A compact property of linear operators.

Question

Let $\Omega\subset\mathbb{R}^{n}$ be a smoothly, open, bounded set and let $1<p<q<\infty$.

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

Answer 1

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.

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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 $L^q$.

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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.