Theorem 2.9.12CAB0E7754072

Proof. Assume first that $1<p<2$, and let $f\in L^ p\cap L^2$. Fix $\lambda >0$ and split $f=f_1+f_2$ with $f_1=f\mathbf1_{\{ |f|>\lambda \} }$ and $f_2=f\mathbf1_{\{ |f|\leq \lambda \} }$. Since $|f_1|\leq \lambda ^{1-p}|f|^ p$, we have $f_1\in L^1\cap L^2$, and $f_2\in L^2$. Apply lemma 2.9.11 to $f_1$, and Chebyshev's inequality with the $L^2$ bound to $f_2$, both with threshold $\lambda /2$. This gives

For a measurable function $h$, Tonelli's theorem applied to $|h(x)|^ p=\int _0^{|h(x)|}p\lambda ^{p-1}\, \mathrm d\lambda $ expresses $\| h\| _ p^ p$ as the integral of $p\lambda ^{p-1}$ times the distribution function $\lambda \mapsto |\{ |h|>\lambda \} |$, also when the norm is infinite. We apply this to $h=T_{ab}f$: multiply the last inequality by $p\lambda ^{p-1}$ and integrate in $\lambda $. By Tonelli's theorem again, the first term on the right contributes $C_ d\int |f|\int _0^{|f|}p\lambda ^{p-2}\, \mathrm d\lambda \, \mathrm d x=C_ dp\| f\| _ p^ p/(p-1)$, and the second contributes $C_ d\int |f|^2\int _{|f|}^\infty p\lambda ^{p-3}\, \mathrm d\lambda \, \mathrm d x=C_ dp\| f\| _ p^ p/(2-p)$; the inner integrals converge precisely because $p>1$ and $p<2$. Therefore

For $p=2$ the bound is Plancherel's theorem.

Now let $2<p<\infty $, put $p'=p/(p-1)\in (1,2)$, let $f\in L^ p\cap L^2$, and write $h=T_{ab}f\in L^2$. For integers $N\geq 1$, put $E_ N=B(0,N)\cap \{ |h|\leq N\} $ and $g_ N=\mathbf1_{E_ N}h|h|^{p-2}$. Then $|g_ N|\leq N^{p-1}\mathbf1_{E_ N}$, so $g_ N\in L^2\cap L^{p'}$, and $|g_ N|^{p'}=\mathbf1_{E_ N}|h|^ p$. Since $T_{ab}$ is self-adjoint on $L^2$, the case $p'$ already proved and Hölder's inequality give

The integral on the left is finite. If it is nonzero, dividing by its $(1/p')$th power gives $(\int _{E_ N}|h|^ p)^{1/p}\leq C_{d,p'}\| f\| _ p$; if it is zero, this bound is trivial. The sets $E_ N$ increase to $\{ |h|<\infty \} $, which has full measure since $h\in L^2$, so monotone convergence gives $\| T_{ab}f\| _ p\leq C_{d,p'}\| f\| _ p$.

By lemma 2.5.1(5), $C_ c^\infty \subset L^ p\cap L^2$ is dense in $L^ p$, and $L^ p$ is complete, so $T_{ab}$ has a unique bounded extension from $L^ p\cap L^2$ to $L^ p$. Finally, for $v\in C_ c^\infty $, lemma 2.9.10(3) gives $\| \partial _ a\partial _ bv\| _ p=\| T_{ab}\Delta v\| _ p\leq C_{d,p}\| \Delta v\| _ p$, and summing over $a$ and $b$ proves (2.9.13). $\square$