Given the two differential operators in the Hilbert space $\mathcal{H}=L_2(0,1)$:
\begin{align*}
Af&=f'',\quad &D(A)=\left\{f:f\in H^2_2(0,1),f(0)=f(1)=0\right\}\\
Bf&=af',\quad &D(B)=\left\{f:f\in H^1_2(0,1),f(0)=f(1)=0\right\}
\end{align*}
Is $B$ relatively compact with respect to $A$ (in the sense of Kato)? Note that $a\geq 0$ is just some constant.
It is clear that $D(A)\subset D(B)$. I suppose the proof is complete if one further shows that $B$ is a compact operator acting on the graph of $A$ with graph norm $\|f\|=\|f\|_\mathcal{H}+\|Af\|_\mathcal{H}$.
On the other hand $B$ can be shown to have a compact inverse, so it is closed. Can't we then just verify $\|Bf\|_\mathcal{X}\leq c\|f\|_\mathcal{X}$, for some $c\geq 0$, thereby directly deducing relative compactness? I might be mixing up things here.
I am rather interested in proving the result for the particular example above, rather than considering the problem for general operators $A$, $B$.
Many thanks in advance!
Best Answer
Yes, $B$ is relatively compact with respect to $A$. We have to show that the embedding $(D(A),\|\cdot\|_A)\hookrightarrow (D(B),\|\cdot\|_B)$ is compact. The graph norm of $B$ is clearly equivalent to the $H^1_2$ norm if $a\neq 0$ and to the $L^2$ norm if $a=0$. It is not quite so obvious, but the graph norm of $B$ is equivalent to the $H^2_2$ norm: $$ \int_0^1 |f'(x)|^2\,dx=\int_0^1\left|\int_0^x f''(t)\,dt\right|^2\,dx\leq \int_0^1x\int_0^x|f''(t)|^2\,dt\,dx\leq\int_0^1 |f''(t)|^2\,dt. $$ Now the compactness of the embedding $D(A)\hookrightarrow D(B)$ follows from the Rellich embedding theorem, which asserts that the embedding $H^2_2\hookrightarrow H^1_2$ is compact.