Cube mean estimate and compact form inclusion

Let $C$ be a cube of side $\delta >0$ and $u\in C_c^{\mathrm{\infty }}({\mathbb{R}}^N;\mathbb{C})$, with $u_C:=\mathrm{\mid }C{\mathrm{\mid }}^{-1}{\int }_{C}u$. Expanding the square and using Fubini on the finite cube gives ${\int }_C\mathrm{\mid }u-u_C{\mathrm{\mid }}^2=\frac{1}{2\mathrm{\mid }C\mathrm{\mid }}{\int }_C{\int }_C\mathrm{\mid }u(x)-u(y){\mathrm{\mid }}^{2}dxdy\mathrm{.}$ Telescope $u(x)-u(y)$ through the $N$ coordinate changes and use $\mathrm{\mid }{\sum }_{j=1}^N{z}_j{\mathrm{\mid }}^2\le N{\sum }_j\mathrm{\mid }{z}_j{\mathrm{\mid }}^2$. For the $j$th change, the fundamental theorem of calculus and Cauchy–Schwarz give $\mathrm{\mid }{D}_j(x,y){\mathrm{\mid }}^2\le \delta {\int }_{\min (x_j,y_j)}^{\max (x_j,y_j)}\mathrm{\mid }{\mathrm{\partial }}_{j}u{\mathrm{\mid }}^2$. For fixed values of the other coordinates and fixed $s\in [0,\delta ]$, the measure of pairs $(x_j,y_j)$ whose segment contains $s$ is $2s(\delta -s)\le {\delta }^2\mathrm{/}2$; integrating the displayed bound over the cube variables and dividing by $2\mathrm{\mid }C\mathrm{\mid }$ therefore yields the safe estimate $\mathrm{\parallel }u-u_C{\mathrm{\parallel }}_{L^2(C)}^2\le N{\delta }^2{\sum }_{j=1}^N\mathrm{\parallel }{\mathrm{\partial }}_{j}u{\mathrm{\parallel }}_{L^2(C)}^2\mathrm{.}$ It applies to complex $u$ because the squared-modulus Cauchy–Schwarz inequality is used.

For $R>0$ and $u\in Q$, the coercive weight gives ${\int }_{\mathrm{\mid }q\mathrm{\mid }>R}\mathrm{\mid }u(q){\mathrm{\mid }}^{2}dq\le \frac{1}{c{R}^2}{\int }_{\mathrm{\mid }q\mathrm{\mid }>R}c\mathrm{\mid }q{\mathrm{\mid }}^2\mathrm{\mid }u(q){\mathrm{\mid }}^{2}dq\le \frac{1}{c{R}^2}\mathrm{\parallel }u{\mathrm{\parallel }}_Q^2\mathrm{.}$ Thus the $L^2$ tails of the $Q$ unit ball are uniformly small for large $R$.

The estimate of step 1.1 passes to every $u\in Q$: approximate $u$ in the $Q$ norm by elements of $C_c^{\mathrm{\infty }}$ (the defining completion), restrict to $C$, and use [F1] to pass the $L^2$ and derivative limits.

Fix $\epsilon >0$. Choose $R$ so the squared tail in step 1.2 is below ${\epsilon }^2\mathrm{/}16$. For the $Q$ unit ball, [F1] gives ${\sum }_j\mathrm{\parallel }{\mathrm{\partial }}_{j}u{\mathrm{\parallel }}_2^2\le {\sum }_j{a}_j^{-1}$. Choose an integer $M$ large enough that $\delta :=2R\mathrm{/}M$ satisfies $N{\delta }^2{\sum }_j{a}_j^{-1}<{\epsilon }^2\mathrm{/}16$, and partition $[-R,R{]}^N$ into the resulting finitely many cubes of side $\delta$. Summing the squared estimates of step 2.1 over these disjoint cubes bounds the total mean-approximation error by $N{\delta }^2{\sum }_j\mathrm{\parallel }{\mathrm{\partial }}_{j}u{\mathrm{\parallel }}_2^2<{\epsilon }^2\mathrm{/}16$. The squared error outside $[-R,R{]}^N$ is the tail from step 1.2; thus replacing $u$ on each cube by its mean and setting it to zero outside the large cube has total squared $L^2$ error below ${\epsilon }^2\mathrm{/}8$, hence norm error below $\epsilon \mathrm{/}2$.

