About the constant from the Marcinkiewicz Interpolation Theorem

Question

The Marcinkiewicz Interpolation Theorem states as following:

Let $T$ be a linear operator of weak type $(p_0,q_0)$ with constant $C_0$ and of weak type $(p_1,q_1)$ with constant $C_1$ where $q_0\neq q_1$. Now for any $t\in (0,1)$, define $p_t,q_t$ by $$ \frac{1}{p_t} = \frac{1-t}{p_0}+\frac{t}{p_1},\quad \frac{1}{q_t} = \frac{1-t}{q_0} + \frac{t}{q_1}. $$ Then $T$ is strong type $(p_t,q_t)$ with constant $C_t$. Moreover, this constant $C_t$ satisfies $$ C_t\leq C C_0^{1-t} C_1^t, $$ where $C=C(p_0, p_1, q_0, q_1, t)$ is bounded for $0<\epsilon \leq t\leq 1-\epsilon<1$ but tends to infinity as $t\to 0$ or $t\to 1$.

When we say an operator $T$ is of strong type $(p,q)$ with constant $C$, it means $|| Tf||_q \leq C||f||_p$.

When we say an operator $T$ is of weak type $(p,q)$ with constant $C$, it means $|| T f ||_{q,w}\leq C||f||_p $.

Every proof of this Theorem (for instance in Folland's "Real Analysis" or more accessibly in this writeup) that I have found has worked by obtaining an inequality involving the distribution function of $Tf$ of the form $$ \lambda_{Tf}(2\alpha)\leq \lambda_{Tg_A}(\alpha)+\lambda_{Th_A}(\alpha). $$ Then by applying the weak type estimates of $T$ and some clever integral manipulation, one obtains a final inequality like: $$ ||Tf||_{q_t}\leq 2 q_t^{\frac{1}{q_t}}\left[C_0^{q_0}(p_0/p_t)^{\frac{q_0}{p_0}}|q_t-q_0|^{-1}+C_1^{q_1}(p_1/p_t)^{\frac{q_1}{p_1}}|q_t-q_1|^{-1}\right]^{\frac{1}{q_t}}=:C_t, $$

for all $f$ with $||f||_p=1$. Then the general inequality follows from the linearity of $T$ because $|T(cf)|=c|T(f)|$ for $c>0$.

My question is as follows: I don't see how we can obtain from this that $C_t\leq C C_0^{1-t} C_1^t$. For example, if we take $q_0=3$, $q_1=6$, $t=\frac{1}{2}$, and $q_t=4$ which satisfies the hypothesis $$ \frac{1}{q_t} = \frac{1-t}{q_0} + \frac{t}{q_1}, $$ then (for some given choice of $p_0, p_1, p_t$) the quantity $$ \frac{\left[C_0^3 +C_1^6 \right]^{\frac{1}{4}}}{C_0^\frac{1}{2}C_1^\frac{1}{2}} $$ should remain bounded with respect to $C_0,C_1$. But it clearly doesn't (Just take $C_0=1$ or something and ramp $C_1\to \infty$).

So how do we obtain the inequality $C_t\leq C C_0^{1-t} C_1^t$? Is there an alternative proof of this?

Answer 1

This posting deals with off diagonal case of the Marcinkiewicz interpolation for $L_p$ spaces. Using stander methods from $L_p$ spaces (first year graduate course in Integration) I show that the method, of introducing a control parameter in the size of the Marcikiewicz decomposition of a function (is I did in my posting on the classical diagonal Marcinkiewicz interpolation) yields an operator bounds that of the sort $C(q_0,q_1,q) A^{t-1}_0A^t_1$.

Though the this posting, the measure involved ($\mu$ and $\nu$) are assumed to be $\sigma$-finite so that Fubini-Tonelli's theorem is applicable (in particular the generalized version of Minkowski's inequality).

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Suppose that $T:L_{p_0}(\mu)+L_{p_1}(\mu)\rightarrow L_0(\nu)$ is sub additive, and that $T$ is of weak type $(p_0,q_0)$ and $(p_1,q_1)$ with $1\leq p_j\leq q_j<\infty$, $q_0\neq q_1$, that is \begin{align} \nu\big(|Tg|>\alpha\big)\leq\left(\frac{A_j\|g\|_{p_j}}{\alpha}\right)^{q_j}\tag{0}\label{zero} \end{align} for $g\in L_{p_j}$, $j\in\{0,1\}$. Let \begin{align} \frac{1}{p}&=\frac{1}{p_0}+t\big(\frac{1}{p_1}-\frac{1}{p_0}\big)\tag{1}\label{one} \end{align} \begin{align} \frac{1}q& =\frac{1}{q_0}+t\big(\frac{1}{q_1}-\frac{1}{q_0}\big)\tag{2}\label{two} \end{align} where $0<t<1$. We show that $T$ is of strong-type $(p,q)$ and that $$\|Tf\|_q\leq 2\Big(\frac{q}{q-q_0}+\frac{q}{q_1-q}\Big)^{\frac1q}A^{1-t}_0A^t_1$$

Without loss of generality assume $p_0<p_1$.

