# The Hölder inequality for fractional order Sobolev seminorm?

This question is post on MSE a week ago. I move it here to draw more attention.

Let $u\in C^\infty(\bar I)$ be given where $I=(0,1)$. Define $$t(\alpha):=\left(\int_I\int_I \frac{|u(x)-u(y)|^\alpha}{|x-y|^{1+s\alpha}}\right)^{\frac1\alpha}$$ where $1<\alpha<2$, $0<s<1$ is fixed. Note the $t(\alpha)$ above defines the fractional order sobolev seminorm. See the Wikipedia article on "Sobolev space", section "Sobolev spaces with non-integer $k$".

We know that for usual $L^p$ space, we have, for $p<q$, $\|u\|_{L^p(I)}\leq \|u\|_{L^q(I)}$ by using Hölder inequality. So I am wondering whether similar properties hold for sobolev fractional seminorm with non-integer $k$.

That is, I am wondering for $1<\alpha_1<\alpha_2<2$, do we have $$t(\alpha_1)\leq Ct(\alpha_2)$$ hold, where $C$ is a constant does not depends on $u$. Just like what we usually have for $L^p$ norm. However, I tried to prove it by using Minkowski inequality or something like Hölder, but I can't do it… I think the domain $I=(0,1)$ would be important and I also tried to use Jensen inequality, but no lucky…

Any ideas? Thank you!

• sorry, you are right, there was something wrong with that – user83457 Nov 10 '16 at 16:03
• The embedding properties mentioned in the wikipedia article should give something like this, though one probably has to replace $t(\alpha)$ by $\|u\|_{\alpha}+t(\alpha)$. – Christian Remling Nov 11 '16 at 18:54

Your question can be rephrased by asking whether one has a Hölder estimate $$|u|_{W^{s, p}} \le C |u|_{W^{s, q}},$$ when $p < q$ or whether $W^{s, q} \subset W^{s, p}$.
One way to see this is to take $$u(x) = \sum_{j \in \mathbb{N}} c_j e^{i 2^j x},$$ and to observe that $$|u|_{W^{s, p}}^p \simeq \sum_{j \in \mathbb{N}} |2^{js} c_j|^p,$$ and then choose suitably $c_j$.
• I understand. Btw, would it make any difference if I restrict my $p$ and $q$ is a small interval? For example, if I assume that $1<p<q<1+\epsilon$ where $\epsilon>0$ could be arbitrary small. Would this make any different? – JumpJump Nov 17 '16 at 0:02
• No, in fact. The integral can be finite for some $q$ and infinite for every $p < q$. The only possibility is to decrease slightly the smoothness in the left-hand side of the inequality by taking a $W^{r, p}$ norm with $r <s$ and $q < p$. – Jean Van Schaftingen Nov 17 '16 at 8:23