Does the Riesz-Thorin interpolation theorem hold for real-linear operators?
The standard proof uses complex linearity, but maybe another proof avoids the assumption of complex-linearity?
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- $\begingroup$ What is the problem. You can complexify a real-linear operator. Perhaps the non-obvious (though true) fact is that the complexification does not change the $L^p$-norm. $\endgroup$Denis Serre– Denis Serre2020-07-15 15:15:28 +00:00Commented Jul 15, 2020 at 15:15
- 2$\begingroup$ The norm of the complexification can change if the operator maps $L^p$ to $L^q$ with $q<p$. When such a situation occurs one needs a constant (2 suffices) in the real Riesz-Thorin, I guess. $\endgroup$Giorgio Metafune– Giorgio Metafune2020-07-15 15:39:51 +00:00Commented Jul 15, 2020 at 15:39
- $\begingroup$ @Denis Serre. Assume that $Tf(z)= \int_U (A(z,w) f(z)+B(z,w) \overline{f(z)})dudv, w=u+iv, z=x+i y$, $U$ is the unit disk. This operator is real-linear, but the function $f:U\to U$ need not be real. So what is the complexification of this operator. $A$ and $B$ are certain kernels? $\endgroup$Lira– Lira2020-07-18 14:50:15 +00:00Commented Jul 18, 2020 at 14:50
- $\begingroup$ @Lira. You should not focus on the fact that $f$ is a complex-valued function. Your operator is real-linear, so just write it in terms of a pair of unctions $(g,h)$ (real and imaginary parts). Then you may complexify :). $\endgroup$Denis Serre– Denis Serre2020-07-18 20:18:59 +00:00Commented Jul 18, 2020 at 20:18
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1 Answer
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No.
Let $T:\mathbb R^2\to\mathbb R^2$ be $T(x_1,x_2)=(x_1-x_2,x_1+ax_2)$, where $1<a<\frac54$.
For $\|x\|_\infty=\|y\|_\infty=1$, we have $$\begin{aligned}\langle Tx,y\rangle&=|(x_1-x_2)y_1+(x_1+ax_2)y_2|\\&\le|y_1+y_2|+|y_1-y_2|+a-1\\&\le a+1\end{aligned}$$ where equality holds when $x=y=(1,1)$. So $\|T\|_{\infty\to1}=a+1$.
We have $$\begin{aligned}\|Tx\|_2^2&=(x_1-x_2)^2+(x_1+ax_2)^2\\&=r\|x\|_2^2-(px_1+qx_2)^2\\&\le r\|x\|_2^2\end{aligned}$$ where $p^2-q^2=a^2-1$, $pq=1-a$, and $r=p^2+2$. Solving the quadratic equation, we have $r=\frac12(a^2+3+(a+1)\sqrt{a^2+2a+5})$. Equality holds when $px_1+qx_2=0$. So $\|T\|_{2\to2}=\sqrt r$.
For $x=(t,1)$, we shall show that $f(t)=\frac{1+a}{2-a}(|t|^{3/2}+1)^2-(|t-1|^3+|t+a|^3)\ge0$, so $\|T\|_{\frac32\to3}=\sqrt[3]{\frac{1+a}{2-a}}$.
For $t\ge1$, we have $f(1)=\frac{(a+1)(a+2)(a-1)^2}{2-a}>0$ and $f'(t)=\frac{3(a-1)}{2-a}(3t^2+2(a-2)t-1)+\frac{3(a+1)}{2-a}(\sqrt t-1)+(a-1)(a^2-a+2)>0$.
For $t\le-a$, we have $f(t)-f(-t)=((a-t)^3-(1-t)^3)+((t+a)^3-(t+1)^3)>0$.
For $-a<t<1$, we have $$\begin{aligned}f(t)&=\textstyle\frac{1+a}{2-a}(|t|^{3/2}+1)^2-(1-t)^3-(t+a)^3\\&=\textstyle\frac{1+a}{2-a}(|t|-\sqrt{|t|}+a-1)^2(|t|+2\sqrt{|t|}+a-1)+3(a^2-1)(|t|-t)\ge0.\end{aligned}$$ Equality holds when $t=\frac12(3-2a\pm\sqrt{5-4a})$.
Therefore $$\begin{aligned}\|T\|_{\infty\to1}\|T\|_{\frac32\to3}^3-\|T\|_{2\to2}^4&=a+1+\tfrac{1+a}{2-a}-\tfrac14(a^2+3+(a+1)\sqrt{a^2+2a+5})^2\\&<\tfrac{21}4-4(1+\sqrt2)^2<0\end{aligned}$$ while Riesz–Thorin says this is $\ge0$.
I wrote this example in my notes a few years ago but I don't remember where it came from.
