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world:minkovski [2025/01/29 09:34]
rdouc [Proof]
world:minkovski [2025/01/29 09:44] (current)
rdouc [Proof]
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 ===== Proof ===== ===== Proof =====
-Without loss of generality, we assume that $f, g \in \lp{p}(\mu)$ and $f, g \geq 0$ (the general case can be handled using the inequality $|f+g| \leq |f| + |g|$). For $s > 0$, define ​+Some alternative proofs can be found [[https://​en.wikipedia.org/​wiki/​Minkowski_inequality|here]].  
 + 
 +Without loss of generality, we assume that $p>​1$, ​$f, g \in \lp{p}(\mu)$ and $f, g \geq 0$ (the general case can be handled using the inequality $|f+g| \leq |f| + |g|$). For $s > 0$, define ​
 $$\varphi(s) = \lr{\int (f + s g)^p \rmd \mu}^{1/​p}.$$ ​ $$\varphi(s) = \lr{\int (f + s g)^p \rmd \mu}^{1/​p}.$$ ​
 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.1738139640.txt.gz · Last modified: 2025/01/29 09:34 by rdouc