I am thinking again the accepted answer which is found here:
When viewing $\delta: \mathbf{S} \to \mathbf{R}$ (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 changing the spaces.
Assume that viewing $\delta : \mathcal{S}(\mathbf{R}) \to \mathbf{C}$ with respect to the usual semi-norms on the Schwartz space. Is usual seminorm enough to keep $\delta$ continuous wit new selection of domain and range?
I think the usual seminorm is a function on a vector space $V$, denoted $\lVert v \rVert$, such that the following conditions hold for all $v$ and $w$ in $V$, and any scalar $c$.
- $\lVert v \rVert \geq 0$
- $\lVert c v \rVert = |c| \lVert v \rVert$, and
- $\lVert v + w \rVert \leq \lVert v \rVert + \lVert w \rVert$.
It is possible for $\lVert v \rVert = 0$ for nonzero $v$. The functional $\lVert f \rVert = |f(0)|$ for continuous functions is a seminorm which is not a norm.
Let $\mathbb R$ be the real line. Let $\mathcal S(\mathbb R)$ be the set of (complex-valued) Schwarz functions, that is $g \in \mathcal S(\mathbb R)$ iff $g$ is infinitely differentiable and all derivatives of $g$ are rapidly decreasing at $\pm \infty$. For nonnegative integers $n,m$, seminorms are defined by $$ \|g\|_{n,m} = \sup\left\{|x^n g^{(m)}(x)|\;:\; x \in \mathbb R\right\} \tag{*}$$ Define functional $\delta : \mathcal S(\mathbb R) \to \mathbb C$ by $$ \delta(g) = g(0). $$ TO SHOW: $\delta$ is continuous, in the topology defined by the seminorms (*).
PROOF. Let $\epsilon > 0$ be given, and let $f \in \mathcal S(\mathbb R)$. To show that $\delta$ is continuous at $f$, it suffices to show: there exist $n,m \in \mathbb N$ and $\delta > 0$ so that, for all $g \in \mathcal S(\mathbb R)$, $$ \text{if } \quad \|g - f\|_{n,m} < \delta \quad\text{then}\quad |\delta(g) - \delta(f)|<\epsilon. $$ In fact, we can take $n=0, m=0, \delta=\epsilon$. If $\|g-f\|_{0,0} < \epsilon$, then $\sup \{|g(x)-f(x)| : x \in \mathbb R\} < \epsilon$, so in particular $|g(0)-f(0)|<\epsilon$, which is to say $|\delta(g) - \delta(f)|< \epsilon$.