1
$\begingroup$

I was reading the proof of Riesz-Thorin interpolation theorem in http://www.math.kit.edu/iana3/lehre/fourierana2014w/media/rieszthorinproof.pdf and get stuck at the last step. We construct the complex function \begin{equation*} F(z)=\int g_zTf_z=\sum_{i,j}|a_i|^{\frac{p}{p(z)}}\frac{a_i}{|a_i|}|b_j|^{\frac{q}{q(z)}}\frac{b_j}{|b_j|}\int_{B_j}T\chi_{A_i} \end{equation*} and finally we have \begin{equation*} \sup_{y\in\mathbb{R}}F(iy)=\sup_{y\in\mathbb{R}}\int g_{yi}Tf_{yi}\leq \|T\|_{L_{p_0}(\mathbb{C}^n)\rightarrow L_{q_0}(\mathbb{C}^n)}\|g_{yi}\|_{L_{q_0}}\|f_{yi}\|_{L_{p_0}} \end{equation*} However, if we come back to the theorem, what we actually want to show is \begin{equation} \sup_{y\in\mathbb{R}}F(iy)\leq N_0\|g_{yi}\|_{L_{q_0}} \|f_{yi}\|_{L_{p_0}}=\|T\|_{L_{p_0}(\mathbb{R}^n)\rightarrow L_{q_0}(\mathbb{R}^n)}\|g_{yi}\|_{L_{q_0}} \|f_{yi}\|_{L_{p_0}} \end{equation} I understand that $\|g_{yi}\|_{L_{q_0}}=\|g_0\|_{L_{q_0}}$ and $\|f_{yi}\|_{L_{p_0}}=\|f_0\|_{L_{p_0}}$, but as we know the complex norm is never smaller than real norm. So we need to show \begin{equation*} \|T\|_{L_{p_0}(\mathbb{C}^n)\rightarrow L_{q_0}(\mathbb{C}^n)}=\|T\|_{L_{p_0}(\mathbb{R}^n)\rightarrow L_{q_0}(\mathbb{R}^n)} \end{equation*} which is not obvious. Moreover, if we have to add a constant $C$ such that \begin{equation*} \|T\|_{L_{p_0}(\mathbb{C}^n)\rightarrow L_{q_0}(\mathbb{C}^n)}\leq C\|T\|_{L_{p_0}(\mathbb{R}^n)\rightarrow L_{q_0}(\mathbb{R}^n)} \end{equation*} Then the statement of this theorem must be modified accordingly. I tried to simplify it into matrix case and regard the linear map $T$ as an $n\times n$ matrix, then $T:\mathbb{C}^n\rightarrow\mathbb{C}^n$ can be looked as another $2n\times 2n$ matrix $T\otimes I_2$, however, the $L_p$ norm in $\mathbb{R}^{2n}$ is different from that in $\mathbb{C}^n$ when $p\neq 2$.

In conclusion, my questions are:

(1) Is the statement of Riesz-Thorin theorem in the above website true for real linear map $T$? If it is true, how to prove it?

(2) In fact I mainly care about how to prove it when $T$ is a symmetric matrix, so is there an easier way to prove the matrix version of this theorem?

(3) If possible, I hope some one can provide me with some reference about Riesz-Thorin theorem and some related topics so that I can learn more. (matrix version is enough but I would be glad to learn the results in more general version)

$\endgroup$
  • $\begingroup$ I am not finding there any statement that the real and complex versions of the norm are the same. $\endgroup$ – Iosif Pinelis Apr 4 '19 at 12:39
  • $\begingroup$ @losifPinelis Me too, so I am a little doubt about this proof... or are there easier way to prove this theorem for the case $T$ is a symmetric matrix? $\endgroup$ – aurora_borealis Apr 4 '19 at 12:49
  • 1
    $\begingroup$ If you don't see such a statement there (in the lecture notes?), then what is your question? $\endgroup$ – Iosif Pinelis Apr 4 '19 at 15:03
  • $\begingroup$ @IosifPinelis Sorry I didn't state it clearly. I have changed my statement and listed my questions above. $\endgroup$ – aurora_borealis Apr 5 '19 at 5:33
  • $\begingroup$ The formulation of the question seems to indicate somekind of misunderstanding: the space $\mathbb{R}^n$ in the linked notes is the underlying measure space of $L_p$. I can't see why the measure space $\mathbb{R^n}$ should be changed to $\mathbb{C}^n$ anywhere in the argument. $\endgroup$ – Jochen Glueck Apr 5 '19 at 15:54
2
$\begingroup$

