TITLE: The Minkowski inequality for fractional order?
QUESTION [6 upvotes]: 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 here, roll down for the section "Sobolev spaces with non-integer $k$". 
We know that for $L^p$ space, if $p<q$ then $\|u\|_{L^p(I)}\leq \|u\|_{L^q(I)}$ by using holder inequality. So I am wondering whether similar properties hold for sobolev space with non-integer $k$.(of course, for integer case it is again a Holder inequality)
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 sth like holder, 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!

REPLY [2 votes]: Call $t(\alpha,s)$ what you define as $t(\alpha)$. If $s_{1}-\frac{1}
{\alpha_{1}}=s_{2}-\frac{1}{\alpha_{2}}$ with $s_{1}\geq s_{2}$ (or
equivalently $\alpha_{2}\geq\alpha_{1}$), then you should have
$$
t(\alpha_{2},s_{2})\leq ct(\alpha_{1},s_{1}).
$$
There is a paper of Simon that you can find online at
http://math.unice.fr/\symbol{126}jsimon/pubs/Simon-E4.pdf
Take a look at Theorem 10.
If you don't change $s$ as you are trying to do, then there are
counterexamples in Stein's book "Singular integrals and differentiability properties of functions". For $s<1/\alpha$, the
function
$$
f(x)=|x|^{s-1/\alpha}(\log1/|x|)^{-\delta}
$$
for $|x|<1/2$ extended smoothly and with compact support to all of
$\mathbb{R}$ should satisfy
$$
t(\alpha,s)<\infty
$$
if and only if $\delta\alpha>1$. It's an exercise in the book.