Equations \eqref{one} and \eqref{two} imply that \begin{align} \frac{p_0(q-q_0)}{q_0(p-p_0)}=\frac{\frac1p\big(\frac1q-\frac1{q_0}\big)}{\frac1q\big(\frac1p-\frac1{p_0}\big)}= \frac{\frac1p\big(\frac1q-\frac1{q_1}\big)}{\frac1q\big(\frac1p-\frac1{p_1}\big)}=\frac{p_1(q_1-q)}{q_1(p_1-q)}\tag{3}\label{three} \end{align} Let us denote by $\sigma$ the value of the expression in \eqref{three}. Let $f\in L_p(\mu)$. For $\alpha>0$ and $\delta>0$ , the latter to be determined later, we have that $$f=f\mathbb{1}_{\{|f|>\alpha^\sigma\delta\}}+ f\mathbb{1}_{\{|f|\leq \alpha^\sigma\delta\}}$$ and $g=f\mathbb{1}_{\{|f|>\alpha^\sigma\delta\}}\in L_{p_0}(\mu)$ and $h=f\mathbb{1}_{\{|f|\leq\alpha^\sigma\delta\}}\in L_{p_1}(\mu)$. Then \begin{align} \nu(|Tf|>\alpha)&\leq \nu(|Tg|>\alpha/2)+\nu(|Th|>\alpha/2)\\ &\leq \Big(\frac{2A_0}{\alpha}\Big)^{q_0}\Big(\int_{\{|f|>\alpha^\sigma\delta\}}|f|^{p_0}\,d\mu\Big)^{\tfrac{q_0}{p_0}}+\Big(\frac{2A_1}{\alpha}\Big)^{q_1}\Big(\int_{\{|f|\leq \alpha^\sigma\delta\}}|f|^{p_1}\,d\mu\Big)^{\frac{q_1}{p_1}} \end{align} Consequently \begin{align} \|Tf\|^q_q&=q\int^\infty_0\alpha^{q-1}\nu(|Tf|>\alpha)\,dt\\ &\leq q(2A_0)^{q_0}\int^\infty_0\alpha^{q-q_0-1}\Big(\int_{\{|f|>\alpha^\sigma\delta\}}|f|^{p_0}\,d\mu\Big)^{\tfrac{q_0}{p_0}}\,dt \quad +\\ &\qquad\qquad q(2A_1)^{q_1}\int^\infty_0\alpha^{q-q_1-1}\Big(\int_{\{|f|\leq\alpha^\sigma\delta\}}|f|^{p_1}\,d\mu\Big)^{\tfrac{q_1}{p_1}}\,dt\tag{4}\label{four} \end{align}

