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cbarker1

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- TL;DR Summary
- I am working on proving the L^p of measureable space (X,S,u) is a vector space. I am lost on why the the triangle intequalty is greater than the 2 max{f(x),g(x)} for a fix x in X.

I am reading Sheldon's Axler Book on Measure theory. He is proving that ##L^p(\mu)## is a Vector space over ##\mathbb{R}.## He claims that if ##\|f+g\|_{p}^p\leq 2^p(\|f\|_{p}^p+\|g\|_{p}^{p})## and nonzero homogenity holds true, then ##L^p{\mu}## is true with the standard addition and scalar multiplication. He starts with the following assumption:

Suppose that ##f,g\in L^p(\mu)## are arbitrary. Then if ##x\in X## is an arbitrary fix element of ##X,## then

##\begin{align*} |f(x)+g(x)|^p&\leq_{\text{triangle inequality}} (|f(x)|+|g(x)|)^p\\ &\leq_{\text{why?}} (2\max{|f(x)|,|g(x)|})^p\\ &\leq2^p(|f(x)|^p+|g(x)|^p)\end{align*}##

If you can explain whys in this proof then I will be able to understand the proof.

Thanks,

Carter Barker

Suppose that ##f,g\in L^p(\mu)## are arbitrary. Then if ##x\in X## is an arbitrary fix element of ##X,## then

##\begin{align*} |f(x)+g(x)|^p&\leq_{\text{triangle inequality}} (|f(x)|+|g(x)|)^p\\ &\leq_{\text{why?}} (2\max{|f(x)|,|g(x)|})^p\\ &\leq2^p(|f(x)|^p+|g(x)|^p)\end{align*}##

If you can explain whys in this proof then I will be able to understand the proof.

Thanks,

Carter Barker

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