world:minkovski
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world:minkovski [2025/01/29 09:39] rdouc [Proof] |
world:minkovski [2025/01/29 09:44] (current) rdouc [Proof] |
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| Then, | Then, | ||
| $$\varphi'(s) = \lr{\int (f + s g)^p \rmd \mu}^{\frac{1}{p} - 1} \int (f + s g)^{p-1} g \, \rmd \mu.$$ | $$\varphi'(s) = \lr{\int (f + s g)^p \rmd \mu}^{\frac{1}{p} - 1} \int (f + s g)^{p-1} g \, \rmd \mu.$$ | ||
| - | Using Hölder’s inequality, we can bound $\varphi'(s)$ as follows: | + | Using Hölder’s inequality for the second term, we can bound $\varphi'(s)$ as follows: |
| $$\varphi'(s) \leq \lr{\int (f + s g)^p \rmd \mu}^{\frac{1}{p} - 1} \lr{\int (f + s g)^p \rmd \mu}^{\frac{p-1}{p}} \lr{\int g^p \rmd \mu}^{1/p} = \lr{\int g^p \rmd \mu}^{1/p}.$$ | $$\varphi'(s) \leq \lr{\int (f + s g)^p \rmd \mu}^{\frac{1}{p} - 1} \lr{\int (f + s g)^p \rmd \mu}^{\frac{p-1}{p}} \lr{\int g^p \rmd \mu}^{1/p} = \lr{\int g^p \rmd \mu}^{1/p}.$$ | ||
| - | Integrating $\varphi'(s)$ from $0$ to $1$, we obtain $$\lr{\int (f + g)^p \rmd \mu}^{1/p} - \lr{\int f^p \rmd \mu}^{1/p} = \varphi(1) - \varphi(0) = \int_0^1 \varphi'(s) \, \rmd s \leq \lr{\int g^p \rmd \mu}^{1/p}.$$ This completes the proof. | + | Hence, |
| + | $$\lr{\int (f + g)^p \rmd \mu}^{1/p} = \varphi(1) = \varphi(0) + \int_0^1 \varphi'(s) \, \rmd s \leq \lr{\int f^p \rmd \mu}^{1/p}+ \lr{\int g^p \rmd \mu}^{1/p}.$$ This completes the proof. | ||
world/minkovski.1738139952.txt.gz · Last modified: 2025/01/29 09:39 by rdouc
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