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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.

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  • $\begingroup$ The notation everybody use is $\langle T,\varphi \rangle$ with $T$ a distribution and $\varphi \in C^\infty_c$ or Schwartz or $C^\infty$. When $T$ is represented by a function, call it $f$. $\endgroup$
    – reuns
    Nov 23, 2017 at 3:34
  • $\begingroup$ I just followed the original notation used in the lecture and in the task. $\endgroup$
    – Cornman
    Nov 23, 2017 at 3:36
  • $\begingroup$ I'm quite sure you exchanged $f,\varphi$, and again use $\langle f, \varphi \rangle$ $\endgroup$
    – reuns
    Nov 23, 2017 at 3:37
  • $\begingroup$ I did not, and why does that matter? $\endgroup$
    – Cornman
    Nov 23, 2017 at 3:41

1 Answer 1

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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 $\left<f,\varphi_{n}'\right>=-\left<f',\varphi_{n}\right>\rightarrow-\left<f',\varphi\right>=\left<f,\varphi'\right>$.

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  • $\begingroup$ What do you note by $\langle f_n, \varphi\rangle$? A net? $\endgroup$
    – Cornman
    Nov 23, 2017 at 3:16
  • $\begingroup$ That is simply $\varphi(f_{n})$. $\endgroup$
    – user284331
    Nov 23, 2017 at 3:17
  • $\begingroup$ So this is the answer on why $\varphi'$ is continuous, right? Can you also help me with the part, that $D$ is continuous regarding the weak-$\ast$-topology? $\endgroup$
    – Cornman
    Nov 23, 2017 at 3:19
  • $\begingroup$ Yes, and I have edited. $\endgroup$
    – user284331
    Nov 23, 2017 at 3:48

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