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The question is from Rudin's Functional Analysis chapter 1 number 17. It is stated as follows.

Show f$ \rightarrow D^\alpha f$ is a continuous mapping of $C^{\infty}(\Omega)$ into $C^{\infty}(\Omega)$ and also $\mathscr{D}_K$ into $\mathscr{D}_K$, for every multi-index $\alpha$.

I tried gathering all the relevant information I could below, but don't know where to go with it. Any and all help is appreciated. Thanks in advance.

$\alpha = (\alpha_1, \ldots, \alpha_n)$ is an ordered n-tuple of nonnegative integers.

$D^\alpha = \displaystyle \left(\frac{\partial}{\partial x_1}\right)^{\alpha_1} \dots \left(\frac{\partial}{\partial x_n}\right)^{\alpha_n}$

whose order is $ |\alpha | = \alpha_1 + \dots + \alpha_n$. If $|\alpha| = 0,$ $ D^\alpha f = f$.

I know the space $C(\Omega)$ is the vector space of all complex-valued continuous functions on $\Omega$ and $\Omega$ is the union of countably many compact sets $K_n \neq \emptyset$ which can be chosen so that $K_n$ lies in the interior of $K_{n+1}$ $(n = 1,2,3,\ldots)$.

$C^\infty(\Omega)$ is the set of all complex functions $f$ such that $D^\alpha f \in C(\Omega)$ for every multi-index $\alpha$.

If $K$ is a compact set in $\mathbb{R}^n$, then $\mathscr{D}_K$ denotes the space of all $f \in C^\infty(\mathbb{R}^n)$ whose support lies in $K$.

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It is a matter of the topology on $C^\infty(\Omega)$ which is given by the family of semi-norms $\|f\|_{K,m}=\sup\lbrace |D^\beta f(x)|: x\in K,\ |\beta|\le m\rbrace$ where $K\subseteq \Omega$ is compact, $n\in\mathbb N$, and $|\beta|=\beta_1+\cdots\beta_n$. Since $D^\alpha$ is linear it is enough to show continuity at $0$ which follows from $\|D^\alpha f\|_{K,m} \le \|f\|_{K,m+|\alpha|}$.

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  • $\begingroup$ Thank you. After getting a chance to talk with my professor he had walked me through the concept of the proof which agrees with what you have stated. $\endgroup$ – Ben Feb 4 '15 at 3:59

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