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 $s$.


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, $$ \exists C>0, \forall f\in H_s, \Vert f\Vert_{BC^k}= \sup_{|\alpha|\leq k}\Vert\partial^\alpha f \Vert_\infty\leq C\Vert f\Vert_s $$

  • $\begingroup$ 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

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