world:minkovski
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world:minkovski [2025/01/29 09:43] rdouc [Proof] |
world:minkovski [2025/01/29 09:44] (current) rdouc [Proof] |
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| $$\varphi'(s) \leq \lr{\int (f + s g)^p \rmd \mu}^{\frac{1}{p} - 1} \lr{\int (f + s g)^p \rmd \mu}^{\frac{p-1}{p}} \lr{\int g^p \rmd \mu}^{1/p} = \lr{\int g^p \rmd \mu}^{1/p}.$$ | $$\varphi'(s) \leq \lr{\int (f + s g)^p \rmd \mu}^{\frac{1}{p} - 1} \lr{\int (f + s g)^p \rmd \mu}^{\frac{p-1}{p}} \lr{\int g^p \rmd \mu}^{1/p} = \lr{\int g^p \rmd \mu}^{1/p}.$$ | ||
| Hence, | Hence, | ||
| - | $$\lr{\int (f + g)^p \rmd \mu}^{1/p} - \lr{\int f^p \rmd \mu}^{1/p} = \varphi(1) - \varphi(0) = \int_0^1 \varphi'(s) \, \rmd s \leq \lr{\int g^p \rmd \mu}^{1/p}.$$ This completes the proof. | + | $$\lr{\int (f + g)^p \rmd \mu}^{1/p} = \varphi(1) = \varphi(0) + \int_0^1 \varphi'(s) \, \rmd s \leq \lr{\int f^p \rmd \mu}^{1/p}+ \lr{\int g^p \rmd \mu}^{1/p}.$$ This completes the proof. |
world/minkovski.txt · Last modified: 2025/01/29 09:44 by rdouc
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