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It is easy to check that the differential operator $\partial^a$ (where $\alpha\in \mathbb{N}_0^n$) is continuous between the Sobolev spaces (with usual norms) $W^{m,p}(U)\to W^{m-|\alpha|,p}(U)$, where $p\in [1,+\infty]$, and $U$ is an open subset of $\mathbb{R}^n$.

My question is : do we know exactly the value of the norm of such (bounded) operator (in this generality, or with conditions on $U$ or the other parameters). (At least this norm is less than one, it is equal to one ?).

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Any chance you could say a little about why you care about the operator norm? I've never seen anyone need this before. I haven't tried to work out anything at all, but one obvious thing to try is rescaling the support of a compactly supported smooth function down to a single point. –  Deane Yang Jan 17 '13 at 17:44
Also, if you really want the exact operator norm, you need to say exactly which "usual" norms you're using on the Sobolev spaces, since there isn't universal agreement among various equivalent options. –  Mark Meckes Jan 17 '13 at 18:49
And of course just do everything in $R^1$. –  Deane Yang Jan 17 '13 at 19:46
If you are talking about $U = (0,1)$, $m=1$, and $\alpha=1$ then the answer is yes, as well as more generally (I believe) if $|\alpha|=1$. Consider the following - as you suggest, in general $||f^\prime||_{L^p} \leq ||f||_{W^{1,p}}$, but if you consider a sequence $f_n$ such that $||f_n^\prime||_{L^p}=1$ and $f_n \to 0$ strongly in $L^p$ (consider a variant of $\frac{sin(nx)}{n}$), then the left hand side of the inequality is one while the right hand side tends to one from above. –  Daniel Spector Jan 18 '13 at 17:47
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1 Answer

Indeed, the norm is one.

To see this, fix a cutoff function $\phi \in C^\infty_c(U)$ (which we only need if $U$ is unbounded, to make sure the constructed functions are integrable) and define

$f_n(x):=\phi(x) \frac{sin(nx_1)}{n^m}$.

Then $f_n \to 0$ strongly in $W^{m-1,p}(U)$ and

$||\nabla^m f_n||_{L^p}= ||\phi(x) sin(nx_1)||_{L^p} + o(\frac{1}{n})$.

Therefore, $||\partial^\alpha f||_{W^{m-|\alpha|,p}} = ||\phi(x) sin(nx_1)||_{L^p} + o(\frac{1}{n})$,

and so in the computation of

$\sup \frac{||\partial^\alpha f||}{||f||}$,

plugging in $f_n$ we have the following lower bound

$\sup \frac{||\partial^\alpha f||}{||f||} \geq \frac{||\phi(x) sin(nx_1)||_{L^p}+o(\frac{1}{n})}{||\phi(x) sin(nx_1)||_{L^p}+o(\frac{1}{n})}$.

Combining this with the upper bound mentioned in the question, we obtain that the norm is in fact one.

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