# Is $\delta$ in $L^\infty$?

I think the question title says is all. I am wondering, is the Dirac delta in the Lebesgue space $L^\infty$?

• No, it is not a function, and even if it were, it would be unbounded. Commented Mar 22, 2017 at 13:03
• To be fair-if it were, it would be bounded. And continuous. And yellow. Commented Mar 22, 2017 at 13:07
• @UmbertoP.: I have added an answer to specify what I mean. Commented Mar 22, 2017 at 13:16

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.

$L^{\infty}$ is a space of functions. Every element in $L^\infty$ is a function, and the Dirac delta is not. So the answer is no.

• Isn't the Dirac delta a generalized function? Does this not count for membership in $L^\infty$? Commented Mar 22, 2017 at 13:05
• @Wapiti It's a generalized function, which means it's not a function. Because you have to generalize the definition in order for $\delta$ to fit into it. It's like a circle is a generalized $n$-gon, but it is not a $n$-gon.
– 5xum
Commented Mar 22, 2017 at 13:06
• Heh...technically it's a space of equivalence classes of functions ;) Commented Mar 22, 2017 at 13:46
• The name "generalized function" is probably the biggest abomination in mathematical analysis. Is like saying that functions are "generalized real numbers". Commented May 8, 2018 at 10:33