Carleson restricted weak interpolation

For q>1 define the testing functional $N_q(h)={\sup }_{B}m(B{)}^{-1+1\mathrm{/}q}{\int }_B\mathrm{\mid }h\mathrm{\mid }$, where the supremum is over measurable B of finite positive measure. It is homogeneous and subadditive by the integral triangle inequality. If $m\{\mathrm{\mid }h\mathrm{\mid }>t\}\le (K\mathrm{/}t{)}^q$, F1 with exponent one, applied on B, gives ${\int }_B\mathrm{\mid }h\mathrm{\mid }\le {\int }_0^{\mathrm{\infty }}\min (m(B),(K\mathrm{/}t{)}^q)dt=\frac{q}{q-1}Km(B{)}^{1-1\mathrm{/}q}$. For K>0 split the integral at $t=Km(B{)}^{-1\mathrm{/}q}$; for K=0 every positive level set is null and the bound is zero. Thus $N_q(h)\le q^{\mathrm{\prime }}K$, where $q^{\mathrm{\prime }}=q\mathrm{/}(q-1)$. Conversely, if $N_q(h)=D<\mathrm{\infty }$, apply the definition to $B=\{\mathrm{\mid }h\mathrm{\mid }>t\}\cap [-R,R]$. When its measure is positive, $tm(B{)}^{1\mathrm{/}q}\le D$; when zero the same measure bound holds. Increasing R gives $m\{\mathrm{\mid }h\mathrm{\mid }>t\}\le (D\mathrm{/}t{)}^q$. This proves both comparisons even if the original level set could have infinite measure.

If u is a nonnegative simple function with $0\le u\le a{\mathbf{1}}_E$, list its finitely many positive values in increasing order. It is the sum of their successive nonnegative differences times the indicators of the corresponding superlevel sets. Each such set lies in E and the differences sum to at most a. Linearity and step 1.1 therefore give $N_q(Au)\le q^{\mathrm{\prime }}{K}_{q}am(E{)}^{1\mathrm{/}q}$ for q=r,s. A complex simple u with $\mathrm{\mid }u\mathrm{\mid }\le a{\mathbf{1}}_E$ is the sum of the positive and negative real parts and i times the positive and negative imaginary parts, each bounded by $a{\mathbf{1}}_E$. Hence $N_q(Au)\le 4q^{\mathrm{\prime }}{K}_{q}am(E{)}^{1\mathrm{/}q}$. The same result is zero for null E because its indicator has zero output almost everywhere by the assumed restricted bound and linearity of the finite coefficient formula.

Let f be a complex simple function supported on a set of finite measure. Put $E_k=\{2^k<\mathrm{\mid }f\mathrm{\mid }\le 2^{k+1}\}$, $m_k=m(E_k)$ and $f_k=f{\mathbf{1}}_{E_k}$. There are only finitely many nonempty E_k, they are disjoint, and $f={\sum }_k{f}_k$ almost everywhere. For each integer n, split $f=f_n^{\mathrm{h}\mathrm{i}\mathrm{g}\mathrm{h}}+f_n^{\mathrm{l}\mathrm{o}\mathrm{w}}$ by summing over k>=n and k<n respectively. Subadditivity of the testing functional and step 2.1 give $N_r(A{f}_n^{\mathrm{h}\mathrm{i}\mathrm{g}\mathrm{h}})\le C_r{\sum }_{k\ge n}{2}^k{m}_k^{1\mathrm{/}r}$ and $N_s(A{f}_n^{\mathrm{l}\mathrm{o}\mathrm{w}})\le C_s{\sum }_{k<n}{2}^k{m}_k^{1\mathrm{/}s}$, with $C_q=8q^{\mathrm{\prime }}{K}_q$. Since $\mathrm{\mid }Af\mathrm{\mid }>2^n$ implies that one of the two pieces has magnitude greater than $2^{n-1}$, the converse comparison in step 1.1 bounds its level-set measure by $m\{\mathrm{\mid }Af\mathrm{\mid }>2^n\}\le (2C_r{)}^r{2}^{-nr}({\sum }_{k\ge n}{2}^k{m}_k^{1\mathrm{/}r}{)}^r+(2C_s{)}^s{2}^{-ns}({\sum }_{k<n}{2}^k{m}_k^{1\mathrm{/}s}{)}^s\mathrm{.}$

