0
$\begingroup$

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

enter image description here

enter image description here

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

$\endgroup$
  • $\begingroup$ The paragraph before last shows how Cauchy sequences converge, so the completeness is right there. $\endgroup$ – Henno Brandsma Jan 12 at 17:52
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.