1
$\begingroup$

The problem is:

If $H_s\subset BC^k$, show that $\sup_{|\alpha|\leq k}|\partial^\alpha f|\leq C\|f\|_s$ using the closed graph theorem.

As I understand, the linear operator/map $\partial^\alpha$ maps $H_s$ to $C^1$, and it is a bounded operator if the linear map is closed.

I am a bit confused about how to use the fact that $H_s\subset BC^k$ there?

P.S. The $BC^k$ is the space of $C^k$ functions $f$ such that $\partial^\alpha f$ is bounded for $\alpha \leq k$. $H_s$ is Sobolev space of order $s$.

$\endgroup$
2
$\begingroup$

Note that if $H_s\subset BC^k$ then the map $I:H_s\to BC^k$ s.t. $f\mapsto I(f)=f$ is a well defined linear operator. Let's prove it has a closed graph.

Let $((f_n,f_n))_n$ be a sequence in the graph of $I$, s.t. it converges to some $(f,g)\in H_s\times BC^k$. We have to prove $g=f$. If $g\neq f$, then there are $\epsilon,r>0$ and $x_0$ s.t. $\Vert f(x)-g(x)\Vert>\epsilon$ if $x\in B(x_0,r)$. Hence $$ \forall n, \exists m>n, x\in B(x_0,r)\implies \Vert f_m(x)-f(x)\Vert>\epsilon/2 $$ from which we deduce there is some $\delta=\delta(\epsilon)>0$ s.t. $$ \forall n, \exists m>n, \Vert f_m-f\Vert_s>\delta $$ which is a contradiction as $f_m\to f$ in $H_s$.

By closed graph theorem, this implies $I$ is continuous, ie, bounded, ie, $$ \exists C>0, \forall f\in H_s, \Vert f\Vert_{BC^k}= \sup_{|\alpha|\leq k}\Vert\partial^\alpha f \Vert_\infty\leq C\Vert f\Vert_s $$

$\endgroup$
  • $\begingroup$ Thank you! just few questions, $I$ is just identity operator, right? $\endgroup$ – Jane Apr 10 '16 at 21:55
  • $\begingroup$ and what is the necessity of $BC^k$ there? why we can not do the same thing for just $C^k$? $\endgroup$ – Jane Apr 10 '16 at 21:56
  • $\begingroup$ @Jane. You are welcome. As for the first question, indeed, $I$ it is the identity operator. As for the second question, I think this is more a technicality than anything else, as there are functions in $C^k$ whose norm is infinite (like $f(x)=x$) $\endgroup$ – Nate River Apr 10 '16 at 22:01
  • $\begingroup$ ah, you mean for $C^k$ functions the left side of last line inequality will be infinite? $\endgroup$ – Jane Apr 10 '16 at 22:03
  • $\begingroup$ Not exactly. The problem is that $C^k$ with that norm is not really a normed vector space (the norm is not defined on all that space). Therefore, if we replace $BC^k$ with $C^k$, then we can't really apply the closed graph theorem. $\endgroup$ – Nate River Apr 10 '16 at 22:13

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.