $f*varphiin I$ for any pair $(f,varphi)in L_1times I$.

Correct me if my interpretation was wrong.

FIRST APPROACH: I know that if $gin L_1$ 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

$$lim_{epsilonto 0} g_epsilon* f=|g|_1 f=ftag1$$

in $L_1$ for any chosen $fin L_1$.

Then my first idea was construct some approximation to the unity from a function $f*varphiin I$ (where $(f,varphi)in L_1times I$), that is, I want to show that if $f*varphiin I$ then $(f*varphi)_epsilonin I$ for any $epsilon>0$, from here it would be easy to see that

$$h_epsilon:=frac{(f*varphi)_epsilon}{|f*varphi|_1}in Itag2$$

and $lim_{epsilonto 0}h_epsilon*r=r$ for any chosen $rin L_1$, what would imply that $I$ is dense in $L_1$. Then note that

$$(f*varphi)_epsilon(x)=epsilon^{-n}int f(x/epsilon-y)varphi(y), dy=epsilon^{-n}int f(x-y+K)varphi(y), dytag3$$

for $K:= x/epsilon-x$. Then also note that $epsilon^{-n} f(,cdot+K)in L_1$ for any chosen $KinBbb R^n$, so an heuristic argument make me think from $(3)$ that $(f*varphi)_epsilonin I$. 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

$$(f*varphi)_epsilon(x)=epsilon^{-n}int f(x/epsilon-y)varphi(y), dy=epsilon^{-n}int tau_{x/epsilon}check f(y)varphi(y), dy\
=epsilon^{-n}int(tau_{x/epsilon}check fcdot varphi)(y), dy
=epsilon^{-2n}inttau_{x/epsilon}check f(y/epsilon)varphi(y/epsilon), dy
\=epsilon^{-2n}int fleft(frac{x-y}epsilonright)varphi(y/epsilon)=(f_epsilon*varphi_epsilon)(x)tag4$$

where $tau_a f(x)=f(x+a)$ and $check f(x)= f(-x)$. Then it would be enough to show that if $varphiin I$ then $varphi_epsilonin I$, 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 $psiin Isetminus{0}$. Then, by the definition of ideal, we knows that $varphi_epsilon*psiin I$ 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.