Compatible extensions from the finite simple core

Fix theta, put $p=p_{\theta }$ and $q=q_{\theta }$, and denote its finite core bound by K (the stated endpoint bound when theta is an endpoint). Density and countable choice give finite simple $s_n$ for $n\ge 1$ with $\mathrm{\parallel }{s}_n-f{\mathrm{\parallel }}_p<1\mathrm{/}n$ for any fixed $f\in L^p$. The bound $\mathrm{\parallel }T{s}_n-T{s}_m{\mathrm{\parallel }}_q\le K\mathrm{\parallel }{s}_n-s_m{\mathrm{\parallel }}_p$ makes $T{s}_n$ Cauchy. Completeness, with its stated countable-choice hypothesis, gives a limit; define $T_{\theta }f$ to be that limit.

To compare parameters a,b, fix a finite-valued measurable representative $f\in L^{p_a}\cap L^{p_b}$. For $n\ge 1$, set $E_n=\{1\mathrm{/}n\le \mathrm{\mid }f\mathrm{\mid }\le n\}$. This set has finite measure because $n^{-p_a}\mu (E_n)\le \int \mathrm{\mid }f{\mathrm{\mid }}^{p_a}$. On $E_n$, round each real and imaginary component toward zero to a multiple of $1\mathrm{/}{n}^2$, and put $s_n=0$ elsewhere. The rounding has finitely many values because the components are bounded by n; its fibers are measurable intervals, and its support lies in $E_n$. Also $\mathrm{\mid }{s}_n\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }$. At a point with f nonzero, it eventually belongs to $E_n$ and the rounding error is at most $\sqrt{2}\mathrm{/}{n}^2$; at a zero of f every $s_n$ is zero. Thus $s_n\to f$ pointwise and $\mathrm{\mid }{s}_n-f{\mathrm{\mid }}^{p_j}\le 2^{p_j}\mathrm{\mid }f{\mathrm{\mid }}^{p_j}$ for j=a,b. Dominated convergence applied to these errors gives simultaneous convergence in both source norms.

If $(t_n{)}_{n\ge 1}$ is a second core approximation converging to f, $\mathrm{\parallel }T{s}_n-T{t}_n{\mathrm{\parallel }}_q\le K(\mathrm{\parallel }{s}_n-f{\mathrm{\parallel }}_p+\mathrm{\parallel }{t}_n-f{\mathrm{\parallel }}_p)\to 0$, so the definition is independent of approximation. Approximate f and g separately; $\alpha s_n+\beta t_n$ approximates $\alpha f+\beta g$, and core linearity gives linearity of the limits. Norm continuity gives $\mathrm{\parallel }{T}_{\theta }f{\mathrm{\parallel }}_q\le K\mathrm{\parallel }f{\mathrm{\parallel }}_p$. Any bounded extension has the same limit on a dense core, proving uniqueness. This includes K=0.

By step 2.1, the one sequence $T{s}_n$ converges in $L^{q_a}$ to $T_{a}f$ and in $L^{q_b}$ to $T_{b}f$. The a.e.-subsequence theorem first gives a subsequence converging a.e. to a representative of $T_{a}f$; apply it again to that subsequence in $L^{q_b}$ to get a further subsequence converging a.e. to $T_{b}f$. Choosing representatives and taking the countable union of their measurable null discrepancies makes the two pointwise limits comparable on one conull set. Uniqueness of complex pointwise limits gives $T_{a}f=T_{b}f$ as classes. The supplier covers q=infinity as well.

If $f_0+f_1=g_0+g_1$ with endpoint components, then $h=f_0-g_0=g_1-f_1$ lies in $L^{p_0}\cap L^{p_1}$. Agreement gives $T_{0}h=T_{1}h$, so $T_0{f}_0+T_1{f}_1=T_0{g}_0+T_1{g}_1$. Componentwise addition and scalar multiplication prove linearity of this sum map.

If $p_0\le p_1$ and $f\in L^{p_{\theta }}$, split $u=f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }>1\}}$, $v=f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\le 1\}}$. For $t>0$, $t^s=\exp (s\log t)$; since exp is increasing and $\exp (0)=1$, $\log t$ has the sign of $t-1$. Thus powers increase with the exponent for $t>1$ and decrease for $0<t\le 1$; at $t=0$ all positive powers are zero. On the first set $\mathrm{\mid }f{\mathrm{\mid }}^{p_0}\le \mathrm{\mid }f{\mathrm{\mid }}^{p_{\theta }}$, and on the second $\mathrm{\mid }f{\mathrm{\mid }}^{p_1}\le \mathrm{\mid }f{\mathrm{\mid }}^{p_{\theta }}$. Thus $u\in L^{p_0}\cap L^{p_{\theta }}$ and $v\in L^{p_1}\cap L^{p_{\theta }}$. Pairwise agreement and linearity give $T_{\theta }f=T_{0}u+T_{1}v$. If $p_1<p_0$, reverse the endpoint labels in this split. If they coincide, f is already in both endpoint spaces. This proves the restriction assertion for all theta, including the endpoints.