Does delta distribution remain continuous with respect to quasinorm?

Question

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).

Answer 1

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

Answer 2

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 $\delta(\varphi_n)=1$.