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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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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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