# Does delta distribution remain continuous with respect to quasinorm?

I am thinking the accepted answer which is found here:

When viewing $\delta: \mathbb{S} \to \mathbb{R}$ (linear and continuous with respect to the usual semi-norms on the Schwartz-space – or similar on the space of test functions), it makes sense to say that $\delta$ is continuous.

and its extension to quasinorms.

Does the $\delta$ distribution with respect to quasinorm remain remain continuous?

A quasinorm is a nonnegative functional $|| \cdot ||$ on a vector space $X$ that satisfies $||x+y||_{X} \leq K( ||x||_{X} + ||y||_{X})$ for some $K \leq 0$ and all $x,y \in X$ and also $||\lambda x||_{X} = |\lambda| ||x||_{X}$ for all scalars $\lambda$. When $K=1$, then the quasinorm is called a norm. (Loukas Grafakos, Classical Fourier Analysis, 2009).

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Which quasinorms are you refering to? – Yiorgos S. Smyrlis Dec 21 '13 at 20:39
Probably $K\ge 0$? But the $L^p$ spaces' (from the questioner's comments after the answer below) norms are literal norms, with "$K$" $=1$. Dirac $\delta$ definitely does not extend continuously from test functions or Schwartz to $L^p$. – paul garrett Dec 21 '13 at 21:51
@paulgarrett Actually it does extend continuously for $p=\infty$. – Vobo Dec 21 '13 at 21:57
@Vobo, heh! Indeed. Nevertheless, I suspect this is not what the questioner had in mind. (And one should note that test functions are not dense in $L^\infty$, but only in the space of continuous functions going to $0$ at infinity.) – paul garrett Dec 21 '13 at 21:58
I am thinking the problem actually in the dense subset $D \subset L^{2}$ where test functions are dense. – Masi Dec 21 '13 at 22:34

$\delta\in {\mathscr S}'(\mathbb R)$, and hence it is a continuous linear functional on Schwartz class.

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So it does not matter if the delta distribution is with respect to the seminorm xor quasinorm. It remains continuous because $\delta \in \mathcal{S}'(\mathbb{R})$. Is this the case? – Masi Dec 21 '13 at 20:57
Do we have to define $\delta' = 0$ to remove the discontinuity? Or is it enough to say $\delta \in \mathcal{S}'(\mathbb{R})$ for also differentiability? To say that $\delta$ is differentiable, the notation of differentiability has to be defined by duality, I think. – Masi Dec 21 '13 at 21:09
Could you specify which quasinorms are you refering to? Dirac function is definitely continuous w.r.t. the seminorms which determine the topology of the Schwartz class. (Locally convex TVS.) – Yiorgos S. Smyrlis Dec 21 '13 at 21:09
$\delta'\ne 0$. In fact, it is defined on the elemements of $mathscr S$ as $\delta'(\varphi)=-\varphi'(0)$, and it is also continuous! – Yiorgos S. Smyrlis Dec 21 '13 at 21:12
This is an excellent answer! See Grafakos book about Tempered distributions and Fourier transform on $L^1 \cap L^2$ for clarification. – Masi Jan 9 '14 at 2:35

The answer is probably no, as the OP doesn't specify a (quasi-)norm. Let $\varphi$ be any non-negative test function with $\varphi(0)=1$ and consider $\varphi_n(x)=\varphi(nx)$.

For every $p<\infty$ you have $||\varphi_n||_p\to 0$, but $\delta(\varphi_n)=1$.

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The answer should be yes. I do not know the exact selected quasinorm. Feel free to choose one which satisfies the continuity. I do not follow the logic behind Yiorgos' answer. The quasinorm, I think, should be selected appropriately. – Masi Dec 22 '13 at 16:06
As I said, $\delta$ is only continuous for $p=\infty$. – Vobo Dec 22 '13 at 21:44
Assume p = 2 and you choose seminorm instead of quasinorm. Is there any chance that delta distribution is continuous? I know that it should be. I can also confirm now the continuity of delta distribution with $p = \infty$ and with quasinorm. – Masi Dec 23 '13 at 14:45
Hey, in your mentioned book for the cas $p=2$ the seminorm and the quasinorm coincide in the norm $||f||=(\int |f|^p)^{1/p}$. $\delta$ is not continuous for this norm as you see by the example above. – Vobo Dec 23 '13 at 15:06
Does quasinorm and seminorm also coincide for $p=\infty$? I am thinking if the delta distribution is continuous with some seminorm when $p = \infty$. – Masi Dec 23 '13 at 15:16