Translation invariant operators in $L^p$ spaces I am reading the following paper:
Hormander, L., $\textit{Estimates for translation invariant operators in }L^p\textit{ spaces}$, Acta Math. $\textbf{104}$ (1960), no. 1-2, 93-140.
In this paper, the following is the first theorem.
Let $T:L^p(\mathbb{R}^n)\to L^q(\mathbb{R}^n)$ be a translation invariant bounded linear operator ; where $p>q$. Then we have $T=0$ if $p<\infty$ and if $p=\infty$ then the restriction of $T$ to $L_0^{\infty}(\mathbb{R}^n)$ is $0$. (Here, $L_0^{\infty}(\mathbb{R}^n)$ is the space all functions in $L^{\infty}(\mathbb{R}^n)$ which vanishes at infinity.)
I understood the proof for $p<\infty$ but I am stuck for $p=\infty$
.
Here is the sketch of the proof.
Let $f\in L^{p}(\mathbb{R}^n)$. The $\textbf{key step}$ is to prove the following for $p<\infty$.
$$ \|f+\tau_yf\|_p\to2^{\frac{1}{p}}\|f\|_p \text{ as }y\to\infty.$$
In order to establish the above claim, we uses the fact that $C_c(\mathbb{R}^n)$ is dense in $L^{p}(\mathbb{R}^n)$ and so that $f$ can be written as $g+h$ ; where $g\in C_c(\mathbb{R}^n)$ and $\|h\|_p$ is very small. Once we establish the above claim, we can get the desired result very easily by using the fact that the operator norm $\|T\|$ is the smallest such non-negative constant for which $T$ is bounded.
Hormander wrote that the same key step is true for $p=\infty$ when we take $f\in L_0^{\infty}(\mathbb{R}^n)$, i.e.,
$$ \|f+\tau_yf\|_{\infty}\to\|f\|_{\infty} \text{ as }y\to\infty\text{ for }f\in L_0^{\infty}(\mathbb{R}^n).$$
I am unable to prove this key step for $p=\infty$. We don't have this fact that $C_c(\mathbb{R}^n)$ is dense in $L^{\infty}(\mathbb{R}^n).$ But do we say that $C_c(\mathbb{R}^n)$ is dense in $L_0^{\infty}(\mathbb{R}^n)$ ?
Thanks in advance.
 A: As a commenter noted, one way to prove this is to observe that $L^{\infty}_c$ is dense in $L_0^{\infty}$. To prove this density take $f\in L^{\infty}_0$ and approximate it by $f_n:=f\cdot \chi_{|x|\leq 2^n}$, where $\chi_{A}$ is the indicator function of the set.
Another way is to argue directly, noting that if $f\in L^{\infty}$ for every $\epsilon>0$ there is an $R$ such that $|f(x)|\leq \epsilon$ for $x\geq R$. Then, if $|y|\geq 2R$, for any point $x$, either $x$ or $x+y$ will be outside the ball of radius $r$ centered at the origin, so
$$
|f(x)+\tau_y f(x)|=|f(x)+f(x+y)|\leq ||f||_{L^{\infty}}+\epsilon.
$$
Thus, $\limsup_{|y|\to \infty}||f+\tau_y f||_{L^{\infty}}\leq ||f||_{L^{\infty}}$.
Also, for any $\epsilon>0$ there exists a point $x$ s.t. $|f(x)|\geq ||f||_{\infty}-\epsilon$. Thus, for $R$ as above, if $|y|\geq R$ then $|\tau_y f(x)|\leq \epsilon$, so
$$
||f(x)+\tau_y f(x)||\geq ||f(x)|-|\tau_y f(x)||\geq ||f||_{L^{\infty}}.
$$
Because of this, $\liminf_{|y|\to \infty}||f+\tau_y f||_{L^{\infty}}\geq ||f||_{L^{\infty}}$ which finishes the proof.
