I was reading section 1.46 form Rudin Functional analysis enter image description here

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I had following doubt 1) Why linear functional $\phi_x(f)=f(x)$ is continous ?

2)Why $D_k$ is closed in $C^\infty$? (I do not understand the intersection of null space argument

3) why $C^\infty $ is complete?I know that over compact set it is complete using uniform convergence but i do not understand reasioning as set is open?

Please help me to understand above question

Any Help will be appreciated

  • $\begingroup$ The paragraph before last shows how Cauchy sequences converge, so the completeness is right there. $\endgroup$ – Henno Brandsma Jan 12 at 17:52

If $x\in K_n$ (every $x\in\Omega$ lies in some $K_n$), the continuity of $\phi_x$ follows from the fact that $|\phi_x(f)|\leq p_n(f)$ (considering the multi-index $\alpha=(0,...,0)$).

Then, $D_K=\cap_{x\in K^c} \ker(\phi_x),$ and as we've argued, $\phi_x$ is continuous for every $x$, so $\ker(\phi_x)$ is closed. The interesction of closed sets is closed.

For the completeness of $C^{\infty},$ Rudin has all the details in the text above. I'd recommend reading it again.


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