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)

  • $\begingroup$ I am not finding there any statement that the real and complex versions of the norm are the same. $\endgroup$ Apr 4, 2019 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? $\endgroup$ Apr 4, 2019 at 12:49
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    $\begingroup$ If you don't see such a statement there (in the lecture notes?), then what is your question? $\endgroup$ Apr 4, 2019 at 15:03
  • $\begingroup$ @IosifPinelis Sorry I didn't state it clearly. I have changed my statement and listed my questions above. $\endgroup$ Apr 5, 2019 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. $\endgroup$ Apr 5, 2019 at 15:54

1 Answer 1


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 $1.00041>1$.

  • $\begingroup$ 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. $\endgroup$ Apr 5, 2019 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. $\endgroup$ Apr 5, 2019 at 21:13
  • $\begingroup$ @IosifPinelis Thank you! $\endgroup$ Apr 6, 2019 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]. $\endgroup$ Apr 6, 2019 at 8:11
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    $\begingroup$ @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. $\endgroup$ Apr 6, 2019 at 13:08

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