# Norm of differential operator between Sobolev spaces

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 ?).

• 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. 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. Jan 17 '13 at 18:49
• And of course just do everything in $R^1$. 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. Jan 18 '13 at 17:47

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.