The assumption $p_j<q_j$, $j\in\{0,1\}$ allows for and application of Minkowski's inequality to estimate the integrals on the right-hand-side of \eqref{four}. \begin{align} \left(\int^\infty_0\alpha^{q-q_0-1}\Big(\int_{\{|f|>\alpha^\sigma\delta\}}|f|^{p_0}\,d\mu\Big)^{\tfrac{q_0}{p_0}}\,dt\right)^{\frac{p_0}{q_0}} & \leq\int_X|f|^{p_0}\Big(\int^{\frac{|f|^{1/\sigma}}{\delta^{1/\sigma}}}_0\alpha^{q-q_0-1}\,d\alpha\Big)^{\frac{p_0}{q_0}}\,d\mu\\ &=\frac{\delta^{-\frac{q-q_0}{\sigma}\frac{p_0}{q_0}}}{(q-q_0)^{\frac{p_0}{q_0}}}\int_X|f|^{p_0} |f|^{\frac{q-q_0}{\sigma}\frac{p_0}{q_0}}\,\mu\\ &=\frac{\delta^{-(p-p_0)}}{(q-q_0)^{\frac{p_0}{q_0}}}\int_X|f|^p\,d\mu \end{align} where the last equalities follows from \eqref{three}. Similarly \begin{align} \left(\int^\infty_0\alpha^{q-q_1-1}\Big(\int_{\{|f|\leq\alpha^\sigma\delta\}}|f|^{p_1}\,d\mu\Big)^{\tfrac{q_1}{p_1}}\,dt\right)^{\frac{p_1}{q_1}} &\leq\int_X|f|^{p_1}\Big(\int^\infty_{\frac{|f|^{1/\sigma}}{\delta^{1/\sigma}}}\alpha^{q-q_1-1}\,d\alpha\Big)^{\frac{p_1}{q_1}}\,d\mu\\ &=\frac{\delta^{\frac{q_1-q}{\sigma}\frac{p_1}{q_1}}}{(q_1-q)^{\frac{p_1}{q_1}}}\int_X|f|^{p_1}|f|^{\frac{q-q_1}{\sigma}\frac{p_1}{q_1}}\,d\mu\\ &=\frac{\delta^{p_1-p}}{(q_1-q)^{\frac{p_1}{q_1}}}\int_X|f|^p\,d\mu \end{align} where, again, the last equalities follows from \eqref{three}. Putting things together, we obtain that for all $f\in L_p(\mu)$ with $\|f\|_p=1$ \begin{align} \|Tf\|^q_q\leq q\left(\frac{(2A_0)^{q_0}\delta^{-\frac{q_0}{p_0}(p-p_0)}}{q-q_0} + \frac{(2A_1)^{q_1}\delta^{\frac{q_1}{p_1}(p_1-p)}}{q_1-q}\right)\tag{5}\label{five} \end{align} Here is where the control parameter $\delta$ plays a role. Choosing $\delta$ so that \begin{align} (2A_0)^{q_0}\delta^{-\frac{q_0}{p_0}(p-p_0)}= (2A_1)^{q_1}\delta^{\frac{q_1}{p_1}(p_1-p)}\tag{6}\label{six} \end{align} we obtain that \begin{align}\|Tf\|^q_q\leq \Big(\frac{1}{q-q_0}+\frac{1}{q_1-1}\Big)(2A_1)^{q_1}\delta^{\frac{q_1}{p_1}(p_1-p)}\tag{7}\label{seven} \end{align} Equation \eqref{six}, along with \eqref{three}, implies that \begin{align} \delta=\frac{(2A_0)^{\frac{q_0\sigma}{q_1-q_0}}}{(2A_1)^{\frac{q_1\sigma}{q_1-q_0}}}\tag{8}\label{eight} \end{align} Substituting \eqref{eight} in \eqref{seven} yields \begin{align} \|Tf\|^q_q\leq\Big(\frac{q}{q-q_0}+\frac{q}{q_1-1}\Big)(2A_0)^{q\big(\frac1q-\frac1{q_1}\big)}(2A_1)^{q\big(\frac1{q_0}-\frac1q\big)}=2^q\Big(\frac{q}{q-q_0}+\frac{q}{q_1-1}\Big)(A_0)^{q(1-t)}(A_1)^{qt} \end{align} Since $|Tcf|=|c||Tf|$ for all $c\in\mathbb{C}$ by assumption, it follows that $$\|Tf\|_q\leq 2\Big(\frac{q}{q-q_0}+\frac{q}{q_1-q}\Big)^{\frac1q}A^{1-t}_0A^t_1$$

Answer 2

Bounds of the form $C(p_0,q_0) C^t_0 C^{1-t}_1$ are covered in Grafakos, L., Classical Fourier Analysis, Third Edition, Springer, Section 1.4.4.

For the diagonal version: $T:L_s(\mu)+L_r(\mu)\rightarrow L_0(\nu)$ whit $T$ of weak-$(s, s)$ and weak-$(r,r)$ type, on can obtain bounds of the for. stated by the OP by keeping some control constant (to be determined) on the size of a function $f\in L_p$ in its Marcinkiewics decomposition as follows:

Suppose $0<s<p<r<\infty$ and that $T:L_s(\mu)+L_r(\mu)\rightarrow L_0(\nu)$ is a sub additive operator such that \begin{align} \nu(|Tg|>\alpha)&\leq\Big(\frac{A_s}{\alpha}\|g\|_s\Big)^s,\qquad g\in L_s(\mu)\\ \nu(|Th|>\alpha)&\leq\Big(\frac{A_r}{\alpha}\|h\|_r\Big)^r,\qquad h\in L_r(\mu) \end{align}

