# Why the triangle inequality is greater than the 2 max{f(x),g(x)}

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• cbarker1
Then, |f+g|^p<=(|f|+|g|)^p<=2^p|f|^p+2^p|g|^p.In summary, Sheldon is proving that ##L^p(\mu)## is a Vector space over ##\mathbb{R}## by showing that if a certain condition holds true, then ##L^p(\mu)## is true with standard addition and scalar multiplication. He begins by assuming that ##f,g\in L^p(\mu)## and uses the triangle inequality and the fact that ##|f|+|g
cbarker1
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MHB
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

Last edited:
Because if, e.g., |f|>=|g|, then |f|+|g|<=|f|+|f|=2|f|

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