Spectral interpolation and lower-order absorption

Put $a=\rho \mathrm{/}\tau \in (0,1)$ and $w_s(\lambda )=\mathrm{\mid }{h}_s(\lambda ){\mathrm{\mid }}^2=\max (1,\lambda {)}^s$. For $f\in H_D^{\tau }$, F2 makes both $1$ and $w_{\tau }$ integrable against the finite positive measure ${\mu }_f$, with $\int 1d{\mu }_f=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$ and $\int w_{\tau }d{\mu }_f=\mathrm{\parallel }f{\mathrm{\parallel }}_{H_D^{\tau }}^2$. On all of $\mathbb{R}$ one has $w_{\rho }=w_{\tau }^a$. Hölder with exponents $1\mathrm{/}a$ and $1\mathrm{/}(1-a)$ applied to $w_{\tau }^a$ and $1$ gives $\int w_{\rho }d{\mu }_f\le (\int w_{\tau }d{\mu }_f{)}^a(\int 1d{\mu }_f{)}^{1-a}$. The domain formula puts $f$ in $H_D^{\rho }$, and the norm identity and square roots prove the first estimate. For $0\le s\le \tau$, the pointwise inequality $w_s\le w_{\tau }$ on all of $\mathbb{R}$ also gives the continuous inclusion $H_D^{\tau }\hookrightarrow H_D^s$ with norm at most one.

Let $g\in L^2(0,T;L^2(\mathrm{\Omega }))$ be strongly measurable. After setting it and its defining simple approximants $s_m$ to zero on their common exceptional time-null set, take $s_m(t)\to g(t)$ in spatial $L^2$ almost everywhere. Put $h_m(t)=s_m(t){\mathbf{1}}_{\{\mathrm{\parallel }{s}_m(t){\mathrm{\parallel }}_2\le 2\mathrm{\parallel }g(t){\mathrm{\parallel }}_2\}}$. The set in this expression is measurable, and $h_m$ is still finite-valued and measurable. If $g(t)=0$, then $h_m(t)=0$; otherwise the cutoff eventually equals one. Thus $h_m(t)\to g(t)$ in $L^2$ almost everywhere and $\mathrm{\parallel }{h}_m(t)-g(t){\mathrm{\parallel }}_2^2\le 9\mathrm{\parallel }g(t){\mathrm{\parallel }}_2^2$. F7 gives $h_m\to g$ in Bochner $L^2$. By F10 choose measurable scalar representatives for all spatial $L^2$ values of all the $h_m$, a countable family of finite sets of values. On the measurable time partition defining $h_m$, use those representatives to define a jointly measurable function $U_m(t,x)$. F6 gives $\mathrm{\parallel }{U}_m{\mathrm{\parallel }}_{L^2((0,T)\times \mathrm{\Omega })}^2={\int }_0^T\mathrm{\parallel }{h}_m(t){\mathrm{\parallel }}_2^{2}dt$ and $\mathrm{\parallel }{U}_m-U_l{\mathrm{\parallel }}_{L^2((0,T)\times \mathrm{\Omega })}^2={\int }_0^T\mathrm{\parallel }{h}_m(t)-h_l(t){\mathrm{\parallel }}_2^{2}dt$. Hence $U_m$ is Cauchy in complex product-space $L^2$. F8 gives a measurable norm limit $U$ and a subsequence $U_{m_k}\to U$ product-almost everywhere. The chosen representatives may be used: their countably many disagreements with representatives from F8 have null union.