Let $f \in L_p$ and define the function $\lambda(\alpha ):= \nu(|T\, f| > \alpha )$. For $\alpha>0$ and $\delta>0$ we have $$f=f\mathbb{1}_{\{|f|>\alpha\delta\}} + f\mathbb{1}_{\{|f|\leq \alpha\delta\}}$$ Clearly, $f_{ 1}=f\mathbb{1}_{\{|f|>\alpha\delta\}} \in L_s(\nu)$ and $f_{ 2}=f\mathbb{1}_{\{|f|\leq\alpha\delta\}} \in L_r(\nu)$. From \begin{align} \left\{ |T\, f|> \alpha \right\} \subset \left\{ |T\, f_{ 1}|> \alpha / 2 \right\} \cup \left\{ |T\, f_{ 2}|> \alpha / 2 \right\}, \end{align} we have that \begin{align} \lambda (\alpha) = \nu\big(|T\, f|> \alpha \big) \leq \nu\big( |T\, f_{ 1} | > \alpha / 2 \big) + \nu \big( |T\, f_{ 2}| > \alpha / 2 \big) \end{align} Then \begin{align} \lambda (\alpha ) \leq \frac{(2 A_s)^s}{\alpha^s} \,\int |f_{1}|^s \, d\mu + \frac{(2 A_{ r})^{r}}{\alpha^{ r}} \, \int |f_{2}|^{r}\, d\mu. \end{align} and so, \begin{align} \lambda (\alpha ) \leq \frac{(2 A_s)^s}{\alpha^s} \, \int_{\{|f|> \alpha\delta\}} |f|^s \, d\mu + \frac{(2A_{r})^{ r}}{\alpha^{ r}} \, \int_{\{|f|\leq \alpha\delta\}} |f|^{ r}\, d\mu\tag{1}\label{one} \end{align} By Fubini's theorem $$\int |T\, f|^{ p}\, d\nu = p\, \int_0^{\infty} \alpha^{{p-1}} \lambda (\alpha )\,d \alpha.$$ Multipying both sides of \eqref{one} by $\alpha^{{p-1}}$ and integrating with respect to $\alpha$ yields \begin{eqnarray*} \int_0^{\infty} \alpha^{{p-1}} \alpha^{{-s}} \int_{\{|f|> \alpha\delta\}} |f|^s \, d\mu \, d \alpha &=& \int|f|^s \int_0^{|f|/\delta} \alpha^{{p-s-1}}\, d \alpha\, d\mu\\ &=& \frac{\delta^{-(p-s)}}{p-s}\, \int |f|^s|f|^{{p-s}}\, d\mu \end{eqnarray*} Similarly, \begin{eqnarray*} \int^\infty_0 \alpha^{p-1} \alpha^{-r} \int_{\{|f|\leq \alpha\delta\}} |f|^r \, d\mu \, d \alpha &=& \int |f|^{ r} \int_{|f|/\delta}^{\infty} \alpha^{{p-1-r}}\, d \alpha\, d\mu\\ &=& \frac{\delta^{r-p}}{r-p}\, \int |f|^{ r} |f|^{{p-r}}\, d\mu \end{eqnarray*} Consequently, \begin{align} \|T\, f\|_p \leq A_p \|f \|_p, \qquad\qquad \left( A_p \right)^p = \left( \frac{(2\, A_s)^s\delta^{-(p-s)}}{p-s} + \frac{ \left( 2\, A_r \right)^{ r}\delta^{r-p}}{r-p} \right)p. \end{align} Choosing $\delta>0$ such that $$(2\, A_s)^s\delta^{-(p-s)} = ( 2\, A_r )^{ r}\delta^{r-p}$$ yields $$A_p=2\Big(\frac{p}{p-s}+\frac{p}{r-p}\Big)^{1/p}A^{\frac{\frac1p-\frac1r}{\frac1s-\frac1r}}_r A^{\frac{\frac1s-\frac1p}{\frac1s-\frac1r}}_s$$ If $\frac1p=\frac{1}{r}+t\big(\frac1s-\frac1r\big)$, one gets that $$A_p=2\Big(\frac{p}{p-s}+\frac{p}{r-p}\Big)^{1/p}A^t_rA^{1-t}_s$$

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Comment: Choosing $\delta=1$ gives a bound for the operator norm $T$, but does gives a convexity relation between $A_s$ and $A_r$ to get the desired form of the operator bound.

The off diagonal version requires much more work.