Is $\delta$ in $L^\infty$? I think the question title says is all. I am wondering, is the Dirac delta in the Lebesgue space $L^\infty$?
 A: $L^{\infty}$ is a space of functions. Every element in $L^\infty$ is a function, and the Dirac delta is not. So the answer is no.
A: In no way you can say that $\delta \in L^\infty(\mathbb R)$. The $\delta$ is not a function, but let us sweep this "detail" under the rug. We can realize $\delta$ as a limit (in some sense that is not necessary to specify): 
$$\tag{1}
\delta(x)=\lim_{n\to \infty} n \zeta(nx), $$ 
where $\zeta\colon \mathbb R \to \mathbb R$ is a nonnegative function that integrates to $1$: 
$$
\int_{-\infty}^\infty \zeta(x)\, dx = 1.$$ 
(This is the familiar "concentrating spike" construction). Now, the sequence of functions $(n\zeta(n\cdot))_{n\in\mathbb N}$ is not bounded in $L^\infty$. Therefore, in no way we can expect that its limit belongs to $L^\infty$, not even in a "weak" or "generalized" way. 
Note. On the other hand, we remark that 
$$
\|n\zeta(n\cdot)\|_{L^1(\mathbb R)} = 1$$ 
for all $n$. This suggests that "$\delta \in L^1(\mathbb R)$" in some generalized sense. And this is true. Indeed, there is the isometric embedding
$$L^1(\mathbb R)\subset M(\mathbb R), $$ 
where $M(\mathbb R)$ denotes the space of (signed) measures with finite total mass, equipped with the total variation norm. The $\delta$ is a perfectly good element of $M(\mathbb R)$.
Note 2. The same argument shows that $\delta\notin L^p(\mathbb R)$, for all $p>1$, not even in a generalized way.
