Is $\varphi\in\mathcal{S}(\mathbb{R})'$, we define $\varphi'$ by $\varphi'(f):=-\varphi(f'), f\in\mathcal{S}(\mathbb{R})$.
For every $\varphi\in\mathcal{S}(\mathbb{R})'$ is $\varphi'$ a tempered distribution and the function $D:\mathcal{S}(\mathbb{R})'\to\mathcal{S}(\mathbb{R})', D\varphi=\varphi'$ is continuous regarding the weak-$\ast$-topology.
$\mathcal{S}(\mathbb{R})$ is the schwartz space, and $\mathcal{S}(\mathbb{R})'$ the set of continuous, linear functionals on $\mathcal{S}(\mathbb{R})$.
I have to show, that $\varphi'$ is a tempered distribution. Hence $\varphi'\in\mathcal{S}(\mathbb{R})'$.
We choose an arbitrary $\varphi\in\mathcal{S}(\mathbb{R})'$ and $f,g\in\mathcal{S}(\mathbb{R}), \lambda\in\mathbb{K}$.
I show, that $\varphi'$ is linear.
$\varphi'(f+\lambda g)=-\varphi((f+\lambda g)')=-\varphi(f'+\lambda g')\stackrel{\varphi~~ linear}{=}-\varphi(f')-\lambda\varphi(g')=\varphi'(f)+\lambda\varphi'(g)$
Question: How can I show, that $\varphi'$ is continuous?
Since $\varphi'(f)=-\varphi(f')$ and $-\varphi$ is continuous, there is nothing to show.
Now I also need to show, that $D:\mathcal{S}(\mathbb{R})'\to\mathcal{S}(\mathbb{R})', D\varphi=\varphi'$ is continuous regarding the weak-$\ast$-topology.
What do I have to show, that it is continuous regarding the weak-$\ast$-topology? Let $(\varphi_\lambda)_\lambda$ be a net in $\mathcal{S}(\mathbb{R})'$, $\varphi\in\mathcal{S}(\mathbb{R})'$. I have to show, that for every
$(\varphi_\lambda)_\lambda\to\varphi$ we have $D((\varphi_\lambda)_\lambda)\to D(\varphi)$ and convergence in the weak-$\ast$-topology is pointwise convergence.
Is that correct? I appreciate every kind of feedback.
Thanks in advance.
For $(f_{n})$ in the Schwartz class and $f_{n}\rightarrow f$ in Schwartz, then so is $f_{n}'\rightarrow f'$ in Schwartz, so $\left<f_{n}',\varphi\right>\rightarrow\left<f',\varphi\right>$ as $\varphi$ is continuous, multiply with negative sign we get $\left<f_{n},\varphi'\right>\rightarrow\left<f,\varphi'\right>$.
For $(\varphi_{n})$ in the dual of Schwartz, and $f$ a Schwartz function, then $\left<f,\varphi_{n}'\right>=-\left<f',\varphi_{n}\right>\rightarrow-\left<f',\varphi\right>=\left<f,\varphi'\right>$.