For $0<\theta <\tau$ put $a=\theta \mathrm{/}\tau$, $X=\mathrm{\parallel }f{\mathrm{\parallel }}_{H_D^{\tau }}^2$ and $Y=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$. Step 1.1 gives $\mathrm{\parallel }f{\mathrm{\parallel }}_{H_D^{\theta }}^2\le X^a{Y}^{1-a}$. Apply Young with $p=1\mathrm{/}a$, $q=1\mathrm{/}(1-a)$ to $u=(\epsilon \mathrm{/}a{)}^a{X}^a$ and $v=(a\mathrm{/}\epsilon {)}^a{Y}^{1-a}$. This yields $X^a{Y}^{1-a}\le \epsilon X+(1-a)(a\mathrm{/}\epsilon {)}^{a\mathrm{/}(1-a)}Y$. For $\theta =0$, the desired inequality holds with $C_{\epsilon }=1$. Thus the absorption estimate holds, including $\theta =1$, $\tau =2$.

The continuous inclusions in step 1.1 carry strong $H_D^{\tau }$ measurability into strong $H_D^{\rho }$ and $L^2$ measurability, by applying the bounded inclusion to the defining simple approximants. Applying step 1.1 almost everywhere and then Hölder in time gives ${\int }_0^T\mathrm{\parallel }{f}_j(t){\mathrm{\parallel }}_{H_D^{\rho }}^{2}dt\le ({\int }_0^T\mathrm{\parallel }{f}_j(t){\mathrm{\parallel }}_{H_D^{\tau }}^{2}dt{)}^a({\int }_0^T\mathrm{\parallel }{f}_j(t){\mathrm{\parallel }}_2^{2}dt{)}^{1-a}$. The right side tends to zero by the stated uniform bound and convergence. An essential uniform bound in $H_D^{\tau }$ gives the required $L^2$ bound because $T<\mathrm{\infty }$. This proves the Bochner convergence claims; the space-time formulation will follow from the identification proved next.

By F6, outside a single time-null set, $U(t,\cdot )$ belongs to spatial $L^2$, all relevant sections are measurable, and $U_{m_k}(t,x)\to U(t,x)$ for almost every $x$. Enlarge this exceptional set by the one where $h_m(t)\to ̸g(t)$ in $L^2$. Fix a remaining time $t$ and an integer $r$. Apply F9 in space to $\mathrm{\mid }{U}_{m_k}(t,x)-U_r(t,x){\mathrm{\mid }}^2$. Since the corresponding $L^2$ classes are $h_{m_k}(t)-h_r(t)$, this gives $\mathrm{\parallel }U(t,\cdot )-U_r(t,\cdot ){\mathrm{\parallel }}_2^2\le {\mathrm{lim\hspace{0.17em}inf}}_k\mathrm{\parallel }{h}_{m_k}(t)-h_r(t){\mathrm{\parallel }}_2^2=\mathrm{\parallel }g(t)-h_r(t){\mathrm{\parallel }}_2^2$. Letting $r\to \mathrm{\infty }$ shows that the class of $U(t,\cdot )$ equals $g(t)$. A final application of F6 proves $\mathrm{\parallel }U{\mathrm{\parallel }}_{L^2((0,T)\times \mathrm{\Omega })}^2={\int }_0^T\mathrm{\parallel }g(t){\mathrm{\parallel }}_2^{2}dt$. Any two jointly measurable representatives of $g$ agree product-almost everywhere, by F6 applied to their disagreement set. Thus this product norm identity is independent of the representative.

In the final assertion, the finite interval and the uniform $L^{\mathrm{\infty }}(0,T;H_D^{\tau })$ bound give the uniform Bochner $L^2(0,T;H_D^{\tau })$ bound. Step 1.1 and the simple-approximation argument in step 2.2 make every $f_j$ strongly measurable into spatial $L^2$ with finite Bochner $L^2$ norm. Steps 1.2 and 2.3 therefore supply its jointly measurable representative and identify its squared product-space norm with ${\int }_0^T\mathrm{\parallel }{f}_j(t){\mathrm{\parallel }}_2^{2}dt$. Space-time $L^2$ convergence to zero is consequently exactly the Bochner $L^2(0,T;L^2(\mathrm{\Omega }))$ convergence required in step 2.2. This proves the final convergence claim without an additional measurability hypothesis.