TITLE: A compact property of linear operators.
QUESTION [7 upvotes]: Let $\Omega\subset\mathbb{R}^{n}$ be a smoothly, open,
bounded set and let $1<p<q<\infty$.

Is it true that any (linear) continuous operator $T:L^{p}\left(\Omega\right)\rightarrow L^{q}\left(\Omega\right)$
is a compact operator?
Is it true that any (linear) continuous operator $T:W^{2,p}\left(\Omega\right)\rightarrow W^{2,q}\left(\Omega\right)$
is a compact operator?

Here $W^{2,p}$ denotes the usual Sobolev space.
Thanks.
I think, in general, 1. is false and 2. is true. 
Any reference is enough for my purpose.

REPLY [5 votes]: Let me proof that (1) is not true for $2\le p<q$, $\Omega=(0,1)$. The trick is to use the so-called Rademacher functions 
$$
r_n: [0,1]\to \mathbb R, \ r_n(t) = sgn(\sin(2^n\pi t)).
$$
Then Khintchin's inequality tells us
$$
\|\sum_{n=1}^N a_n r_n \|_{L^q} \le C_q\|\sum_{n=1}^N a_n r_n \|_{L^2}
$$
where $C_q$ is independent of $N$. (It says even more: the $L^p$-norms are equivalent on the span of such functions.) If the right-hand side is bounded, then we can pass to the limit $N\to\infty$.
Now define $T$ to be the projection onto these functions
$$
Tx = \sum_{n=1}^\infty r_n \left(\int_0^1 r_n x \right).
$$
Then 
$$
\|Tx\|_{L^q} \le C_q\|Tx\|_{L^2}  \le c \|x\|_{L^2} \le \|x\|_{L^p},
$$
where I used Khintchine's inequality, the fact that $(r_n)$ are orthonormal on $L^2$, and $p\ge 2$. This proves that $T$ is continuous from $L^p$ to $L^q$. Moreover, $T$ is not compact: it holds $Tr_n=r_n$ for all $n$, and $(r_n)$ is orthonormal, hence does not contain a strongly converging subsequence.

If linearity of $T$ is not required, here is a much easier construction:
For (1), consider the mapping
$$
(T(f))(x):= |f(x)|^{p/q}.
$$
Then $T$ maps from $L^p$ to $L^q$. Moreover, $T$ acts as identity for functions that take only values in $\{0,1\}$. The set of all such functions is closed but not compact in $L^q$.

Assume (1) is not true. 
Let $\Omega$ have $C^{1,1}$ boundary. 
Then define
$$
T_2(f):= (-\Delta)^{-1}T(-\Delta f),
$$
where $(-\Delta)^{-1}: H^1_0(\Omega)^* \to H^1_0(\Omega)$ maps $L^q$ functions to $W^{2,q}(\Omega)$. Now $T_2$ is the identity on the set of functions with $-\Delta f(x) \in \{0,1\}$, which is not compact in $W^{2,q}$.
Hence (2) does not hold.