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I am trying to understand how to start with this exercise.

Let $\Omega \subset \mathbb{R}^n$, $n\ge 3$, open and bounded and $$ C^{1,b}(\Omega)= \{\,f\in C^1(\Omega): \text{$f$ and all its partial derivatives $D_if$ are bounded} \} $$ with the norm $\,\|f\|=\|f\|_{\infty} + \sum_i \|D_if\|_\infty$.

Now let $g:\Omega \times \Omega \rightarrow \mathbb{R}$, such that, $$ x \neq y \,\,\Longrightarrow\,\, |g(x,y)| \le C\,|x-y|^2 $$ for some constant $C>0$, and $$ \int _{\Omega} g(.,y)\,f(y)\, dy \in C^{1,b}(\Omega)\quad \text{for all $\,f\in C^b(\Omega)$.} $$ Then show that there is a constant $A$ such that $$ \Big|\,D_i \int_{\Omega} g(x,y) \,f(y)\, dy\,\Big|\le A\|f\|_{\infty}, \quad\text{for all $\,\,f \in C^b(\Omega),\, x \in \Omega$ and $i=1,...,n$.} $$ Okay let's summarize this. What I probably need to show is that the map $V: C^b(\Omega) \rightarrow C^{1,b}(\Omega)$, $f \mapsto \int_{\Omega} g(.,y)f(y) dy$ is a continuous map(it is linear). If I have this, then this solves the excercise. What's the problem here?- The thing is, we do not know anything about the nature of $g$. Furthermore, I notice that all spaces that appear here are Banach spaces!

By the way, in this excercise is a hint given that we shall use the following theorem:

$X,Y,Z$ Banach spaces and $T:X\rightarrow Y$ linear and $J:Y \rightarrow Z$ linear, injective and continuous as well as $J \circ T$ be continuous, then $T$ is also continuous.

I do not see how this is helpful cause I do not see any injective operator that appears here.

If anything is unclear, please let me know.

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  • $\begingroup$ I wonder if you can choose $J \colon C^{1,b}(\Omega) \to C^b(\Omega)$ as the natural injection map. $\endgroup$ – Siminore Jan 9 '14 at 9:04
  • $\begingroup$ but in that case you would have to show that the map $J \circ T$ is continuous too, which appears to be somewhat hard. $\endgroup$ – user66906 Jan 9 '14 at 19:19
  • $\begingroup$ Here is the proof of that lemma $\endgroup$ – Norbert Jan 11 '14 at 13:21
  • $\begingroup$ @Siminore Then what to take for $T$ ? $\endgroup$ – DiffeoR Jan 14 '14 at 11:13
  • $\begingroup$ @Lipschitz Can you explain "What I probably need to show is that the map $V : C^b(\Omega) \rightarrow C^{1,b}(\Omega)$,$$ f \rightarrow \int_{\Omega} g(\cdot,y)f(y)\;dy $$ is a continuous map(it is linear). If I have this, then this solves the excercise. " How ? I don't see $D_i$ which is appearing in the last inequality we are trying to answer. $\endgroup$ – DiffeoR Jan 14 '14 at 11:21
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Following Siminore's idea, and your comment, it seems that we need only show $$ \Big|\int_{\Omega} g(x,y) \,f(y)\, dy\,\Big|\le A\|f\|_{\infty}$$

But $$ \Big|\int_{\Omega} g(x,y) \,f(y)\, dy\,\Big|\le\int_{\Omega} C|x-y|^2 \,f(y)\, dy \le A\|f\|_{\infty},$$ where $A=\sup_{x\in \Omega}\int_{\Omega} C|x-y|^2 dy$.

Now the bound on g was never fully used, so let me know if there is a mistake here.

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  • $\begingroup$ What are you trying to show here ? Because $D_i$ is not coming into picture. $\endgroup$ – DiffeoR Jan 14 '14 at 10:45
  • $\begingroup$ @DiffeoR see Siminore's comment. We need only work in $C^b$. $\endgroup$ – Brian Rushton Jan 14 '14 at 12:02
  • $\begingroup$ What I understand you are showing the boundedness of the operator $V$ defined by @Lipschitz in the problem. $T$ has to be $$D_i \circ V$$ or may be $$ f + \sum_{i = 1}^{n} D_i \circ V, $$ otherwise what is the use of $J$ and $T$ because the $D_i$ is not appearing either in $T$ (which I think you have assumed as $V$) or $T \circ J$ which would be continuous again if we proved $T$ is continuous. Please also see my question to Lipschitz in the comment to the problem, how just continuity of $V$ is sufficient to arrive at the solution of the problem ? Siminore, and Lipschitz also see. $\endgroup$ – DiffeoR Jan 14 '14 at 12:28
  • $\begingroup$ @DiffeoR I am using inclusion, not differentiation, as my $J$. Differentiation is not injective. $\endgroup$ – Brian Rushton Jan 14 '14 at 12:34
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    $\begingroup$ @DiffeoR did you see the theorem at the end of the post? If we prove that the map to $C^b$ is continuous, the intermediate map to $C^{1,b}$ is also continuous and thus bounded, EVEN THOUGH the norms are different. It is an indirect proof using a difficult theorem. $\endgroup$ – Brian Rushton Jan 14 '14 at 12:42

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