Does $ W_1\hookrightarrow W_2 \hookrightarrow\hookrightarrow W_3$ imply that $W_1 \hookrightarrow\hookrightarrow W_3$?

Question

Let $(W_1, \|\cdot\|_1)$, $(W_2, \|\cdot\|_2)$ and $(W_3, \|\cdot\|_3)$ be three Banach spaces such that $$ W_1\hookrightarrow W_2 \hookrightarrow\hookrightarrow W_3, $$

where $W_1\hookrightarrow W_2$ means that $W_1\subset W_2$ and $\exists c_1>0$ such that $\|u\|_{W_2}\le c_1\|u\|_{W_1}$ for all $u\in W_1$. The definition of compact embedding $W_2 \hookrightarrow\hookrightarrow W_3$ is the one stated in here: https://en.wikipedia.org/wiki/Compact_embedding.

I am trying to understand if I can state that $$W_1 \hookrightarrow\hookrightarrow W_3.$$

I am pretty sure that the $W_1 \hookrightarrow W_3$ because $u\in W_1\implies u\in W_2$ and then $$\|u\|_{W_3}\le c_2 \|u\|_{W_2}\le c_1 c_2\|u\|_{W_1},$$ for a positive constant $c_2$. Hence the first property in https://en.wikipedia.org/wiki/Compact_embedding.

Anyone could help to understand if also the second property in https://en.wikipedia.org/wiki/Compact_embedding is satisfied and then $W_1 \hookrightarrow\hookrightarrow W_3$ is true?

Answer 1

If you use an equivalent formulation of compactness of an operator, I think the answer is relatively simple.

You just need to take a bounded set $A\subseteq W_1$ and you need to check whether its image under the continuous embedding $i_{1,3}:W_1\to W_3$ is relatively compact.

Since you took the continuous embedding to be the composition of the continuous embeddings, $i_{1,2}:W_1\to W_2$ and $i_{2,3}:W_2\to W_3$, you can check in steps what happens to a bounded set in $W_1$.

Take $A\subseteq W_1$ bounded. Hence, there exists an $M>0$ such that $A\subseteq B_{\Vert \cdot \Vert_1}(0;M)$, a ball around $0$ at radius $M$ in $W_1$. Since $i_{1,2}$ is continuous, you see that

$$i_{1,2} ( A ) \subseteq B_{\Vert \cdot \Vert_2}(0;c_2M). $$

Note also that $i_{1,3}(A)= i_{2,3} \big( i_{1,2}(A) \big)$. Since $i_{2,3}$ is compact and $C:=i_{1,2}(A)$ is bounded, you get that $i_{2,3}(C)$ is relatively compact in $W_3$.

This shows that $i_{1,3}$ maps bounded sets to relatively compact sets. You could reformulate this logic to any of the equivalent definitions of a compact operator.

Answer 2

It is enough to prove that the identity operator $i : W_{1} \to W_{3}$ is a compact operator (see the wiki page you linked). But note that $i = j \circ k$, where $j : W_{2} \to W_{3}$ and $k : W_{1} \to W_{2}$ are the identity maps. Since $W_{2} \hookrightarrow \hookrightarrow W_{3}$ we know that $j$ must be compact. And certainly $k$ is a continuous map by the continuous embedding. Since $i$ is a composition of a continuous operator with a compact operator then $i$ is compact and we are done.

Answer 3

If $(v_k)_{k \in \mathbb{N}}$ is a bounded sequence in $W_1$, then it is also bounded in $W_2$, because of the continuous embedding $W_1 \hookrightarrow W_2$. This follows directly from the definition that you correctly pointed out.

But according to the compact embedding $W_2 \hookrightarrow \hookrightarrow W_3$, the sequence $v_k$ has a convergent subsequence in $W_3$. So indeed $W_1 \hookrightarrow \hookrightarrow W_3$.