Let $X=(X, \|\cdot\|_X)$ and $Y=(Y, \|\cdot\|_Y)$ be Banach spaces such that $X \hookrightarrow Y$, that is, $X$ is continuous embedding in $Y$. In other words, the inclusion map $i: X \longrightarrow Y$ is continous. Suppose further that $i$ is compact operator, that is, $X$ is compactly embedded in $Y$.
Let $A \subset X$ be a closed subspace of $X$. Thus, $A=(A, ||\cdot||_X)$ is a Banach space.
Question. It's true that $A$ is compactly embedded in $Y$?
I think it's true, since $A \subset X \subset Y$, $i: A \longrightarrow Y$ is continuous and a compact operator. My reasoning is right?
Best Answer
Clearly the inclusion map $\iota:A\hookrightarrow X$ is bounded and thus continuous, by the ideal property of the compact operators $I^{\infty}(X,Y)=\{T\in\mathcal{L}(X,Y): T \text{ compact}\}$, which says that for a bounded linear operator $S\in\mathcal{L}(Z,X)$ we have $TS\in I^{\infty}(Z,Y)$ you see that $$A\overset{\iota}{\hookrightarrow}X\overset{i}{\hookrightarrow}Y$$ yields a compact embedding $i\circ \iota.$