User Tools
world:minkovski
Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision | |||
|
world:minkovski [2025/01/29 09:43] rdouc [Proof] |
world:minkovski [2025/01/29 09:44] (current) rdouc [Proof] |
||
|---|---|---|---|
| Line 15: | Line 15: | ||
| $$\varphi'(s) \leq \lr{\int (f + s g)^p \rmd \mu}^{\frac{1}{p} - 1} \lr{\int (f + s g)^p \rmd \mu}^{\frac{p-1}{p}} \lr{\int g^p \rmd \mu}^{1/p} = \lr{\int g^p \rmd \mu}^{1/p}.$$ | $$\varphi'(s) \leq \lr{\int (f + s g)^p \rmd \mu}^{\frac{1}{p} - 1} \lr{\int (f + s g)^p \rmd \mu}^{\frac{p-1}{p}} \lr{\int g^p \rmd \mu}^{1/p} = \lr{\int g^p \rmd \mu}^{1/p}.$$ | ||
| 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.1738140208.txt.gz · Last modified: 2025/01/29 09:43 by rdouc
Page Tools
