Upper Bound for Operator Norm in Marcinkiewicz Interpolation Theorem

Question

Exercise 1.3.3(c) Let $0<p_0<p<p_1<\infty$ and let $T$ be an operator as in Theorem 1.3.2($\|T(f)\|_{L^{p_0,\infty}(Y)}\leq A_0\|f\|_{L^{p_0}(X)}$ for all $f\in L^{p_0}(X)$ and $\|T(f)\|_{L^{p_1,\infty}(Y)}\leq A_1\|f\|_{L^{p_1}(X)}$ for all $f\in L^{p_1}(X)$) that also satisfies $|T(f)|\leq T(|f|)$, for all $f\in L^{p_0}+L^{p_1}$.

(c) When $0<p_0<p_1<\infty$, then the norm of $T$ from $L^p$ to $L^p$ is at most

\[\min_{0<\lambda<1}p^{\frac{1}{p}}\left(\frac{B(p-p_0,p_0+1)}{(1-\lambda)^{p_0}}+\frac{\frac{p_1-p+1}{p_1-p}}{\lambda^{p_1}}\right)^{\frac{1}{p}}A_0^{\frac{1/p-1/p_1}{1/p_0-1/p_1}}A_1^{\frac{1/p_0-1/p}{1/p_0-1/p_1}}\]

where $B(s,t)$ is the Beta function. [Hint: When $p_1<\infty$ write $f=f_0+f_1$, where $f_0=f-\delta\alpha$ when $f\geq\delta\alpha$ and zero otherwise. Use that $|\{|T(f)|>\alpha\}|\leq|\{|T(f_0)|>(1-\lambda)\alpha\}|+|\{|T(f_1)|>\lambda\alpha\}|$ and optimize over $\delta>0$.]

MY ATTEMPT: From the hint, $f_0=\max(f-\delta\alpha,0)$ and $f_1=\min(f,\delta\alpha)$. We have \begin{align*} & d_{T(f)}(\alpha)\leq d_{T(f_0)}((1-\lambda)\alpha)+d_{T(f_1)}(\lambda\alpha)\\ \leq&\left(\frac{A_0}{(1-\lambda)\alpha}\right)^{p_0}\|f_0\|_{L^{p_0}}^{p_0}+\left(\frac{A_1}{\lambda\alpha}\right)^{p_1}\|f_1\|_{L^{p_1}}^{p_1}\\ =&\left(\frac{A_0}{(1-\lambda)\alpha}\right)^{p_0}\int_{f\leq\delta\alpha}(f-\delta\alpha)^{p_0}d\mu+\left(\frac{A_1}{\lambda\alpha}\right)^{p_1}\left[\int_{f>\delta\alpha}(\delta\alpha)^{p_1}d\mu+\int_{f\leq\delta\alpha}f^{p_1}d\mu\right] \end{align*} Hence, \begin{align*} &\|T(f)\|_{L^p}^p=\int_0^\infty p\alpha^{p-1}d_{T(f)}(\alpha)d\alpha\\ \leq&\left(\frac{A_0}{1-\lambda}\right)^{p_0}\int_0^\infty p\alpha^{p-p_0-1}\int_{f\leq\delta\alpha}(f-\delta\alpha)^{p_0}d\mu d\alpha\\ &+\left(\frac{A_1}{\lambda}\right)^{p_1}\left[\int_0^\infty p\alpha^{p-1}\int_{f>\delta\alpha}\delta^{p_1}d\mu d\alpha+\int_0^\infty p\alpha^{p-p_1-1}\int_{f\leq\delta\alpha}f^{p_1}d\mu d\alpha \right]\\ =&\frac{pA_0^{p_0}}{(1-\lambda)^{p_0}}\int_X\int_0^{f/\delta}f^{p_0}(1-\alpha\cdot\frac{\delta}{f})^{p_0}(\alpha\cdot\frac{\delta}{f})^{p-p_0-1}d(\alpha\cdot\frac{\delta}{f})(\frac{f}{\delta})^{p-p_0}d\mu\\ &+\left(\frac{A_1}{\lambda}\right)^{p_1}\left[\int_X\int_{f/\delta}^\infty p\alpha^{p-1}\delta^{p_1}d\alpha d\mu+\int_X f^{p_1}\int_{f/\delta}^\infty p\alpha^{p-p_1-1}d\alpha d\mu\right]\\ =&\frac{pA_0^{p_0}}{(1-\lambda)^{p_0}}\delta^{p_0-p}\|f\|_{L^p}^pB(p_0+1,p-p_0)+\frac{A_1^{p_1}}{\lambda^{p_1}}\left[\delta^{p_1-p}\|f\|_{L^p}^p+\frac{p}{p_1-p}\delta^{p_1-p}\|f\|_{L^p}^p\right]\\ =&\frac{pA_0^{p_0}}{(1-\lambda)^{p_0}}\delta^{p_0-p}\|f\|_{L^p}^pB(p_0+1,p-p_0)+\frac{p_1A_1^{p_1}}{(p_1-p)\lambda^{p_1}}\delta^{p_1-p}\|f\|_{L^p}^p \end{align*} when optimizing $\delta$ by taking derivatives of $\delta$, I cannot conclude the result.