For the finite partition in step 3.1, the piecewise-constant mean function $m_u$ satisfies $\mathrm{\parallel }{m}_u{\mathrm{\parallel }}_2^2={\sum }_C\mathrm{\mid }C\mathrm{\mid }\mathrm{\mid }{u}_C{\mathrm{\mid }}^2\le {\sum }_C\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(C)}^2\le \mathrm{\parallel }u{\mathrm{\parallel }}_2^2\le 1$ by Cauchy--Schwarz [F3]. Thus each mean coordinate has modulus at most $\mathrm{\mid }C{\mathrm{\mid }}^{-1\mathrm{/}2}$, so the vectors of means lie in a bounded box in a finite-dimensional complex coordinate space. Subdivide its real-and-imaginary coordinates into finitely many boxes of sufficiently small side length so that the associated piecewise-constant functions approximate every mean vector to $L^2$ error below $\epsilon \mathrm{/}2$. Combining with step 3.1 gives a finite $\epsilon$-net for the image of the $Q$ unit ball. Scaling shows that the image of every bounded $Q$ set is totally bounded.

Let $K$ be the $H_N$ closure of the image of the $Q$ unit ball. It is complete because it is closed in the complete space $H_N$ [F1]. Suppose an open cover of $K$ has no finite subcover. For each $n\ge 1$, step 4.1 supplies a finite $2^{-n-2}$-net for the image of the $Q$ unit ball with centers in $H_N$; the corresponding closed balls of radius $2^{-n-1}$ cover its closure $K$. Starting with $C_0=K$, choose the least ball index for which $C_n:=C_{n-1}\cap \bar{B}(p_n,2^{-n-1})$ has no finite subcover. Such an index exists, since otherwise finitely many finite subcovers would cover $C_{n-1}$. Each $C_n$ is nonempty, closed, nested, and has diameter at most $2^{-n}$. For $m>n$, any point of nonempty $C_m$ lies in both selected balls, so $\mathrm{\parallel }{p}_m-p_n{\mathrm{\parallel }}_2\le 2^{-m-1}+2^{-n-1}$; hence $(p_n)$ is Cauchy and converges to some $p\in H_N$. Since $C_n$ is nonempty and is contained in both $K$ and $\bar{B}(p_n,2^{-n-1})$, the distance from $p_n$ to $K$ is at most $2^{-n-1}$. Therefore $\mathrm{dist}(p,K)\le \mathrm{\parallel }p-p_n{\mathrm{\parallel }}_2+2^{-n-1}\to 0$; closedness of $K$ gives $p\in K$. For fixed $n$ and every $m>n$, nonempty $C_m\subseteq C_n$ has a point $y\in C_m$ with $\mathrm{\parallel }{p}_m-y{\mathrm{\parallel }}_2\le 2^{-m-1}$. Thus there are points of $C_n$ arbitrarily close to $p$, and closedness gives $p\in C_n$. An open set of the cover containing $p$ contains $K\cap B(p,r)$ for some $r>0$; for large $n$, $\mathrm{diam}(C_n)<r$, so that one open set covers $C_n$, a contradiction. Thus $K$ is compact. All recursive ball choices are least admissible indices in finite lists.

Step 5.1 proves that the closed-unit-ball image under $J$ has compact closure; applying the equivalent criterion in [F2] to the linear inclusion proves that $J:Q\hookrightarrow H_N$ is compact.