Set $b_k=2^{pk}{m}_k$. The sum over n of $2^{np}$ times the first term in step 3.1, apart from its constant, is ${\sum }_{n\in \mathbb{Z}}({\sum }_{j\ge 0}{2}^{-(p-r)j\mathrm{/}r}{b}_{n+j}^{1\mathrm{/}r}{)}^r\mathrm{.}$ For nonnegative numbers x_j and summable nonnegative weights w_j, finite Hölder gives $({\sum }_j{w}_j{x}_j{)}^r\le ({\sum }_j{w}_j{)}^{r-1}{\sum }_j{w}_j{x}_j^r$. Here $w_j=2^{-(p-r)j\mathrm{/}r}$ has finite sum W_r because p>r. Applying this inequality, then summing over n, bounds the display by $W_r^r{\sum }_k{b}_k$. One may first use finite sets of n,j and then increase them: every summand is nonnegative and each shifted sum of the finitely supported b is at most ${\sum }_k{b}_k$. Likewise the second term gives ${\sum }_{n\in \mathbb{Z}}({\sum }_{j\ge 1}{2}^{-(s-p)j\mathrm{/}s}{b}_{n-j}^{1\mathrm{/}s}{)}^s\le W_s^s{\sum }_k{b}_k,$ where $W_s={\sum }_{j\ge 1}{2}^{-(s-p)j\mathrm{/}s}<\mathrm{\infty }$ since s>p. These are two explicit geometric sums, not an interpolation theorem used as an undeclared supplier.

Apply F1 to Af and split the positive t-axis into $[2^n,2^{n+1})$. Monotonicity of its level-set measure yields $\mathrm{\parallel }Af{\mathrm{\parallel }}_p^p\le (2^p-1){\sum }_n{2}^{np}m\{\mathrm{\mid }Af\mathrm{\mid }>2^n\}$. The nonnegative sum is valid whether or not finiteness is known initially. Steps 3.1 and 4.1 bound it by $C{\sum }_k{2}^{pk}{m}_k\le C\int \mathrm{\mid }f{\mathrm{\mid }}^p$, because $\mathrm{\mid }f\mathrm{\mid }>2^k$ on E_k. Thus the strong bound is established for every finite-support simple f, with C depending only on r,p,s and the two restricted constants, not on the finite family or selector.

For general $f\in L^p(\mathbb{R})$, the same finite model formula makes sense: each packet belongs to $L^{p^{\mathrm{\prime }}}$ and $L^p$ by its Schwartz decay, so F4 makes the coefficients absolute. More explicitly an exponent M with Mp>1 and Mp'>1 gives integrable powers of the packet bound. For every integer $j\ge 1$ define $Q_j(t)=\mathrm{sgn}(t)\lfloor j\min (\mathrm{\mid }t\mathrm{\mid },j)\rfloor \mathrm{/}j$, with $\mathrm{sgn}(0)=0$, and put $f_j={\mathbf{1}}_{[-j,j]}(Q_j(\mathrm{Re}f)+i{Q}_j(\mathrm{Im}f))$. Each component of $f_j$ takes only the $2j^2+1$ values $k\mathrm{/}j$ with $\mathrm{\mid }k\mathrm{\mid }\le j^2$, so $f_j$ is a simple measurable function supported on a finite-measure set. Componentwise rounding toward zero gives $\mathrm{\mid }{f}_j\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }$, while $f_j\to f$ almost everywhere. Hence $\mathrm{\mid }{f}_j-f{\mathrm{\mid }}^p\le 2^p\mathrm{\mid }f{\mathrm{\mid }}^p$, and F5 gives $\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_p\to 0$. Hölder gives convergence of every packet coefficient; more strongly, the triangle inequality and selector bound give $\mathrm{\parallel }A{f}_j-Af{\mathrm{\parallel }}_p\le {\sum }_{u\in S}\mathrm{\parallel }{\varphi }_u{\mathrm{\parallel }}_p\mathrm{\parallel }{\varphi }_u{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\mathrm{\parallel }{f}_j-f{\mathrm{\parallel }}_p⟶0.$ This finite auxiliary sum may depend on S; it is used only to identify the limit, not in the uniform estimate. Pass to the limit in step 5.1 using Minkowski's norm continuity to obtain the stated uniform strong bound. Empty finite families, zero simple functions and zero endpoint constants are covered without division by them. The strict inequalities r<p<s are exactly what makes the two geometric weights summable; no endpoint weak-to-strong inference is made. AC is inherited as identified in F6. This conditional interpolation proof does not assume that the Hunt restricted estimates have already been proved.