# Tempered distribution, weak-$\ast$-topology, functional analysis

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.

• 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 $f$. Nov 23, 2017 at 3:34
• I just followed the original notation used in the lecture and in the task. Nov 23, 2017 at 3:36
• I'm quite sure you exchanged $f,\varphi$, and again use $\langle f, \varphi \rangle$ Nov 23, 2017 at 3:37
• I did not, and why does that matter? Nov 23, 2017 at 3:41

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

• What do you note by $\langle f_n, \varphi\rangle$? A net? Nov 23, 2017 at 3:16
• That is simply $\varphi(f_{n})$. Nov 23, 2017 at 3:17
• 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? Nov 23, 2017 at 3:19
• Yes, and I have edited. Nov 23, 2017 at 3:48