Answer 1

So, as indicated in my earlier comments, I end up with the same expression as you involving $\lambda$ and $\delta$; however, I couldn't arrive at the expression in the exercise statement. Here's my attempt at the optimization step.

Define a function of $\varphi=\varphi(\delta)$ by $$\varphi(\delta):=\dfrac{pA_{0}^{p_{0}}}{(1-\lambda)^{p_{0}}}B(p_{0}+1,p-p_{0})\delta^{p_{0}-p}+\dfrac{p_{1} A_{1}^{p_{1}}}{(p_{1}-p)\lambda^{p_{1}}}\delta^{p_{1}-p}$$ For convenience, lets write $B=B(p_{0}+1,p-p_{0})$. Differentiating and setting equal to zero, $$0=\varphi'(\delta)=\dfrac{p(p_{0}-p) A_{0}^{p_{0}}}{(1-\lambda)^{p_{0}}}B\delta^{p_{0}-p-1}+\dfrac{p_{1}A_{1}^{p_{1}}}{\lambda^{p_{1}}}\delta^{p_{1}-p-1}$$ So the critical point, which you can check is a minimum, is $$\delta=\left(\dfrac{p (p-p_{0})A_{0}^{p_{0}}B\lambda^{p_{1}}}{p_{1}A_{1}^{p_{1}}(1-\lambda)^{p_{0}}}\right)^{1/(p_{1}-p_{0})}$$

Using the fact that we chose $\delta$ so that $$\dfrac{p(p-p_{0})A_{0}^{p_{0}}B}{(1-\lambda)^{p_{0}}}\delta^{p_{0}-p}=\dfrac{p_{1}A_{1}^{p_{1}}}{\lambda^{p_{1}}}\delta^{p_{1}-p},$$ we see that \begin{align*} \varphi(\delta)&=\dfrac{p(p-p_{0})A_{0}^{p_{0}}B}{(1-\lambda)^{p_{0}}}\delta^{p_{0}-p}\left[\dfrac{1}{p-p_{0}}+\dfrac{1}{p_{1}-p}\right]\\ & \\ &=A_{0}^{p_{0}\frac{p_{1}-p}{p_{1}-p_{0}}}A_{1}^{p_{1}\frac{p-p_{0}}{p_{1}-p_{0}}}p^{\frac{p_{1}-p}{p_{1}-p_{0}}}p_{1}^{\frac{p-p_{0}}{p_{1}-p_{0}}}(p-p_{0})^{\frac{p_{1}-p}{p_{1}-p_{0}}}\dfrac{B^{\frac{p_{1}-p}{p_{1}-p_{0}}}}{(1-\lambda)^{p_{0}\frac{p_{1}-p}{p_{1}-p_{0}}}\lambda^{p_{1}\frac{p-p_{0}}{p_{1}-p_{0}}}}\left[\dfrac{1}{p-p_{0}}+\dfrac{1}{p_{1}-p}\right] \end{align*} Taking $p^{th}$ roots, I recognize the desired expressions involving $A_{0}$ and $A_{1}$, but I don't see how to get that the factor involving the sum.

Answer 2

Matt's calculations are right. To get the desired result, we only need to apply the Young's inequaly, i.e., $$ A^{\theta}B^{1-\theta}\le \theta A+(1-\theta)B, $$ when $A,B>0$ and $\theta\in (0,1)$.