Concerning your Question (3): A nice proof of the Riesz--Thorin theorem is given in Section 3.5 in Lax's book. A chapter devoted to interpolation of linear operators, including different versions of the Riesz--Thorin theorem, with further references therein, is contained in Mashreghi's book as Chapter 8.

As for your Question (1), as Jochen Glueck commented, you probably meant $L_p(\mathbb R^n,\mathbb R)$ and $L_p(\mathbb R^n,\mathbb C)$ instead of $L_p(\mathbb R^n)$ and $L_p(\mathbb C^n)$, respectively. If so, the answer to your Question 1 is negative. Indeed, let $\ell_{2,p}$ denote the space $\mathbb R^2$ with the $\ell_p$ norm. Since $\ell_{2,p}$ can be isometrically embedded into $L_p(\mathbb R^n,\mathbb R)$, we may restrict the consideration to $2\times2$ matrices $T$ with the corresponding operator norms $\|T\|_{p,q}$, $\|T\|_{p_0,q_0}$, $\|T\|_{p_1,q_1}$.

Then we find that the inequality $$\|T\|_{p,q}\le\|T\|_{p_0,q_0}^{1-t}\|T\|_{p_1,q_1}^t \tag{1} $$ fails to hold if $$T=\left(\begin{array}{cc} 0.448531 & 0.812143 \\ -0.772457 & 0.469272 \\ \end{array} \right) $$ and $$(p_0, q_0, p_1, q_1, t)=(3.99136, 1.32751, 8.82177, 1, 0.854335); $$ then the ratio of the left-hand side of (1) to its right-hand side is about $1.00041>1$.

$\endgroup$
  • $\begingroup$ I'm not sure I follow your answer to Question (1): If $T$ is an operator on a real-valued $L^p$-space, then the operator norm of $T$ coincides with the operator norm of the canonical extension of $T$ to the complex-valued $L^p$-space. Thus, the Riesz-Thorin theorem is certainly also true on real-valued $L^p$-spaces. $\endgroup$ – Jochen Glueck Apr 5 '19 at 17:49
  • $\begingroup$ @JochenGlueck : "If $T$ is an operator on a real-valued $L^p$-space, then the operator norm of $T$ coincides with the operator norm of the canonical extension of $T$ to the complex-valued $L^p$-space." Why would that be so? In fact, for the operator (given by matrix) $T$ in my answer, its "real" norm is about $0.903636$ of its "complex" norm -- this is how that $T$ was found. Also, you are welcome to check the calculations. $\endgroup$ – Iosif Pinelis Apr 5 '19 at 21:13
  • $\begingroup$ @IosifPinelis Thank you! $\endgroup$ – aurora_borealis Apr 6 '19 at 7:41
  • $\begingroup$ @IosifPinelis: A proof for the assertion I mentioned can, for instance, be found in Proposition 2.1.1 of these notes. Alternatively, you can also refer to this preprint. [to be continued]. $\endgroup$ – Jochen Glueck Apr 6 '19 at 8:11
  • 1
    $\begingroup$ @JochenGlueck Thanks for your reference which help me understand what I originally want to know(when $T$ is a symmetric matrix and $p$ equals to $q$). To sum up your two links, they tell us the real norm and complex norm are coincide when $p=q$, and $\|T\|_{\mathbb{C}}\leq 2\|T\|_{\mathbb{R}}$ in general. And this result does not contradict with the numerical example introduced above. $\endgroup$ – aurora_borealis Apr 6 '19 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.