Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question
1  
Which quasinorms are you refering to? –  Yiorgos S. Smyrlis Dec 21 '13 at 20:39
1  
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
1  
@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

2 Answers 2

up vote 1 down vote accepted

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

share|improve this answer
    
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
1  
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
1  
$\delta'\ne 0$. 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 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 $\delta(\varphi_n)=1$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.