We now focus on \begin{gather*} F(p)=p^{\frac{p_{1}-p}{p_{1}-p_{0}}}p_{1}^{\frac{p-p_{0}}{p_{1}-p_{0}}}(p-p_{0})^{\frac{p_{1}-p}{p_{1}-p_{0}}}\dfrac{B^{\frac{p_{1}-p}{p_{1}-p_{0}}}}{(1-\lambda)^{p_{0}\frac{p_{1}-p}{p_{1}-p_{0}}}\lambda^{p_{1}\frac{p-p_{0}}{p_{1}-p_{0}}}}\left[\dfrac{1}{p-p_{0}}+\dfrac{1}{p_{1}-p}\right]. \end{gather*}

Denote $\frac{p_1-p}{p_1-p_0}$ by $s$, then we know that $\frac{p-p_0}{p_1-p_0}=1-s$, and $$ F(p)=p^s p_1^{1-s} (p-p_0)^s \cfrac{B^s}{(1-\lambda)^{p_0s}\lambda^{p_1(1-s)}}\left[\dfrac{1}{p-p_{0}}+\dfrac{1}{p_{1}-p}\right]. $$ When $0<p<1$,by Young's inequality, we have \begin{align*} F(p)&=p\left[\dfrac{1}{p-p_{0}}+\dfrac{1}{p_{1}-p}\right]\left(\cfrac{(p-p_0)B}{(1-\lambda)^{p_0}}\right)^{s}\left(\cfrac{pp_1}{\lambda^{p_1} } \right)^{1-s}\\ &\le p\left( \cfrac{p_1-p_0}{(p-p_0)(p_1-p)}\cfrac{p_1-p}{p_1-p_0}\cfrac{(p-p_0)B}{(1-\lambda)^{p_0}}+\cfrac{p_1-p_0}{(p-p_0)(p_1-p)}\cfrac{p-p_0}{p_1-p_0}\cfrac{pp_1}{\lambda^{p_1}}\right)\\ &\le p\left(\cfrac{B}{(1-\lambda)^{p_0}}+\cfrac{1+\frac p{p_1-p}}{\lambda^{p_1}}\right)\\ &\le p\left(\cfrac{B}{(1-\lambda)^{p_0}}+\cfrac{1+\frac 1{p_1-p}}{\lambda^{p_1}}\right). \end{align*} And when $p\ge 1$, use Young's inequality again, we have \begin{align*} F(p)&=\left[\dfrac{1}{p-p_{0}}+\dfrac{1}{p_{1}-p}\right]\left(\cfrac{p(p-p_0)B}{(1-\lambda)^{p_0}}\right)^{s}\left(\cfrac{p_1}{\lambda^{p_1} } \right)^{1-s}\\ &\le p\cfrac{p_1-p_0}{(p-p_0)(p_1-p)}\cfrac{p_1-p}{p_1-p_0}\cfrac{(p-p_0)B}{(1-\lambda)^{p_0}}+\cfrac{p_1-p_0}{(p-p_0)(p_1-p)}\cfrac{p-p_0}{p_1-p_0}\cfrac{p_1}{\lambda^{p_1}}\\ &=p\cfrac{B}{(1-\lambda)^{p_0}}+\cfrac{1+\frac p{p_1-p}}{\lambda^{p_1}}\\ &\le p\left( \cfrac{B}{(1-\lambda)^{p_0}}+\cfrac{1+\frac 1{p_1-p}}{\lambda^{p_1}}\right). \end{align*}

Hence, we have \begin{align*} \|T\|_{L^p\to L^p}^p&\le p\left( \cfrac{B}{(1-\lambda)^{p_0}}+\cfrac{\frac {p_1-p+1}{p_1-p}}{\lambda^{p_1}}\right)A_{0}^{p_{0}\frac{p_{1}-p}{p_{1}-p_{0}}}A_{1}^{p_{1}\frac{p-p_{0}}{p_{1}-p_{0}}}\\ &=p\left( \cfrac{B}{(1-\lambda)^{p_0}}+\cfrac{\frac {p_1-p+1}{p_1-p}}{\lambda^{p_1}}\right) A_0^{p\left(\frac{\frac 1p-\frac 1 {p_1}}{\frac 1 p_0-\frac 1 {p_1}}\right)}A_1^{p\left(\frac{\frac 1{p_0}-\frac 1 {p}}{\frac 1 p_0-\frac 1 {p_1}}\right)} \end{align*} which is exactly what we want.