Inclusion map between Sobolev space

Question

For $Q=[-\pi,\pi)$ and $s\geq 0$, define the Sobolev space \begin{equation} H^s(Q):=\{f\in L^2(Q):\sum_{k\in\mathbb{Z}}(1+k^2)^s|\hat f(k)|^2<+\infty\} \end{equation} Prove that if $0\leq s <t$ then the inclusion map $I:H^t(Q)\rightarrow H^s(Q)$ is well defined, linear,bounded and compact.

Hint: Exploit that $S_N:H^t(Q)\rightarrow H^s(Q)$ is of finite rank (where $S_Nf(x)= \sum_{k=-N}^{N} \hat{f}(k)e^{ikx}$)

How can I proceed?

Answer 1

The hint is alluding to the following result about compact operators:

Theorem If $X$ is a Banach space, the subspace $\mathcal{K}(X,Y)$ of compact operators $X \rightarrow Y$ is a closed in the space of bounded linear operators $\mathcal{B}(X,Y).$

Proof sketch: We wish to show $TB_X$ is totally bounded, so let $\varepsilon>0.$ Then choose $n$ such that $\lVert T_n - T \rVert < \varepsilon/3$ and let $x_1,\dots,x_k \in B_X$ such that $\{T_nx_1,\dots,T_nx_k\}$ is an $\varepsilon/3$-net for $T_nB_X.$ Then by the triangle inequality, $\{Tx_1,\dots,Tx_k\}$ is an $\varepsilon$-net for $TB_X.$

Since finite rank operators are compact, it suffices to prove that $S_N \rightarrow I$ in $\mathcal{B}(H^t(Q),H^s(Q)).$ Indeed if $f \in H^t(Q)$ we have, \begin{align*} \lVert S_Nf - f \rVert_s^2 &= \sum_{|k|>N} (1+k^2)^s|\hat f(k)|^2 \\ &= \sum_{|k|>N} (1+k^2)^{s-t} (1+k^2)^t |\hat f(k)|^2 \\ &\leq (1+N^2)^{s-t} \sum_{|k|>N} (1+k^2)^t |\hat f(k)|^2 \\ &\leq \frac1{(1+N^2)^{t-s}} \lVert f \rVert_t^2. \end{align*} Note we used the inequality that $(1+k^2)^{s-t} \leq (1+N^2)^{s-t}$ for all $k \geq N,$ which is valid since $s-t < 0.$

This shows that $\lVert S_N - I \rVert \leq (1+N^2)^{-(t-s)/2},$ which tends to $0$ as $N \rightarrow \infty.$