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Apr 6, 2019 at 14:26 comment added Jochen Glueck By the way, I would still be very interested to learn how the complex matrix norms in the above example were computed.
Apr 6, 2019 at 14:21 comment added Jochen Glueck @aurora_borealis: Ah, I see. Somehow, I only had the case $p = q$ in mind, while $p \not= q$ in Iosif Pinelis's example (I actually did not notice that - which indicates that I should have thought more carefuly about the example before posting a comment).
Apr 6, 2019 at 13:08 comment added aurora_borealis @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.
Apr 6, 2019 at 8:11 comment added Jochen Glueck [continuation] I do not have checked your computations numerically, yet - but anyway, may I ask by which algorithm you computed the matrix norms in your example? Even for $2 \times 2$-matrices, computing a good approximation of the complex operator norm by brute force methods seems to demand of a lot of computational resources since you need a very dense grid on the two-dimensional complex unit ball. Or am I just being ignorant, and there is a good way to compute the complex $\ell^p$-operator norm of small matrices analytically (without using that is coincides with the real operator norm)?
Apr 6, 2019 at 8:11 comment added Jochen Glueck @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].
Apr 6, 2019 at 7:41 comment added aurora_borealis @IosifPinelis Thank you!
Apr 6, 2019 at 7:36 vote accept aurora_borealis
Apr 5, 2019 at 21:13 comment added Iosif Pinelis @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.
Apr 5, 2019 at 17:49 comment added Jochen Glueck 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.
Apr 5, 2019 at 17:14 history edited Iosif Pinelis CC BY-SA 4.0
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Apr 5, 2019 at 13:26 history answered Iosif Pinelis CC BY-SA 4.0